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Selected Titles in This Series
Volume
4 W . J. Kaczor and M. T. Nowak
Problems in mathematical analysis I: Real numbers, sequences and
series
2000
3 Roger Knobel
An introduction to the mathematical theory of waves
2000
2 Gregory F. Lawler and Lester N . Coyle
Lectures on contemporary probability
1999
1 Charles Radin
Miles of tiles
1999
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Problems in
Mathematical
Analysis I
Real Numbers,
Sequences and Series
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STUDENT MATHEMATICAL LIBRARY
Volume 4
Problems in
Mathematical
Analysis I
Real Numbers,
Sequences and Series
W J. Kaczor
M.T. Nowak
iAMS
AMERICAN MATHEMATICAL SOCIETY
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Editorial Board
David Bressoud
R o b e r t Devaney, Chair
Carl Pomerance
Hung-Hsi W u
Originally published in Polish, as
Zadania z Analizy M a t e m a t y c z n e j . Cz$sc Pierwsza.
Liczby Rzeczywiste, Ciagi i Szeregi Liczbowe
© 1996, W y d a w n i c t w o U n i w e r s y t e t u Marii Curie-Sklodowskiej, Lublin.
Translated, revised a n d a u g m e n t e d by t h e a u t h o r s .
2000 Mathematics
Subject Classification.
Primary 00A07, 4 0 - 0 1 .
Library of Congress C a t a l o g i n g - i n - P u b l i c a t i o n D a t a
Kaczor, W. J. (Wieslawa J.), 1949[Zadania z analizy matematycznej. English]
Problems in mathematical analysis. I. Real numbers, sequences and series /
W. J. Kaczor, M. T. Nowak.
p. cm. — (Student mathematical library, ISSN 1520-9121 ; v. 4)
Includes bibliographical references.
ISBN 0-8218-2050-8 (softcover : alk. paper)
1. Mathematical analysis. I. Nowak, M. T. (Maria T.), 1951- II. Title.
III. Series.
QA300K32513 2000
515 / .076-dc21
99-087039
Copying and reprinting. Individual readers of this publication, and nonprofit
libraries acting for them, are permitted to make fair use of the material, such as to
copy a chapter for use in teaching or research. Permission is granted to quote brief
passages from this publication in reviews, provided the customary acknowledgment of
the source is given.
Republication, systematic copying, or multiple reproduction of any material in this
publication is permitted only under license from the American Mathematical Society.
Requests for such permission should be addressed to the Assistant to the Publisher,
American Mathematical Society, P. O. Box 6248, Providence, Rhode Island 02940-6248.
Requests can also be made by e-mail to reprint-permissionQams.org.
© 2000 by the American Mathematical Society. All rights reserved.
The American Mathematical Society retains all rights
except those granted to the United States Government.
Printed in the United States of America.
@ The paper used in this book is acid-free and falls within the guidelines
established to ensure permanence and durability.
Visit the AMS home page at URL: http://www.ams.org/
10 9 8 7 6 5 4 3 2
09
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Contents
Preface
Notation and Terminology
Problems
Chapter 1. Real Numbers
1.1. Supremum and Infimum of Sets of Real Numbers.
Continued Fractions
1.2. Some Elementary Inequalities
Chapter 2. Sequences of Real Numbers
2.1. Monotonic Sequences
2.2. Limits. Properties of Convergent Sequences
2.3. The Toeplitz Transformation, the Stolz Theorem and
their Applications
2.4. Limit Points. Limit Superior and Limit Inferior
2.5. Miscellaneous Problems
Chapter 3. Series of Real Numbers
3.1. Summation of Series
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Contents
Vlll
3.2. Series of Nonnegative Terms
72
3.3. The Integral Test
88
3.4. Series of Positive and Negative Terms - Convergence,
Absolute Convergence. Theorem of Leibniz
92
3.5. The Dirichlet and Abel Tests
99
3.6. Cauchy Product of Infinite Series
102
3.7. Rearrangement of Series. Double Series
105
3.8. Infinite Products
112
Solutions
Chapter 1. Real Numbers
1.1. Supremum and Infimum of Sets of Real Numbers.
Continued Fractions
125
1.2. Some Elementary Inequalities
136
Chapter 2. Sequences of Real Numbers
2.1. Monotonic Sequences
151
2.2. Limits. Properties of Convergent Sequences
162
2.3. The Toeplitz Transformation, the Stolz Theorem and
their Applications
181
2.4. Limit Points. Limit Superior and Limit Inferior
189
2.5. Miscellaneous Problems
208
Chapter 3. Series of Real Numbers
3.1. Summation of Series
245
3.2. Series of Nonnegative Terms
269
3.3. The Integral Test
302
3.4. Series of Positive and Negative Terms - Convergence,
Absolute Convergence. Theorem of Leibniz
309
3.5. The Dirichlet and Abel Tests
324
3.6. Cauchy Product of Infinite Series
333
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Contents
IX
3.7. Rearrangement of Series. Double Series
342
3.8. Infinite Products
360
Bibliography - Books
379
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Preface
This book is an enlarged and revised English edition of a Polish
version published in 1996 by the Publishing House of Maria CurieSklodowska University in Lublin, Poland. It is the first volume of a
planned series of books of problems in mathematical analysis. The
second volume, already published in Polish, is under translation into
English. The series is mainly intended for students who take courses
in basic principles of analysis. The choice and arrangement of the
material make it suitable for self-study, and instructors may find it
useful as an aid in organizing tutorials and seminars.
This volume covers three topics: real numbers, sequences, and
series. It does not contain problems concerning metric and topological
spaces, which we intend to present in subsequent volumes.
The book is divided into two parts. The first part is a collection
of exercises and problems, and the second contains their solutions.
Complete solutions are given in most cases. Where no difficulties
could be expected or when an analogous problem has already been
solved, only a hint or simply an answer is given. Very often various
solutions of a given problem are possible; we present here only one,
hoping students themselves will find others.
XI
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xi i
Preface
With the student in mind, we have tried to keep things at an elementary level whenever possible. For example, we present an elementary proof of the Toeplitz theorem about the so-called regular transformation of sequences, which in many texts is proved by methods of
functional analysis. The proof presented is taken from Toeplitz's original paper, published in 1911 in Prace Matematyczno-Fizyczne, Vol.
22. We hope that our presentation of this part of real analysis will
be more accessible to readers and will ensure wider understanding.
All the notations and definitions used in this volume are standard
and commonly used. The reader can find them, for example, in the
textbooks [12] and [23], in which all necessary theoretical background
can be found. However, to make the book consistent and to avoid
ambiguity, a list of notations and definitions is included.
We have borrowed freely from many textbooks, problem books
and problem sections of journals like the American Mathematical
Monthly, Mathematics Today (Russian) and Delta (Polish). A complete list is given in the bibliography. It was beyond the authors'
scope to trace all original sources, and we may have overlooked some
contributions. If this has happened, we offer our sincere apologies.
We are deeply indebted to all our friends and colleagues from the
Department of Mathematics of Maria Curie-Sklodowska University
who offered stimulating suggestions. We have had many fruitful conversations with M. Koter-Morgowska, T.Kuczumow, W. Rzymowski,
S. Stachura and W. Zygmunt. Our sincere thanks are also due to
Professor Jan Krzyz for his help in preparing the first version of the
English manuscript. We are pleased to express our gratitude to Professor Kazimierz Goebel for his encouragement and active interest in
the project. It is our pleasure to thank Professor Richard J. Libera,
University of Delaware, for his invaluable and most generous help
with the English translation and for all his suggestions and corrections which greatly improved the final version of the book.
W. J. Kaczor, M. T. Nowak
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Notation and
Terminology
•
•
•
•
•
•
•
•
•
R - the set of all real numbers
R + - the set of all positive real numbers
Z - the set of all integers
N - the set of all positive integers
Q - the set of all rationals
(a, b) - open interval with the endpoints a and b
[a, b] - closed interval with the endpoints a and b
[x] - the integral part of a real number x
For x e R,
( 1
sgnx = < - 1
0
for x > 0,
for x < 0,
for x = 0.
For n e N,
n! = 1 - 2 - 3 - . . . - n ,
(2n)!! = 2 • 4 • 6 •... • (2n - 2)(2n) and
(2n - 1)!! = 1 • 3 • 5 • ... • (2n - 3)(2n - 1).
If A c K is nonempty and bounded from above, then sup A
denotes the least upper bound of A. If a nonempty set A is
not bounded above, then we assume that sup A = -foo.
xin
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Notation and Terminology
XIV
If A c t is nonempty and bounded from below, then inf A
denotes the greatest lower bound of A. If a nonempty set A
is not bounded below, then we assume that inf A = — oo.
A sequence {an} of real numbers is said to be monotonically
increasing (monotonically decreasing) if a n + i > an for all
n £ N (a n +i < an for all n £ N). The class of monotonic sequences consists of the increasing and the decreasing
sequences.
A number c is a limit point of the sequence {an} if there is a
subsequence {ank} of {an} converging to c.
Let S be the set of all the limit points of {an}. The limit inferior, lim a n , and the limit superior, lim a n , of the sequence
{an} are defined as follows:
+oc
lim an = < —oo
lim an
n—+oc
if {an} is not bounded above,
if {an} is bounded above and S = 0,
supS
if {an} is bounded above and S ^ 0,
-oo
if {an} is not bounded below,
-f oo
if {an} is bounded below and S = 0,
inf S
if {an} is bounded below and S ^ 0.
An infinite product
Yl an is said to be convergent if there
71=1
exists no £ N such that an ^ 0 for n > UQ and the sequence
{a n o a n o + i • ... • anQ+n} converges, as n —> oo, to a limit P0
other than zero. The number P = a ^ • • • • * a>n0-i ' Po is
called the value of the infinite product.
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Problems
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Chapter 1
http://dx.doi.org/10.1090/stml/004/01
Real Numbers
1.1. Supremum and Infimum of Sets of Real
Numbers. Continued Fractions
1.1.1. Show that
sup{x e Q : x > 0 , x2 < 2} = V2.
1.1.2. Let A C M be a nonempty set. Define —A = {x : —x e A } .
Show that
sup(—A) = —inf A,
inf (—A) = - s u p A.
1.1.3. Let A, B e l
be nonempty. Define
A + B = {z = x + y:xeA,
yeB},
A-B
y €B}.
= {z = x-y:xeA,
Show that
sup(A -f B) = sup A + sup B ,
sup( A - B) = sup A - inf B.
Establish analogous formulas for inf (A + B) and inf (A — B).
3
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Problems. 1: Real Numbers
4
1.1.4. Given nonempty subsets A and B of positive real numbers,
define
A • B = {z = x - y : x € A, y eB}
A = {Z = ± : X € A } .
Show that
sup(A • B) = sup A • sup B .
Show also that, if inf A > 0, then
sup
(1) = ex
and, if inf A = 0, then sup (^-) = +oo. Additionally, show that if A
and B are bounded sets of real numbers, then
sup(A • B)
= max{sup A • sup B , sup A • inf B , inf A • sup B , inf A • inf B } .
1.1.5. Let A and B be nonempty subsets of real numbers. Show
that
sup(A U B) = max{sup A, sup B}
and
inf (A U B) = min{inf A, inf B } .
1.1.6. Find the least upper bound and the greatest lower bound of
A i , A2 defined by setting
Ai = J2(-l)"+1 + ( - 1 ) ^ (2 + £) : n €N} ,
fn-1
2717T
1
A2 = <
7 cos —— : n e N > .
\n+l
3
J
1.1.7. Find the supremum and the infimum of the sets A and B ,
where A = {0.2,0.22,0.222,... } and B is the set of decimal fractions between 0 and 1 whose only digits are zeros and ones.
1.1.8. Find the greatest lower and the least upper bounds of the set
of numbers \n^' , where n e N.
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1.1. Supremum and Infimum. Continued Fractions
5
1.1.9. Find the supremum and the infimum of the set of numbers
n+
2~l
2"m , where n,ra G
1.1.10. Determine the least upper and the greatest lower bounds of
the following sets:
(a)
A = < — : m, n G N, m < 2n \ ,
(b)
B = { v ^ ~ [y/n] : n G N} .
1.1.11. Find
(a)
sup{a: e R : x 2 + a ; + l > 0 } ,
(b)
mf{z = x + x~1 : x > 0},
(c)
inf{2 = 2* + 2-
:x>0}.
1.1.12. Find the supremum and the infimum of the following sets:
(a)
A = < — + — : m, n G N > ,
[ n
m
J
(c)
C
(d)
D = i T — ^ — : m G Z, n G N i ,
^ |m| -h n.
J
(e)
E= 4
\ m
: ra,neN>,
[m + n
J
f
^^
T,T1
: m, n G N > .
1.1.13. Let n > 3 be an arbitrarily fixed integer. Take all the possible
finite sequences ( a i , . . . , a n ) of positive numbers. Find the least
upper and the greatest lower bounds of the set of numbers
E:
^ afc + afc
Ofc
+1
-f afc+2
where we put a n +i = a\ and a n +2 = &2 •
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Problems. 1: Real Numbers
6
1.1.14. Show that for any irrational number a and for any positive
integer n there exist a positive integer qn and an integer pn such
that
_Pn\
_1_
qn I nQn
Show also that {pn} and {qn} can be chosen in such a way that we
have
a—
<
qn
1.1.15. Let a be irrational. Show that A = {m + not : m, n G Z}
is dense in R, i.e. in any open interval there is at least one element
of A.
1.1.16. Show that {cosn : n G N} is dense in [—1,1].
1.1.17. Let x G R \ Z. Define the sequence {xn}
x = [x] + — ,
X\
xi = [xi] + — ,...,
X2
by setting
x n _i = [x n _i] -h — .
Xn
Then
X = \X +
N +
[x2] +
•• -f
[x n _i] -h
Show that x is rational if and only if there exists n G N for which
xn is an integer.
Remark. The above representation of x is said to be a finite continued fraction. The expression
1
a0 +
a\ +
a2 +
"• +
Gn-1 +
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1.1. Supremum and Infimum. Continued Fractions
7
will also be written in the more convenient form
1|
II
II
ao + T^ + T^ + .-. + T-1-.
\CL2
\ai
\an
1.1.18. For positive real numbers ai,a2,...,o n set
Po = do ,
Pi = a0ai + 1,
<7o = 1,
^i = ai,
Pk = Pk-i&k + Pk-2,
qk = qk-ia>k + qk-2> with fc = 2,3,...,n,
and define
^0 = 0 0 ,
i?fc = « o + i
hi
h ... + i — ,
k = l,2,...,n .
(jRfc is called the kth convergent to a o + T^ + T^ + --- + -n~JShow that
Rk = —
for k — 0, l,...,n.
1.1.19. Show that if pk.qk are defined as in the foregoing problem
and the ao,oi,...,a n are integers, then
Pk-iQk ~ qk-iPk = (~l) fc
for
k = 1,2,..., n.
Use this equality to conclude that p/- and qk are co-prime.
1.1.20. For an irrational number x we define a sequence {xn}
#i =
1
x - [ xr^5
]'
x
2
1
xi-[xi]'"*'
n
xn-i
by
1
- [xn_i]'""
Moreover, we put ao = [x], an = [x n ], n = 1,2,..., and
Rn = a0 +
T-1
|ai
+ -r-1 + ... + r 1 - .
|a2
|a n
Show that the difference between the number x and its nth convergent
is given by
x
- R =
v
;
where pn,<7n are defined in 1.1.18. Conclude that x is between its
two consecutive convergents.
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Problems. 1: Real Numbers
8
1.1.21. Show that the set {sinn : n G N} is dense in [—1,1],
1.1.22. Apply the result in 1.1.20 to prove that for every irrational
number x there exists a sequence i^}
of rational numbers, with
odd qn, such that
I
Pn
< 2
\X
Qn
I Qn
(Compare with 1.1.14.)
1.1.23. Derive the following formula for the difference of two consecutive convergents:
(-l)n
Rn+i — Rn —
.
QnQn+l
1.1.24. Let x be irrational. Show that its convergents Rn defined
in 1.1.20 are successively closer to x, that is,
\x- Rn+i\ <\x - Rn\,
1.1.25. Prove that
imation of x of all
That is: if r/s is
such that \x — r/s\
n = 0,1,2,....
the convergent Rn = pn/qn is the best approxrational fractions with denominator qn or less.
a rational number with a positive denominator
< \x — i? n |, then s > qn.
1.1.26. Expand each of the following as infinite continued fractions:
1.1.27. For a positive integer &, find the representation of y/k2 + k
by an infinite continued fraction.
1.1.28. Find all the numbers x € (0,1) whose infinite continued
representations have a\ (see Problem 1.1.20) equal to a given positive
integer n.
1.2. Some Elementary Inequalities
1.2.1. Show that if a^ > —1, k = 1,..., n, are all positive or negative,
then
(1 + ai) • (1 + a 2 ) •... • (1 + a n ) > 1 + ai + a 2 + ... + an .
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9
1.2. Some Elementary Inequalities
Remark. Note that if a\ = a 2 = ... = an = a then we get the well
known Bernoulli inequality: (1 + a)n > 1 + n a , a > - 1 .
1.2.2. Using induction, prove the following result: If a i , a 2 , . . . , a n
are positive real numbers such that a\ • a 2 •... • an = 1, then ai + a 2 +
... + an>n.
1.2.3. Let An, Gn and Hn denote the arithmetic, geometric and
harmonic means of n positive real numbers a i , a 2 , ...,a n ; that is,
_ Q>1 + Q>2 + ... + CLn
A
An —
Gn
=
71
,
n
\/0>l ' . . . • CLn j
"" i + i + ... + -^ *
Show that An > Gn > Hn.
1.2.4. Using the result (Gn < An) in the foregoing problem, establish
the Bernoulli inequality
(1 + x)n > 1 + nx for x > 0.
1.2.5. For n €N, verify the following claims:
1
1
1
1
2
(b)
^TI + ^T2 + ^T3 + - + 3 ^ T T > 1 '
W
1
1
<
2 3n + l
(d)
n(v / ^TT-l)<l + i + ... + -
+
1
J_
1
2
+
+
+
<
3n + 2 '" 5n 5 n + l 3'
<n(l
1
;
+
r ) ,n > l .
v^TT n-fiy'
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Problems. 1: Real Numbers
10
1.2.6. Show that for any x > 0 and n G N we have
xn
1
3
2n <
1 + x + x + x + ... + x
- 2n + 1
2
1.2.7. Let {an} be an arithmetic progression with positive terms.
Show that
Vaian < y/a1a2.^an
<
——.
1.2.8. Show that
V ^ < >/n!< ^ 5 ^ ,
neN.
1.2.9. Let dk , k = 1,2,..., n, be positive numbers and let ^ a^ < 1.
fc=i
Show that
n
n2.
1.2.10. Let a/c > 0, A: = 1,2, ...n, where n > 1, and set s = J2 akVerify the following claims:
(b)
(c)
k=\
E - ^ ^ - ^ '
jTT-^-Y'zn + l.
1.2.11. Show that if a^ > 0, fc = 1, ...,n, and a\ • a2 • ... • a n = 1,
then
(l + a 1 )(l + a 2 ) . . . . . ( l + a n ) > 2 n .
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11
1.2. S o m e E l e m e n t a r y I n e q u a l i t i e s
1 . 2 . 1 2 . Prove the following Cauchy
inequality:
n
n
k=i
fc=i
£«*»* <E«*XX
VJk=l
1 . 2 . 1 3 . Show t h a t
E H + SJfe=l
EM
\jfe=i
2\ 2
4
<£(«2+«)
fe=i
1 . 2 . 1 4 . Show t h a t if E afc = E 6fc = *>
/c=l
/c=l
/^Qfcbfc
fc=l
then
< 1.
1 . 2 . 1 5 . For afc > 0, k = 1,2, ...,n, verify the following claims:
n
n
(a)
(b)
1
a f e
E
E ^ 7 ^ n 2 '
u
fc=i fe=i
k
n
n 1 -a
afe _ Ji k n
1_afe
E
E
^
)'
Ofe ^ E(
fc=i fc=i
*:=!
(c)
2
2
( l o g a a ! ) + ( l o g a a 2 ) + ... + ( l o g a a n ) 2 > - ,
n
provided
a\ • a<i •... • an = a 7^ 1.
1 . 2 . 1 6 . For a > 0, show t h a t
-,
y^Qfcfrfc
fc=i
n
n
fc=l fc=l
1.2.17. Establish t h e following inequalities:
^ M
k=l
< y/n
X^aM
Kk=l
-
Vn^2\akl
fe=l
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Problems. 1: Real Numbers
12
1.2.18. Show that
(a)
, z ^ - - i
- ^ - - . ^ f c
fc=i
2
<•»
IE?) ^E^E?fc=l fc=l
Vjfc=l
1.2.19. Show that
5>£ <£ar*£«r>
for any real p,q and any positive ai,a2, ...,a n .
n
1.2.20. Find the minimum of the sum ^T, a\
fer
subject to the con-
n
straint ]T) ak = 11.2.21. Let
PiiP2i-~<>Pn be given positive numbers. Find the mini-
n
n
mum of J ] p/ca| subject to the constraint ^ a^ = 1.
fc=i fc=i
1.2.22. Show that
^ a 2 f c + 2aia 2
£ afc J < ( n - l )
u=i
<fc=i
1.2.23. Verify the following claims:
(a)
(b)
/
n
\
2
2
/
n
\
X > + &*) * 5>M
u=i
u=i
E4 - E<
\Jfe=l
\Jfe=l
2
+
/
n
\
2
U>* '
u=i
< E l ^ "6jfelfc=l
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1.2. Some Elementary Inequalities
13
1.2.24. Let
PiiPii •••iPn be given positive numbers. Find the minimum of
n
( n
\
2_] l + I Z-lak I
n
a
fc=i
\k=i
subject to the constraint
J
/^Pfca/c = 1k=i
1.2.25.. Prove the following Chebyshev inequality: If a\ > a2 >
... > an and bi > 62 > ... > bn (or a\ < a2 < ... < an and
b\ < 62 < ... < 6 n ), then
22^k^2h < ny^akbk.
k=l
k=l
k=l
1.2.26. Assuming ak > 0, A: = 1,2, ...,n, and p € N, show that
\
k=l
fc=l
/
1.2.27. Establish the inequality
(a + 6 ) 2 < ( l + c ) a 2 + ( l + - ) 6 2
for positive c and any real a and 6.
1.2.28. Show that |\/a 2 + b2 - Va2 + c21 < |fe-c|.
1.2.29. For positive a, 6, c, verify the following claims:
(a)
b
(d)
v
'
(e)
be
ac
,
ab ^ ,
— + — + — > (a + b +N c),
a
0
c
1
1
1 ^ 1
a
b
c
- + - + ->-=
2
6+ c
b2-a2
2
y/bc
1
1
+ -= + - 7 =,
c-f-a
a-f
2 2
c -b
+
r +
c+ a
a+b
1(0-6)2
a+ 6
< —
8
a
2
y/ca
2
^
y/ab
9
6
a+ b+c
2 2
a -c
n
-r
^ 0,
b+c
^ ^ l(a-fe)2
Vab < - -—-—!—
8 0
...
provided
b < a.
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Problems. 1: Real Numbers
14
1.2.30. For ak e R, bk > 0, k = 1,2,..., n, set
m = min< — : k = 1,2, ...,n > and M = maxj — : fe = 1,2, ...,n >.
Show that
a\ +a2 + ... + a n
m <
— < M.
- 6 i + 6 2 + ... + 6 n ~
1.2.31. Show that, if 0 < <*i < a 2 < ... < a n < f, n > 1, then
tanai <
sin a i + sina2 + ... + s i n a n
< tanan.
cosOL\ -f cosa 2 + ... + cosa n
1.2.32. For positive Ci,c 2 , ...,c n , and fci, A;2,..., A:n G N, set
5 = max{ ktfc{, k^cT2,..., V ^ } ,
5 = min{ kt/c[, k^fc2,..., V c ^ } .
Show that
5 < [a\ • a2 •... • an)fci+fc2+--+fen < 5.
1.2.33. For afc > 0, bk > 0, A: = 1,2,..., n, set
M = m&x<-—
: fc = l,2,...,n > ,
6
I*
Show that
J
ai +a\ + ... + a£
<M.
6i + M6| + ...+ M n - ! ^
1.2.34. Show that, if # is greater than any of the numbers ai,a 2 ,...,
a n , then
1
x-a\
1
+ x-a2
+ "'...+
1
x-an
>
x -
Q
n
i+a2+---+an '
1.2.35. Let Cfc = (£), A: = 0,1,2, ...,n, be the binomial coefficients.
Establish the inequality
v/cT+V/C2 + - + V / ^ < Vn(2n ~ !)•
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1.2. Some Elementary Inequalities
15
1.2.36. For n > 2, show that
/ 2 „_ 2 y-i
fc=0
x
'
1.2.37. Let a,k > 0, fc = 1,2, ...,n, and let A n be their arithmetic
mean. Show that for any integer p > 1,
fc=i
^
fc=i
1.2.38. For positive a^, fc = 1,2,..., n, we set a = ai + c&2 + ... -f a n .
Show that
n—1
2
fc=l
1.2.39. Show that for any rearrangement 6i, 6 2 ,..., 6n of the positive
numbers oi,a2, ...,o n ,
T-n'
r + h~ + "' +
1.2.40. Prove the Weierstrass inequalities: If 0 < a& < 1, fc =
1,2,..., n, and a\ + c&2 + ... + a n < 1, then
n
(a)
n
1
1
afc
1 + J2 * < I I t + ) <
/c=i
n
(b)
a
fc=i
n
..
1 fl
1 - J2 "* < II^ - *)<
/s=i
n >
1 — ^ a^
fc=i
fc=i
n •
i + J2 ak
fc=i
1.2.41. Assume that 0 < a^ < 1, fc = 1,2, ...,n, and set ai 4- a2 +
... + o n = o. Show that
ak
Err a
fc
na
n-a
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Problems. 1: Real Numbers
16
1.2.42. Let 0 < Ofc < 1, fc = 1,2,..., n, and n > 2. Verify the inequality
71
n
n Y, ak
fc=i
Y-±—<
—
Z-f1 1 ++ aa* - «
1
fc=i
*
Eajb + nfta*
fc=i fc=i
1.2.43. For nonnegative a^, A; = 1,2, ...,n, such that a\ + 0,2 + ... 4a n = 1, show that
n(1 + afc )^( n+1 ) n II afc '
(a)
n
n
n^ - *)^^- )"!!0*-
(b)
1 0
1
fc=l fc=l
1.2.44. Show that if a& > 0, A; = 1,2, ...,n, and ^
k=l-'lk
then
j ^ ^ - = n — 1,
ni-> ( n-ir.
k=lak
1.2.45. Prove that under the assumptions of 1.2.43, we have
n
nu+afc)
na-flfc)
^ (n
r—
n > ^r—vx—
+ ^l ) —
"
(n-l)n >
n>l.
1.2.46. Show that for positive ai, 02,..., a n ,
a i
02 + ^3
1
a
2
03+^4
,
,
a
n-2
an_i+an
fln-1
an-hai
,
Qn
a\ + 02
. ™
4
1.2.47. Let £ and a i , a 2 , . . . , a n be any real numbers. Establish the
inequality
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1.2. Some Elementary Inequalities
17
1.2.48. Show that for positive ai,a2, ...,an> and 61,62, ...,6 n , we
have
y/fa
+ 6i)(a 2 + 6 2 )...(a n + 6n) > ^aia 2 ...a n + ^ i ^ . . . 6 n .
1.2.49. Assume that 0 < a\ < 0,2 < ... < a n and Pi,P2>---?Pn are
n
nonnegative such that Yl Vk = 1- Establish the inequality
where A = ^(ai + a n ) and G = ^/ala^.
1.2.50. For a positive integer n, let cr(n) and r(n) denote the sum
of all the positive divisors of n and the number of these divisors,
respectively. Show that ^ y > y/n.
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Chapter 2
http://dx.doi.org/10.1090/stml/004/02
Sequences of Real
Numbers
2.1. Monotonic Sequences
2.1.1. Show that
(a) if a sequence {an} is monotonically increasing, then lim an =
n—>oc
sup{a n : n G N};
(b) if a sequence {an} is monotonically decreasing, then lim an =
inf{a n : n G N}.
n—>oo
2.1.2. Let ai,a2,...,a p be fixed positive numbers. Consider the sequences
a7t + a2 + ... + a%
and
xn = \fs^ , n G
P
Show that the sequence {xn} is monotonically increasing.
Hint. First establish monotonicity of the sequence -J -^- >, n > 2.
2.1.3. Show that the sequence {a n }, where an = ~ , n > 1, strictly
decreases and find its limit.
2.1.4. Let {an} be a bounded sequence which satisfies the condition
o>n+i > o n — T^r, n G N. Show that the sequence {a n } is convergent.
Hint. Consider the sequence {an — ^r=r} •
19
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Problems. 2: Sequences of Real Numbers
20
2.1.5. Prove the convergence of the sequences:
(a)
an = _ 2 ^ + (i ! +
-L + ... + -L);
(b)
6n = - 2 v ^ + ( ^ + - l +
... + -I=).
Hint. First establish the inequalities
1
2(y/n + 1 - 1) < - = + -7= + ... + - = < 2Vn,
Vl
v2
yn
n €N.
2.1.6. Show that the sequence {an} defined recursively by
a\ = - ,
a n = -y/3an_i — 2 for n > 2,
converges and find its limit.
2.1.7. For c > 2, define the sequence {an} recursively as follows:
a\ = c2,
a n +i = (a n - c) 2 ,
n > 1.
Show that the sequence {an} strictly increases.
2.1.8. Suppose that the sequence {an} satisfies the conditions
0 < an < 1,
an(l - an+i) > ~ for n G N.
Establish the convergence of the sequence and find its limit.
2.1.9. Establish the convergence and find the limit of the sequence
defined by
&i = 0, a n + i = \/6 -I- an for n > 1.
2.1.10. Show that the sequence defined by
ai=0,
a2=2>
an+i = g ( l + an + an-i)
for
n > 1
converges and determine its limit.
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2.1. M o n o t o n i c Sequences
21
2.1.11. Study the monotonicity of the sequence
a
H h
"=(2^TIJ!!'
-
and determine its limit.
2.1.12. Determine the convergence or divergence of the sequence
(2n)!!
.
(2n+l)!!
n> 1.
2.1.13. Prove the convergence of the sequences
(a)
(b)
0
« =
1 +
22
+
3 2 + - + ^>
i + J_ + J_ + ... + J_,
Gn =
n e N
5
neK
2.1.14. Show the convergence of the sequence {a n }, where
1
1
1
an — —====. -\ .
-f ... H ,
=,
y/n(n + 1)
^/(2n - l)2n
x / ( n - h l ) ( n + 2)
n G N.
2.1.15. For p G N , a > 0 and ai > 0, define the sequence {an} by
setting
a n + i = - ((p - l)an + - ^ r r J > ^ G N.
Determine lim a n .
n—•oo
2.1.16. Define {a n } recursively by
a n + i = J 2 -f >/^n
ax =
for n > 1.
Prove the convergence of the sequence {an} and find its limit.
2.1.17. Define the recursive sequence {an} as follows:
2(2a n + 1 )
^_^
—— for n G N.
an + o
Establish the convergence of the sequence {an} and find its limit.
1
ai = 1, a n +i =
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Problems. 2: Sequences of Real Numbers
22
2.1.18. Determine all c > 0 such that the recursive sequence {an}
defined by setting
1
c
a
i = 2'
a
n+i==2^C+
a
^
f o r n
^
N
converges. In case of convergence find lim an.
n—•oo
2.1.19. Let a > 0 be fixed and define the sequence {an} by setting
a
^
i
n + 3°
a\ > 0 and a n + i = OLnTr-^
Sal + a
_
__
lor n G R
Determine all a\ for which the sequence converges and in such a case
find its limit.
2.1.20. Let {an} be defined recursively by
Gn+i = ~A ^— for n > 1.
4 - 6an
Determine for which a\ the sequence converges and in case of convergence find its limit.
2.1.21. Let a be arbitrarily fixed and let {an} be defined as follows:
a\ £ R
and
a n + i = a\ + (1 — 2a)an -f a 2
for
n G N.
Determine all ai such that the sequence converges and in such a case
find its limit.
2.1.22. For c > 0 and 6 > a > 0, define the recursive sequence {an}
by setting
al + ab
^
a\ = c, a n + i = —-—— for n e N.
a-\-b
Determine for which values a, b and c the sequence converges and find
its limit.
2.1.23. Prove the convergence and find the limit of the sequence
{an} defined inductively by
a\ > 0,
an+i = 6-,
7 + an
n € N.
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23
2.1. Monotonic Sequences
2.1.24. For c > 0, define the sequence {an} as follows:
ai = 0, an+\ = y/c + an, n e N.
Show the convergence of the sequence and determine its limit.
2.1.25. Investigate the convergence of the sequence defined by
a\ = y/2, an+\ = \f2a~n~, n e N.
2.1.26. Let k E N be fixed. Study the convergence of the sequence
{an} defined by setting
a1 = >/5, a n + i = >/5a^, n e
2.1.27. Investigate the convergence of the sequence {a n } given by
1 < ai < 2,
a2n+1 = 3a n - 2, n G N.
2.1.28. For c > 1, define the sequences {an} and {6n} as follows:
(a)
ai = \Jc{c-
(b)
6i = \/c,
1),
a n + i = y/c(c-
1) + a n , n > 1;
&n+i = \/c6^, n > 1.
Prove that both sequences tend to c.
2.1.29. Given a > 0 and 6 > 0, define the sequence {an} by setting
0 < ai < 6,
Find
a n + i = A/
-^
V a 4-1
for
n > 1.
lim a n .
n—+oo
2.1.30. Prove the convergence of {an} defined inductively by
a\ = 2,
and find its limit.
a n +i =2-1- ———
3
+f
for
n > 1
On
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Problems. 2: Sequences of Real Numbers
24
2.1.31. The recursive sequence {an} is given by setting
a\ = 1,
a>2 = 2,
a n + i = yJan-\ + yfa^
for
n > 2.
Show that the sequence is bounded and strictly increasing. Find its
limit.
2.1.32. The recursive sequence {an} is given by setting
ai = 9,
a2 = 6,
an+\ = y/an-i
+ ^Ja^
for
n > 2.
Show that the sequence is bounded and strictly decreasing. Find its
limit.
2.1.33. Define the sequences {an} and {bn} as follows:
0<6i<ai,
a n +i =
and
bn+1 = \/anbn
for
n € N.
Show that {a n } and {^)n} both tend to the same limit. (This limit
is called the arithmetic-geometric mean of a\ and b\.)
2.1.34. Show that the sequences {an} and {bn} given by
0<6i<ai,
an+1 =
Qn +
I71
and frn+1 =
Qn
^
6n
for
neN
are both monotonic and have the same limit.
2.1.35. Let the recursive sequences {an} and {bn} be given by setting
(\ ^ u ^
a n + 6n
2a n 6 n
0 < &i < ai, a n +i =
and o n + i =
— for
neN.
2
a n + 6n
Prove the monotonicity of these sequences and show that both of them
tend to the arithmetic-geometric mean of a\ and b\. (See Problem
2.1.33.)
2.1.36. Show the convergence and find the limit of { a n } , where
n + 1 [2
+
22
^=2^r(i T
+
-
+
2n\
T1
r
for
ne
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2.1. Monotonic Sequences
25
2.1.37. Suppose that a bounded sequence {an} is such that
1
2
Gn+2 < o a n + l + o a n
f°r
Tl > 1.
Prove the convergence of the sequence {a n }.
2.1.38. Let {an} and {bn} be defined as follows:
an = f 1 + - ) ,
6n = ( 1 + - J
for
n € N.
Using the arithmetic-geometric-harmonic mean inequalities, show that
(a) an < bn
for n e N.
(b) the sequence {an} is strictly increasing,
(c) the sequence {bn} is strictly decreasing,
Show also that {an} and {bn} both have the same limit, defined to
be Euler's number e.
2.1.39. Let
an = (l -h - )
for
n G N.
(a) Show that if x > 0, then the sequence {an}
strictly increasing.
is bounded and
(b) Let x be any real number. Show that the sequence {an}
bounded and strictly increasing for n > — x.
is
The number ex is defined to be the limit of this sequence.
2.1.40. Suppose that x > 0, I e N and / > x. Show that the
sequence {bn}, where
/
x\ i~^n
bn = ^1 + - \
for
n e N,
is strictly decreasing.
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Problems. 2: Sequences of Real Numbers
26
2.1.41. Establish the monotonicity of the sequences {an} and {bn},
where
an = 1 + - + ... H
Inn for n G N,
2
n- 1
6n = l + - + ... 4- + - - l n n for n G N.
2
n—1 n
Show that both of them tend to the same limit 7, which is known as
Euler's constant.
Hint. Apply the inequality (l + ^ ) n < e < ( l + ^ ) n , which follows from 2.1.38.
2.1.42. Given x > 0, set an = 2 y / x, n G N. Show that the sequence
{an} is bounded. Show also that it is strictly increasing if x < 1 and
strictly decreasing if x > 1. Compute lim an.
Moreover, put
cn = 2 n ( a n - l )
dn = 2n (1 - — )
and
Show that {c n } is decreasing and {dn}
quences have the same limit.
for
n G N.
is increasing and both se-
2.2. Limits. Properties of Convergent Sequences
2.2.1. Calculate:
(a)
lim \ / l 2 + 2 2 + ... + n 2 ,
(b)
lim
(c)
(d)
(e)
n + sin n 2
n-*<x>
n + cos n
n—>oc
lim
n—>oc
l - 2 + 3 - 4 + . . . + (-2n)
/
9
. ,
lim (V2 - ^2)(>/2 - #2) •... • (V2 n
lim
_,
2
"V2),
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2.2.
Limits. P r o p e r t i e s of Convergent Sequences
27
71'
- . , ,.
/
1
2
n
( h ) n->oo
„ l l n L y nr ¥2 -+T T1 + Z
- + n2 + n
n 2T -+T 2^ + '*
'
/.x
/
n
2n
nn
W n™oo
_ h l ?_ V -n T3 T
+ T1 + Z-T^T
n 3 -f 2 + -' " + n 3 -f n
2 . 2 . 2 . Let 5 > 0 and p > 0. Show t h a t
hm —
n->oo (1
2 . 2 . 3 . For a G (0,1), calculate
2 . 2 . 4 . For a £ Q, calculate
2 . 2 . 5 . Show t h a t the limit
— = 0.
+p)n
lim ((n + l ) a n—•oo
na).
lim sin(n!a7r).
n—>-oo
lim s i n n does not exist.
n—>oo
2 . 2 . 6 . Show t h a t for any irrational a the limit
not exist.
lim sinna7r does
n—>-oo
2 . 2 . 7 . For a € R, calculate
hm -
n->oo n VV
a+-
n/
+U+\
nj
2 . 2 . 8 . Suppose an ^ 1 for all n and
integer A:, c o m p u t e
lim
n—>oo
+ ... + a +
\
n
lim an = 1. Given a positive
71—>00
an + QTl + ... + a ^ - f c ^
an — 1
2 . 2 . 9 . Find
,.
1
1
I N
f"
h- m
T-ir-z
+
TT-z—:
+
+
ooVl-2-3
2-3-4
"'
n-(n + l)-(n + 2 ) y
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Problems. 2: Sequences of Real Numbers
28
2.2.10. Calculate
-=-fc3-l
hm TT 7-5—-.
3
n-^00 1 1 fc + 1
k=2
2.2.11. Determine
lim n
i
EE n3
3
t=i j=i
2.2.12. Compute
-3 J-4 I •... • ( 1 -
lim [ 1 - A)(l
n->oc\
2-3/ V
* /
V
(n + l ) - ( n + 2)
2.2.13. Calculate
,.
hm
n
A f c 3 + 6A;2 + llfc + 5
>
TZ
—
.
(fc + 3 ) !
^°°k^i
2.2.14. For x ^ - 1 and x ^ 1, find
n
lim
El
fc=l
2fc_1
2="-
2.2.15. Determine for which x G R the limit
lim 1 7 ( 1 + x 2 * )
-1"-1"
l—•OO
n—•oo
exists and find its value.
2.2.16. Determine all x G R such that the limit
n
,
lim fc=Q
17 Ix 1 + S
n—•oo
A A
\
2
t l
exists and find its value.
2.2.17. Establish for which x G R the limit
lim f](l + x*k +x™k)
n—• oo
A l
fc=l
exists and find its value.
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2.2. Limits. Properties of Convergent Sequences
29
2.2.18. Calculate
,.
l - l ! + 2-2! + ... + n - n !
hm
—
.
n-+oo
(n+1)!
2.2.19. For which x E M does the equality
n 1999
j
lim
n-oo nx - (n - l ) x
2000
hold?
2.2.20. Given a and b such that a > 6 > 0, define the sequence {an}
by setting
ab
a\ = a -f 6, a n = oi
, n > 2.
Gn-l
Determine the nth term of the sequence and compute lim an.
n—•oo
2.2.21. Define the sequence {an} by setting
(H = 0,
a2 = 1 and
a n + i — 2a n + a n _i = 2
for
n > 2.
Determine its nth term and calculate lim o n .
n—+oo
2.2.22. For a > 0 and 6 > 0, consider the sequence {a n } defined by
ai =
a6
y
an =
: and
=,
n > 2.
Determine its nth term and find lim an.
n—>oo
2.2.23. Let {a n } be a recursive sequence defined as follows:
ai = 0,
an =
an_i+3
,
n > 2.
Find the formula for the n t h term of the sequence and find its limit.
2.2.24. Study the convergence of the sequence given by
a\ = a,
a n = 1 + 6o n _i,
n > 2.
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Problems. 2: Sequences of Real Numbers
30
2.2.25. The Fibonacci sequence {an} is defined as follows:
a\ = d2 = 1,
fln+2
= 0"n + G n + l ,
Tl > 1.
Show that
an = a-0 '
where a and /? are roots of x2 = x + 1. Compute lim tfa^.
n—+oo
2.2.26. Define the sequences {an} and {bn} by setting
a\ = a,
a n + bn
h = 6,
bn +1
Show that lim an = lim 6 n .
n—+00
n—>oo
2.2.27. Given a £ {1,2,...,9}, compute
n digits
a 4- aa -f ... + aa...a
lim
10 s
'
n—>oo
2.2.28. Calculate
lim (tfn- l) r
2.2.29. Suppose that the sequence {an} converges to zero. Find
lim a".
n-+oo
n
2.2.30. Given positive Pi, P2> ••• >Pfc a n d a i , a 2 ,... ,a^, find
lim
n^oo
PiaxT +p2a2
+ — +Pk%
pia^ + p 2 a£ + ... + p^a^
2.2.31. Suppose that lim
Qn+1
q. Show that
n—>oc
(a) if q < 1, then lim a n = 0,
n—>oo
(b) if (j > 1, then lim |a n | = oc.
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2.2. Limits. Properties of Convergent Sequences
31
2.2.32. Suppose that lim y/\an\ — q. Show that
n—>oo
(a) if q < 1, then lim an = 0,
n—• oo
(b) if q > 1, then lim |a n | = oo.
n—KX>
2.2.33. Given a real number a and x G (0,1), calculate
lim
naxn.
n—•oc
2.2.34. Calculate
m(m - 1) • ... • (m - n + 1) nn
hm —-^
-x ,
n—*oo
n\
2.2.35. Assume that
Show that
lim anbn
.
for
__
, . , ,
m G N and |z| < 1.
lim an = 0 and {6 n } is a bounded sequence.
n—KX>
=0.
n—>oo
2.2.36. Show that if lim an = a and lim bn = 6, then
n—>oo
n—•oo
lim max{a n , 6 n } = max{a, b}.
n—>oo
2.2.37. Let an > - 1 for n G N and let lim a n = 0. For p G N,
n—KX>
find
lim ^/l -f- an.
n—+oo
2.2.38. Assume that a positive sequence {an}
For natural p > 2, determine
lim
n->oo
^ ^ ~
1
converges to zero.
,
a„
2.2.39. For positive ai,a2, ...,a p , find
lim f y (n + ai)(n + a 2 ) •... • (n + a p ) - n J.
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Problems. 2: Sequences of Real Numbers
32
2.2.40. Calculate
1
,. (
1
1
\
lim
,0
4- , ,
+ ... 4- - = = = =
n-oo V>/n2 + 1 V V T 2
\/n2+n + l/
2.2.41. For positive ai,ei2, ...,a p , find
lim
n/a?+aJ
+ ... + aJ
2.2.42. Compute
n 1999
n 1999
lim \ / 2 sin2
4- cos2
n->oc V
n -f 1
n -f 1
2.2.43. Find
lim (n 4- 1 4- n cos n) 2n+nsin?
2.2.44. Calculate
sz,±{f^-i
2.2.45. Determine
*SX(\KJ-I
2.2.46. For positive a^, fc = 1,2, ...,p, find
p
lim
- Y" ^afc
2.2.47. Given a € (0,1), compute
n-l
lim Y ] ( a + - )
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2.2. Limits. Properties of Convergent Sequences
33
2.2.48. Given real x > 1, show that
lim (2^/x-l)n
= x2.
2.2.49. Show that
Km e ^ : 1 ) " = i.
2.2.50. Which of the following sequences are Cauchy sequences?
.
(a)
tanl
tan 2
tann
2
2n
an = —— + —=+ ... +
2
,
1
2
, 1 1
2 + 3+
= 1+
1
-+n'
(c)
a
<d>
a
(e)
an = a i ^ 1 + a 2 g 2 + ... + an<f\
"
'
n2
2
^l^-2 i 3 + -" + ^ n " 1 ^ T I ) '
for |?| < 1 , |a fc | < M, fc = l,2,...,
W
^ — 2 2 + 32+ "•+ ( n + i ) 2 -
2.2.51. Suppose that a sequence {a n } satisfies the condition
|a n +i - a n + 2 | < A|a
with a A e (0,1). Prove that {an} converges.
2.2.52. Given a sequence {a n } of positive integers, define
1
1
1
S n = — + — + ...+ —
and
a.
Prove that if {Sn} converges, then {ln<7n} also converges.
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Problems. 2: Sequences of Real Numbers
34
2.2.53. Show that the sequence {Rn} of convergents to an irrational
number x (defined in Problem 1.1.20) is a Cauchy sequence.
2.2.54. For an arithmetic progression {an} whose terms are different
from zero, compute
,.
/
1
1
hm \a1a2 + a<2.az4-... +
n-+oo
1
anan+i
2.2.55. For an arithmetic progression {an} with positive terms, calculate
l
,.
(
lim —= - —
l
l
— + ——
— + ... + —=
2.2.56. Find
X
(a)
lim n(ye — l),
(b)
n—>oo
1
lim
n—+ 00
Ik
77,
.
2.2.57. Let {a n } be a sequence defined as follows:
ai = a, 02 = 6, a n + i = pan-\
+ (1 — p)an» n = 2,3,...
Determine for which values a, b and p the sequence converges.
2.2.58. Let {an} and {bn} be defined by setting
a\ = 3 , 61 = 2, a n + i = a n -f 26n
and
6 n + i = a n -f bn.
Moreover, let
Cn = ir,
On
neN.
(a) Show that |c n + i - \ / 2 | < | | c n - \/2|, n G N.
(b) Calculate lim c n .
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2.3. Toeplitz Transformation and Stolz Theorem
35
2.3. The Toeplitz transformation, the Stolz
theorem and their applications
2.3.1. Prove the following Toeplitz theorem on regular transformation of sequences into sequences.
Let {cU)k : 1 < fc < n, n > 1} be an array of real numbers such
that:
(i)
cn k —> 0
'
n
for each fc G N,
n—>oo
(ii)
y^c n ) / c —> 1,
(iii)
there exists C > 0 such that for all positive integers n:
*-—*
fc=l
ro—>oo
n
^|cn,fc|<C.
fc=l
Then for any convergent sequence {an} the transformed sequence
n
{bn} given by 6 n = Yl cnkO>k, n> 1, is also convergent andn lim 6 n =
fc=i '
~"°°
lim an.
n—>oo
2.3.2. Show that if lim an = a, then
n—->oo
ai + a 2 + ... + a n
lim
n—>oo
n
= a.
2.3.3.
(a) Show that the assumption (iii) in the Toeplitz theorem (Problem
2.3.1) can be omitted if all the numbers cn^ are nonnegative.
(b) Let {bn} be the transformed sequence defined in the Toeplitz
theorem with cn^ > 0, 1 < fc < n, ra > 1. Show that if lim an =
n—>oo
-|-oo, then lim bn = +oo.
n—KX)
2.3.4. Show that if lim an — -hoc, then
n—>oo
lim
n—>oo
oi + a 2 + ...+ a n
77,
= -foo.
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36
P r o b l e m s . 2: S e q u e n c e s o f R e a l N u m b e r s
2 . 3 . 5 . Prove t h a t if lim an = a, then
n—»oo
lim
na\ -f (n — l)a2 + ... + 1 • an
a
n—>oo
2.3.6. Show t h a t if a positive sequence {an}
lim i / a i •... • a n = a.
converges t o a, then
n—•oo
2 . 3 . 7 . For a positive sequence { a n } , show t h a t if
then
lim ^ ^
= a,
lim r/a^, — a.
n—>-oo
2.3.8. Let
lim a n = a and
n—>oc
r
lim
2.3.9. Let {an}
(i)
(ii)
^ i 6 n + « 2 ^ n - l + ... +Clnbi
and {bn}
6 n > 0, n e N,
lim
lim 6 n = 6. Show t h a t
n—+00
and
= ab.
be two sequences such t h a t
lim ( 6 i + b 2 + ... + &„) =+cx>,
-^=9-
Prove t h a t
,.
a i + a2 + ... + an
lim
—- = g.
n-+oo b\ + 02 + ... "f 0 n
2 . 3 . 1 0 . Let { a n } and {6 n } be two sequences for which
(i)
bn > 0, n € N,
(ii)
lim a n = a.
and
lim (&i -f b2 + ... 4- 6 n ) = + o o ,
n—•oo
Show t h a t
lim
n-^oo
ai&i + a20 2 + ... + anbn
= a.
6i + D2 + ... + 0 n
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2.3. Toeplitz Transformation a n d Stolz Theorem
37
2.3.11. Using the result in the foregoing problem, prove the Stolz
theorem.
Let {xn},
{yn} be two sequences that satisfy the conditions:
(i)
{yn}
(n)
n
strictly increases to + oo,
lim
= g.
-*°° 2/n - 2/n-l
Then
n—oo yn
2.3.12. Calculate
(a)
lim -i= M + ^ + ... + - L ) ,
(b)
lim
_
(c)
lim -T-T-
(d)
(e)
(f)
(g)
—- [a + — + ... + —
n—*oo n
a > 1,
(fc + n)!
1!
^
W —F=
1 +
lim -7=
1
/
n!
+ ...+
/ceN,
1
..
lfc + 2fc + ... + n*
, ^T
lim
—
, k e N,
fc+i
rv
1 + 1 - a + 2- a 2 + ... + n - a n
, a > 1,
lim
n—>oo
n • an+1
lim
n—>oc
nfc v
2.3.13. Assume that
y
fc + 1
/cGN.
lim an — a. Find
lim —= ai + —7= -f -7= + ... 4- -7=
n->oo ^ y
V2
V3
Vn
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Problems. 2: Sequences of Real Numbers
38
2.3.14. Prove that if {an} is a sequence for which
lim (a n +i - an) = a,
n—>oo
then
lim — = a.
n—>oo 71
2.3.15. Let lim an = a. Determine
n—>oo
an_i
(an
r
ai
\
2.3.16. Suppose that lim an — a. Find
n—>oo
,
v
v
f
a
n
t
a
«1
n-l
\
2.3.17. Let k be an arbitrarily fixed integer greater than 1. Calculate
n->oo V \n
J
2.3.18. For a positive arithmetic progression {a n }, find
l i m n(ai -... »a w )n
n^oo ai + ... + a n
2.3.19. Suppose that {an} is such that the sequence {bn} with
6 n = 2an + a n _i, n > 2, converges to b. Study the convergence of
2.3.20. Suppose that {an} is a sequence such that
for some real x. Prove that
lim nxan = a
n—>oo
lim n x (ai • a<i • ... • a n ) " = ae x .
n—>-oo
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2.3. Toeplitz Transformation and Stolz Theorem
39
2.3.21. Calculate
(a)
lim i ± J + - + n - i + n
Inn
(b)
lim
1+1+I+
+ _J_
Inn
n—>-oo
2.3.22. Assume that {an} tends to a. Show that
— + — + ... + — = a.
hm
n-*oo I n n V 1
2
n /
2.3.23. Find
(a)
v
'
(<0
(e)
iim (-0*—) U ,
n—oo \nne~n
(b) lim ( - ^ 1 - 1 "
n-^oo Y n 3 n e _ n /
J
5
(d) n->oo
.limf^)
,
\(n!)
Jimf^V.
lim - ¥ = , fcGN.
2.3.24. Show that if lim an = a, then
1
lim
1
n
^ afc
> — = a.
n-^oo Inn f—' /c
fc=i
2.3.25. For a sequence {a n }, consider the sequence {An} of arithmetic means, i.e. An = Q i+ a ?+--+ Q n. Show that if lim^oo j4 n = A,
then also
iim J - y ; ^ = A
n->oo I n n k=\
f—' A;
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Problems. 2: Sequences of Real Numbers
40
2.3.26. Prove the converse to the Toeplitz theorem stated in 2.3.1:
Let {cn^k : 1 < fc < n, n > 1} be an array of real numbers. If for
any convergent sequence {an} the transformed sequence {bn} given
by setting
n
bn = ^ c n > f c a f c ,
n > 1
fc=i
is convergent to the same limit, then
(i)
cn,fc —• 0 for each fceN,
n—xx>
n
(ii)
J^Cnt
—• 1,
(iii)
there exists C > 0 such that for all positive integers n
n
2.4. Limit Points. Limit Superior and Limit
Inferior
2.4.1. Let {a n } be a sequence whose subsequences {a2k}, {&2/c+i}
and {03^} are convergent.
(a) Prove that the sequence {an} is convergent.
(b) Does the convergence of any two of these subsequences imply the
convergence of the sequence {a n }?
2.4.2. Does the convergence of every subsequence of {an} of the
form {a 5 . n }, s > 1, imply the convergence of the sequence {an}?
2.4.3. Let {a P n }, {aqn},... ,{a S n } be subsequences of {an} such
that the sequences {p n }, {qn}, •••» {s n } a r e pairwise disjoint and form
the sequence {n}. Show that, if S , S p , S q , ...,S S are the sets of all
the limit points of the sequences {a n }, {aPn }, {aqn },..., {a 5n }, respectively, then
s = spusqu...uss.
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2.4. Limit Points. Limit Superior and Limit Inferior
41
Conclude that, if every subsequence {aPn}, {a q n },..., {aSn} converges
to a, then the sequence {a n } also converges to a.
2.4.4. Is the above theorem (Problem 2.4.3) true in the case of infinitely many subsequences?
2.4.5. Prove that, if every subsequence {anfc} of a sequence {an}
contains a subsequence {ank } converging to a, then the sequence
{an} also converges to a.
2.4.6. Determine the set of limit points of the sequence {a n }, where
(a)
an=
(b)
an =
(c)
(d)
(e)
(f)
v/4(-D"+2)
1 /
[n-l] \ /
- 3 - -3
2\
3
) ( (l-(-l)")2n + l
[n-l]
3
(1 + cos nn) In 3n + In n
/
n-K\n
an = 1 cos — 1 ,
2n 2
«n = - ^
[2n2]
7
2.4.7. Find the set of all the limit points of the sequence {an} defined by
(a)
an = na-
(b)
a n = na - [na], a 0 Q,
(c)
a n = sin7rna,
a € Q,
(d)
a n = sin irna,
a $ Q.
[na], a G Q,
2.4.8. Let {ak} be a sequence arising by an arbitrary one-to-one
indexing of the elements of the matrix { tfn — v^ro}, n, ra E N. Show
that every real number is a limit point of this sequence.
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42
Problems. 2: Sequences of Real Numbers
2.4.9. Assume that {an} is a bounded sequence. Prove that the set
of its limit points is closed and bounded.
2.4.10. Determine lim an and lim an, where
2n 2
(a)
"2n 2 l
.7 J
71 — 1
U7T
—-COS—-,
(b)
0<n =
(c)
an = ( - l ) n n ,
(d)
an = n ( " 1 ) n - ,
(e)
a n = 1 + n si
(f)
an=(l + ^ V ( - i r + sin^,
(g)
an = \ / l + 2»(- 1 ) B ,
(h)
a n = I 2 cos - — I ,
...
W
71+ 1
3
717T
Inn — (1+cosn7r)n
Un
=
h^
•
2.4.11. Find the limit superior and the limit inferior of the following
sequences:
(a)
an — not — [na], a G Q,
(b)
an = na-[na],
a^Q,
(c)
an — sin7rna,
QGQ,
(d)
an = sin7rna,
a ^ Q.
2.4.12. For an arbitrary sequence {a n }, prove that
(a) if there exists k e N such that for any n greater than fc the
inequality an < A holds, then lim an < A,
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2.4. Limit Points. Limit Superior and Limit Inferior
43
(b) if for any k G N there exists n^ greater than k such that
&nk < A, then lim an < A,
n—>oo
(c) if there exists k G N such that for any n greater than k the
inequality an > a holds, then lim an > a,
n—>oc
(d) if for any k G N there exists n/c greater than k such that
&nfc > &> then lim a n > a.
n—>-oo
2.4.13. Assume that for a sequence {a n } the limit inferior and the
limit superior are both finite. Prove that
(a) L = lim an if and only if
n—»oo
(i) for every e > 0 there exists A; G N such that a n < L + £ if n > k
and
(ii) for every e > 0 and /c G N there is n/e > k such that L — e<
(b) / = lim an if and only if
ank.
n—>oo
(i) for every £ > 0 there exists k G N such that Z — £ < an if n > k
and
(ii) for every e > 0 and /c G N there is n^ > fc such that ank <l + e.
Formulate the corresponding statements for infinite limit inferior
and limit superior.
2.4.14. Assume that there is an integer no such that for n > no
the inequality an < bn holds. Prove that
(a)
lim an < lim bn,
n—>oo
(b)
n—>oo
lim an < lim bn.
n—•oo
n—>oo
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Problems. 2: Sequences of Real Numbers
44
2.4.15. Prove that (excluding the indeterminate forms of the type
+oo — oo and — oo -f- oo) the following inequalities hold:
lim an + lim bn < lim (an -f- bn) < lim an + lim bn
n—•oo
n-^oc
n—>oo
n—>oo
n—»oo
< lim (an -f bn) < lim a n -f lim bn.
n—>-oo
n—•oo
n—>-oo
Give examples of sequences for which "<" in the above inequalities
is replaced by " < ".
2.4.16. Do the inequalities
lim an + lim bn < lim (a n + bn),
n—•oo
n—+00
n—*oc
lim (a n + bn) < lim a n 4- lim bn
n—+00
n—^oo
n—»oo
remain valid in the case of infinitely many sequences?
2.4.17. Let {an} and {6 n } be sequences of nonnegative numbers.
Prove that (excluding the indeterminate forms of the type 0 • (-f 00)
and (+00) • 0) the following inequalities hold:
lim an • lim bn < lim (an • bn) < lim an • lim bn
n-^00
n^oo
n—^oo
n—^00
n—+oo
< lim (an • bn) < lim an • lim bn.
n—^oo
n—•oo
n—*oo
Give examples of sequences for which "<" in the above inequalities
is replaced by "<".
2.4.18. Prove that a necessary and sufficient condition for the convergence of a sequence {an} is that both the limit inferior and the
limit superior are finite and
lim an = lim an.
n-^oo
n-+oo
Prove that the analogous theorem is also true for sequences which
properly diverge to — oo or +00.
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2.4. Limit Points. Limit Superior and Limit Inferior
45
2.4.19. Show that, if lim an = a, a G 1R, then
n—>oo
lim (an + bn) = a + lim bn,
n—>-oo
n—»oo
lim (a n -j- 6n) = a + lim bn.
n—>oo
n—•oo
2.4.20. Show that, if lim an = a, aGM, a > 0, and there exists a
n—>oo
positive integer no such that bn > 0 for n > no, then
lim (an • 6n) == a • lim 6n,
n—•oo
n—>oc
n—>oo
n—»oo
lim (a n • bn) = a • lim 6 n .
2.4.21. Prove that
lim (—an) = — lim o n ,
lim (—on) = — lim a n .
2.4.22. Prove that for any positive sequence {a n },
lim
, an
lim a n
n—»oo
lim — =
n-+oc an
(Here ^
lim an
n—>-oo
= 0, £ = +oo).
2.4.23. Prove that, if {a n } is a positive sequence such that
lim an - lim — = 1,
n—>oo
n—+00 a n
then the sequence {an} is convergent.
2.4.24. Show that, if {a n } is a sequence such that for any sequence
{bn},
lim (an + bn) = lim a n -f lim 6 n
or
n—>oo
n—»oo
n—»oo
lim (an + 6n) = lim a n + lim 6 n ,
n—>-oo
then {a n } is convergent.
n—•oo
n—»oo
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Problems. 2: Sequences of Real Numbers
46
2.4.25. Show that, if {an} is a positive sequence such that for any
positive sequence {bn},
lira (a n • bn) = lim a n • lim bn
n—>-oo
n—>oo
n—>oo
or
lim (a n • bn) = lim a n • lim 6 n ,
n—>oo
n—•oo
n—>oc
then {a n } is convergent.
2.4.26. Prove that for any positive sequence { a n } ,
lim
n-^oc
< hm yan < hm >/a n < lim
an
n-^oo
n
-"°°
n
-^°°
.
an
2.4.27. For a given sequence {a n }, define {6 n } by setting
bn = - ( a i + a 2 + ... + a n ),
n
n G N.
Prove that
lim a n < lim 6 n < lim bn < lim an.
n^oo
n^oo
n
-*°°
n >
- °°
2.4.28. Prove that
(a)
lim (max{a n ,6 n }) = m a x ] lim a n , lim bn \ ,
(b)
lim (min{a n ,6 n }) = min < lim a n , lim 6 n > .
n—>-oo
Ln—>oo
n—>oo
)
Are the equalities
(c)
(d)
lim (min{a n ,6 n }) = min \ lim a n , lim 6 n I ,
lim (max{a n ,6 n }) = max < lim a n , lim 6 n >
rc—•oo
Ln—>oo
n—>oo
J
also true?
2.4.29. Prove that every sequence of real numbers contains a monotonic subsequence.
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47
2.5. Miscellaneous Problems
2.4.30. Use the result in the foregoing exercise to deduce the BolzanoWeierstrass theorem:
Every bounded sequence of real numbers contains a convergent
subsequence.
2.4.31. Prove that for every positive sequence {a n },
lim
> 4.
Show that 4 is an optimal estimate.
2.5. Miscellaneous Problems
2.5.1. Show that, if lim an = +oc or lim an = —oo, then
n—•oo
n—• oo
lim
n-+oo \
1+ — )
= e.
a„
2.5.2. For x G l , show that
lim (l + -Y = ex.
2.5.3. For x > 0, establish the inequality
< \n(x -f 1) < x.
v
x+2
'
Prove also (applying differentiation) that the left inequality can be
strengthened to the following:
x
*2iX
x+1
<
x+2 < \n(x + 1),
x > 0.
2.5.4. Prove that
(a)
(b)
lim n{ tya — 1) = In a, a > 0,
n—>-oo
lim n(tfn-l)
= +oo.
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Problems. 2: Sequences of Real Numbers
48
2.5.5. Let {an} be a positive sequence with terms different from 1.
Show that if lim an = 1, then
71—>OC
lim
n—»oo a n — 1
2.5.6. Let
= 1.
*» = 1 + I[ + i + "- + ^
Show that
lim a n = e
and
nGR
0 < e — a n < —-.
n—»oo
77,71!
2.5.7. Prove that
rp
l i m ( 1 +
l!
T
+
rpli
2 !
+
-
+
^!1=:e*-
2.5.8. Show that
lim ( - + —^- + ... + i - J = In2,
(a)
(b)
lim ( .
*-*°°\y/n{n+l)
+
y/{n+l)(n
+ 2)
+ ... + — = = = = = ) = ln2.
^/2n{2n + 1) /
2.5.9. Find the limit of the sequence {an}, where
2.5.10. Let {an} be the recursive sequence defined by
a\ = 1,
a n = n(a n _i + 1)
for
n = 2,3,....
Determine
lim
TT ^
+
Mm
2.5.11. Prove that lim (n!e - [n!e]) = 0.
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2.5. Miscellaneous Problems
49
2.5.12. Given positive a and 6, show that
n—•oo \
s^V-v s
Z
2.5.13. Let {a n } and {&n} be positive sequences such that
lim 0%= a,
n—>oo
lim b™ = 6,
where
a, 6 > 0,
n—>oo
and suppose that positive numbers p and g satisfy p + q = 1. Prove
that
lim (pan + qbn)n = apbq.
n—>oo
2.5.14. Given two real numbers a and 6, define the recursive sequence
{an} as follows:
a\ = a,
a 2 = 6,
an+i =
Find lim an.
n
an + —a n _i,
n
n > 2.
n—+oc
2.5.15. Let {an} be the recursive sequence defined by
a\ = 1,
a 2 = 2,
a n + i = n(a n -f a n _i),
n > 2.
Find an explicit formula for the general term of the sequence.
2.5.16. Given a and b, define {an} recursively by setting
oi = a,
a 2 = 6,
1
2n-l
an+i = —an_i H
an,
2n
2n
n > 2.
Determine lim a n .
n—• oo
2.5.17. Let
°» = 3 - E f c ( i b + 1 j ( f c + 1)I. » €N .
(a) Show that lim a n = e.
n—>oo
(b) Show also that 0 < an - e <
1
(n+1) (n+1)r
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Problems. 2: Sequences of Real Numbers
50
2.5.18. Calculate lim nsin(27ra!e).
n—>oc
2.5.19. Suppose that {an} is a sequence such that an < n, n =
1,2,..., and lim an = -hoc. Study the convergence of the sequence
( ! - £ ) " . n-1,2,....
2.5.20. Suppose that a positive sequence {bn} diverges to -foo.
Study the convergence of the sequence
h \
n
1 + -M , n = l,2,....
nJ
2.5.21. Given the recursive sequence {an} defined by setting
0 < ai < 1,
an+i = an(l - an),
n > 1,
prove that
(a)
lim nan = 1,
n—>oo
lim n ( 1 " n a " ) = 1.
n-+oo
In n
(b)
2.5.22. The sequence {a n } is defined inductively as follows:
0 < a\ < 7T,
Prove that
«n+i = sina n ,
n > 1.
lim y/nan — y/3.
n—*oc
2.5.23. Let
a\ — 1,
a n + i = an + —
,
n > 1.
fc=i
Prove that
n
hm ,/
= 1.
^°° v 2hm
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2.5. Miscellaneous Problems
51
2.5.24. For {an} defined inductively by
a\ > 0,
determine
a n +i = arctana n ,
n > 1,
lim an •
n—>oo
2.5.25. Prove that the recursive sequence defined by
0 < ai < 1,
an+\ = cosa n ,
n > 1,
converges to the unique root of the equation x = cos x.
2.5.26. Define the sequence {an} inductively as follows:
a\ = 0,
a n + i = 1 — sin(a n — 1),
Find
n > 1.
1 n
lim — V a f c .
n—• oo 77, ^ — '
2.5.27. Let {an} be the sequence of consecutive roots of the equation tan a; = x, x > 0. Find lim (a n +i — a n ).
n—»oo
2.5.28. For |a| < |
defined by
and ai £ M, consider the recursive sequence
a
n+i = asinan,
n > 1.
Study the convergence of the sequence.
2.5.29. Given a\ > 0, consider the sequence {an} defined by setting
an+i = ln(l + a n ),
n > 1.
Prove that
(a)
(b
lim nan = 2,
n—»oo
hm
n-^oo
Inn
= -.
3
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52
Problems. 2: Sequences of Real Numbers
2.5.30. Define the recursive sequence {an} by putting
a\ = 0 and
/l\an
an+i = 1 — 1
,
n > 1.
Study the convergence of the sequence.
2.5.31. Given a\ > 0, define the sequence {an} as follows:
an+i = 2 ~ a n ,
n > 1.
Study the convergence of the sequence.
2.5.32. Find the limit of the sequence defined by
ai = >/2,
an+i=2^\
n > 1.
2.5.33. Prove that if lim (an - a n _2) = 0, then
n—>oo
lim
n—•oo
fl
= 0.
2.5.34. Show that if for a positive sequence {an} the limit
lim n I 1 - ^ ± ±
exists (finite or infinite), then
lim — ^
n^oo Inn
also exists and both limits are equal.
2.5.35. Given a\,b\ £ (0,1), prove that the sequences {an} and
{bn} defined by
an+i = a i ( l - a n - bn) + a n ,
&n+i = &i(l - a n - bn) + 6 n ,
n > 1,
converge and find their limits.
2.5.36. For positive a and ai, consider the sequence {an} defined
by setting
an-K = a n(2 - oa n ), n = 1,2,....
Study the convergence of the sequence.
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2.5. Miscellaneous Problems
53
2.5.37. Show that if a\ and a,2 are positive and
«n+2 = \f^a + \J&n+\-> 71 = 1, 2, ...,
then the sequence {an} converges. Find its limit.
2.5.38. Assume that / : R+ —> E + is a function increasing with
respect to every variable and there exists a > 0 such that
f(x,x,...,x)
>x
for
0 < x < a,
f(x,x,...,x)
<x
for
x > a.
Given positive oi,a2, ... ,afc, define the recursive sequence {a n } by
fln = / ( f l n - l , f l n - 2 , . » j f l n - f c )
f°r
n > A.
Prove that lim an = a.
n—*oo
2.5.39. Let a\ and a,2 be given positive numbers. Study the convergence of the sequence {an} defined by the recursive relation
a n + i = anean~Q"n-1
for n > 1.
2.5.40. Given a > 1 and x > 0, define {a n } by setting ai =
a x , a n +i = a a n , n G N. Study the convergence of the sequence.
2.5.41. Show that
Y / 2 + V / 2 + ... + V/2 = 2 c o s - ^ I .
N
v
n roots
'
Use this relation to find the limit of the recursive sequence given by
setting
ai = A/2, a n + i = >/2-fa n , n > 1.
2.5.42. Let {e n } be a sequence whose terms are equal to one of the
three values -1, 0, 1. Establish the formula
ei]J2 + e2^2 + ... + e„V2 = 2sin (^ ] T
£l
g:;£fc j ,
n G N,
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Problems. 2: Sequences of R e a l N u m b e r s
54
and show that the sequence
+ £n V2
an = €iy2 + s2y2-{-...
converges.
2.5.43. Calculate
,.
(
1
1
1 \
lim arctan - + arctan ———72 -f ... + arctan -—7
n^oc y
2
2•2
2n^ /
2.5.44. Find lim sin(7r\/n2 + n).
n—•oo
2.5.45. Study the convergence of the recursive sequence defined as
follows:
ai
\/2,
a 2 - V 2 + v7^,
an+2 = y
2
+ V3 + a n for n > 1.
2.5.46. Show that
Jirr^ J 1 + 2if 1 + 3 y 1 + ...\]l + {n-\)y/Y+n
= 3.
2.5.47. Given a > 0, define the recursive sequence { a n } by putting
a\ < 0,
a n +i =
1 for n € N.
an
Show that the sequence converges to the negative root of the equation
x2 -f x = a.
2.5.48. Given a > 0, define the recursive sequence {an}
by setting
a
an+i =
for n G N.
an + 1
Show that the sequence converges to the positive root of the equation
x2 4* x — a.
Q>\ > 0,
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2.5. Miscellaneous Problems
55
2.5.49. Let {an} be the sequence defined by the recursive formula
2 + an
tor n G N.
1 + an
Show that the sequence is Cauchy and find its limit.
ai = 1,
an+i =
2.5.50. Show that the sequence defined by
ai>0,
an+i=2H
, n € N,
is Cauchy and find its limit.
2.5.51. Given a > 0, define {a n } as follows:
ai = 0,
an+i =
2 4-o n
Study the convergence of the sequence.
for n e N.
2.5.52. Assume that
aiGl
a n + i = \an - 2 1 _ n |
and
for n <E N.
Study the convergence of the sequence and in case of convergence find
its limit .
2.5.53. Show that
(a) if 0 < a < 1, then
n—1
. A
J a3
n—•oo ^—' n — 1
J= l
(b) if 0 < a < 1, then
hm na
n-^oo
> —r =
,
^—' 7a-7
1— a
(c) if 6 > 1, then
,.
n^V~l
1
lim — > —— = i
7n-^oo bn ^—' 7
6-1
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Problems. 2: Sequences of Real Numbers
56
2.5.54. Calculate
lim
sin
n—•oo
n+ 1
-f sin
n+ 2
+ ... + s m — .
2nJ
2.5.55. Find
lim TT [ 1 + —«4 1 ,
n-+oo -LX V
en* 7/
x
(a)
where
c > 0,
k=i
A /
lim TT ( 1
(b)
n-^oo
± ±
k=l
\
A:2 \
=• J ,
cn6
J
where
c > 1.
2.5.56. Determine
lim
n-+oo
— TT sin — = .
n\ x±
nJn
k~\
2.5.57. For the sequence {an} defined by
a =
n
y
v
-1
- S (fcj '
show that
n
^ X'
lim a n = 2.
n—•oo
2.5.58. Determine for which values of a the sequence
converges.
2.5.59. For x G R, define {x} = x - [x]. Find lim {(2 + \/3) n }n—>oo
2.5.60. Let {a n } be a positive sequence and let Sn = a\ -\-a,2 +... +
a n , n > 1. Suppose that
<
*->n+l
((5 n - l)an + a n _ i ) , n > 1.
Determine lim a n .
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2.5. Miscellaneous Problems
57
2.5.61. Let {an} be a positive sequence such that
i-
an
Find
a
r~
A
hm — = 0,
hm
n—»oo 77,
l +a2
n—>oo
lim
+ — + an
ft
^
< oo.
aj + al+ ... + <%
2.5.62. Consider two positive sequences {an} and {bn} such that
lim
= 0
n->oo a\ 4- a<i -f ... + an
and
lim
—
-— = 0.
n^oo 6i + 62 + ... 4- bn
Define the sequence {cn} by setting
cn = aiK + a2bn-i + ... + o n 6i, n G N.
Show that
lim
n^oo
Cl
^
= 0.
+C2 + ... + Cn
2.5.63. Find
e" n .
lim [ 1 + - 1
2.5.64. Suppose that a sequence {an} bounded above satisfies the
condition
a n + i - an >
^,
n G N.
Establish the convergence of {a n }.
2.5.65. Suppose that a bounded sequence {a n } satisfies the condition
> an,
n G N.
Establish the convergence of { a n } .
2.5.66. Let Z and L denote the limit inferior and the limit superior of
the sequence {a n }, respectively. Prove that if lim (a n +i — an) = 0,
n—• oo
then each number in the open interval (Z, L) is a limit point of { a n } .
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58
Problems. 2: Sequences of Real Numbers
2.5.67. Let / and L denote the limit inferior and the limit superior of
the sequence {a n }, respectively. Assume that for any n, a n + i —an>
—an, where an > 0 and lim an = 0. Prove that each number in
n—>-oo
the open interval (/,L) is a limit point of
{an}.
2.5.68. Let {an} be a positive and monotonically increasing sequence. Prove that the set of all limit points of the sequence
n + an
ne N,
is an interval (which is reduced to a singleton in case of convergence).
2.5.69. Given ai GM, consider the sequence {an} defined by
{
-^2
1 + an
2
for even n,
for odd n.
Find the limit points of this sequence.
2.5.70. Is zero a limit point of the sequence { v ^ s i n n } ?
2.5.71. Prove that for a positive sequence {a n },
2.5.72. Prove the following generalization of the foregoing result:
For a positive integer p and a positive sequence { a n } ,
2.5.73. Prove that for a positive sequence {a n },
HE
n
(l±f2±l
- l) > 1.
Show that 1 is the best possible constant.
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59
2.5. Miscellaneous Problems
2.5.74. Let
an = yi + }Jl + ... + Vl.
Find lim a n .
2.5.75. Let {an} be a sequence whose terms are greater than 1 and
suppose that
..
lnlnan
iim
= a.
77,
n—*oo
Consider the sequence {bn} defined by
yai + yjia2
+ ... + y/a^,
n G N.
Prove that if a < In 2, then {bn} converges; and if a > In2, then it
diverges to +oo.
2.5.76. Assume that the terms of the sequence {an} satisfy the condition
0 < a n +m < a<n + a>m for n, m G N.
Show that the limit lim — exists.
n—>oo
n
2.5.77. Assume that the terms of the sequence {an} satisfy the condition
0 < an+m < a>n ' Q>m for n, m G N.
Show that the limit lim j / a ^ exists.
n—>oo
2.5.78. Assume that the terms of the sequence {an} satisfy the conditions
\0>n\ <
a
1,
n + ttm — 1 < a n + m < a n + a m + 1
for n, ra G N.
(a) Show that the limit lim
9LIL
exists.
(b) Show that if lim &*• = g, then
ng — 1 < a n < ng + 1 for
n G N.
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Problems. 2: Sequences of Real Numbers
60
2.5.79. Assume that {an} is a positive and monotonically increasing
sequence that satisfies the condition
> nO>m
Un-m
for
71,171
£N.
Prove that if s u p { ^ : n G N} < +00, then the sequence
converges.
{^}
2.5.80. Given two positive numbers ai, a2, prove that the recursive
sequence {an} defined by
for
a n +2 =
n eN
converges.
2.5.81. For b\>a\>
0, consider the two sequences {an} and {bn}
defined recursively by setting
a n +i =
n
2
n
>
&n+i = \An+i&n
for
n G N.
Show that both sequences converge to the same limit.
2.5.82. Let a^n, 6^,nr n G N, A: = 1,2, ...,n, be two triangular
arrays of real numbers with bk n ¥" 0- Suppose that j ^ - —• 1
uniformly with respect to /c, that is, for any e > 0 there is a positive
integer no such that
|flfc,n _
I bk,n
1
I
< e
for each n > UQ and k = 1,2,..., n. Show that if lim ^ 6/~ n exists,
n
then
lim V V
n
~'OG
k=i
= lim V V n .
fc=l
fc=l
2.5.83. Given a ^ 0, find
n
lim V s i n
k=l
(2fc-l)a
n^
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2.5. Miscellaneous Problems
61
2.5.84. For a > 0, determine
2.5.85. Find
lim f[(l
fc=l
+JL).
x
7
2.5.86. For p y^O and g > 0, determine
2.5.87. Given positive numbers a, b and d such that b > a, calculate
a(a + d)...(a + nd)
n^oo 6(6 + d)...(6 + nd) '
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Chapter 3
http://dx.doi.org/10.1090/stml/004/03
Series of Real Numbers
3.1. Summation of Series
3.1.1. Find the infinite series and its sum if the sequence {Sn} of
its partial sums is given by setting
(a)
(c)
Sn = ^—, n €N,
n
Sn = aretann, n G N,
(b)
Sn = — - i , n €N,
2n n
f-l)
5 n = -——, n G N.
n
(d)
3.1.2. Find the sum of the series
<•>^ n£ (=n +&l ) '•
2
n=l
oo
(C)
2
v
'
/ o
2
w
w^ (£2 n - l ) ( 2 n + l) '
n—\
oo
r
v-^ n - Vn - 1
S^TTT
(e) ~£ (v^+\/n + 1)^71(71+1)
2
<d)
v
^->
2
'
v
'
1
1
S=^.
63
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Problems. 3: Series of Real Numbers
64
3.1.3. Compute the following sums:
X
i
( ^
a
^Sr^
m
fo+l)(3n+l)
()
m
(b)
Sln(n+i)(2n-l)'
7 + /
n=l
77T1 + 4)
T:—~ >
n(3n
"
(2n + l)n
3.1.4. Find the sum of the series
V
(a)
^
L
n ( n + ll)...(n
) . . . + ra)
v
n=l
oo
, m 6 N,
'
1
r, m e N ,
*—' nln-f m)
(b)
V - T —
n=l
'
E- J (n + l)(n + 2)(n + 3)(n + 4)'
(c)
3.1.5. Compute
OO
,
OO
w S>£.
w E;
n=l
3.1.6. Calculate
OO
E
n=l
1
n=l
Inn I
n — In n
j
sin —2^+1
COS
2n+1'
3.1.7. Find
E n!(n -h1n
4
n=0
2
+ l)'
3.1.8. Show that
OO
E. 3 - 5 - . . . - ( 2 n + l )
n=l
_,
n
1
=
2"
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3.1. Summation of Series
65
3.1.9. Suppose that {an} is a sequence satisfying
lim ((en + l)(a 2 + l)...(a n + 1)) = p,
0 < g < +oo.
n—>oo
Prove that
oo
_ 1_ 1
On
n=1(ai
+ l)(a 2 + l)...(an + l) "
p'
where we assume that — = 0.
oo
3.1.10. Using the result in the foregoing problem, find the sum of
the series
Y^ n — 1
(a)
n— 1
y,
2n-l
^ 2 - 4 - 6 - . . . -2n'
(b)
n=l
oo
(c)
i
S(i-^)(i-l)-(i-^)'
3.1.11. Let {an} be a recursive sequence given by setting
a\ > 2,
Show that
y>
a n + i = a^ - 2 for
n € N.
_ ai - v/a? - 4
1
^-J ai • a 2 •... • an
2
n=l
3.1.12. For b > 2, verify that
E
71!
^b(6+l)...(b +
ra-l)
6-2'
3.1.13. For a > 0 and 6 > a -f 1, establish the equality
E
a(a -f l)...(a -f n — 1)
6(6+l)...(6 + n - l )
=
a
6-a-l'
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Problems. 3: Series of Real Numbers
66
3.1.14. For a > 0 and b > a -f 2, verify the following claim:
°°
a(a + l)...(a + n - 1) _
E fc(6+l).»(6 + n - l )
n==i
n
a(6 - 1)
~ (6-a-l)(6-a-2)'
3.1.15. Let J^ — be a divergent series with positive terms. For
71=1
aU
b > 0, find the sum
£~ ( 2 + &)( 3 + b)...(On+l + fc)
ai • a,2 •... • an
a
a
3.1.16. Compute
f(-ir^.
n=0
3.1.17. Given nonzero constants a, 6 and c, suppose that the functions / and g satisfy the condition f(x) = af(bx) -f cg(x).
(a) Show that, if lim anf(bnx)
= L(x) exists, then
n—•oo
oo
Y,ang{bnx)
=
f(x) - L(x)
n=0
(b) Show that, if Jim a~nf(b~nx)
= M(x) exists, then
n—+oo
oo
^a-ng{b-nx)
=
M{x) -
af{bx)
n=0
3.1.18. Applying the identity sinx = 3 sin | - 4 sin 3 | , show that
3
x-sini
x
n=0
oo
i1
n
n=0
3n
sin 3
x
3 . x
-r = - sin —.
3~n+1
4
3
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67
3.1. Summation of Series
3.1.19. Applying the identity cot a; = 2cot(2x) + tan a; for x ^ /c|,
fcGZ, show that
E — tan — =
n=0
l
X
1
2n
2n
x
3.1.20. Using the identity
establish the formulas
^
/^ \
2 cot (2a;).
arctan x = arctan(frr) + arctan
1+bJ?,
n
\~^
(a) > arctan
(11 ~- ub)b
K
)u
xx
————^ = arctanx
n=0
for
0 < b < 1,
for
x ^ 0
n
v~>
(b — l)b x
(b)
>^ arctan
, 0 ,. 0 = arccot x
w
^
1 + '62 n + 1 x 2
n=0
and
6 > 1.
3.1.21. Let {a n } be the Fibonacci sequence defined by setting
a 0 = a\ — 1,
and put 5 n =
a n + i = a n + a n _i, n > 1,
fc=0
v (-ir
^
Sn
3.1.22. For the Fibonacci sequence {an} defined in the foregoing
problem, calculate
y
(-l)n
3.1.23. For the Fibonacci sequence {an} defined in 3.1.21, determine the sum of the series
oo
V^ arctan
1
.
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Problems. 3: Series of Real Numbers
68
3.1.24. Find the sum:
oo
2
y ^ arctan — ,
(a)
°°
1
V^ arctan —x
(b)
n=l
oo
(c)
n=\
> arctan
n=l
,
n 4 - 2 n 2 + 5*
3.1.25. Let {an} be a positive sequence that diverges to infinity.
Show that
oo
E arctan
n
— Q>n
an4-l
-L
= arctan —.
1 + anan+i
ai
3.1.26. Prove that any rearrangement of the terms of an infinite
positive series does not change its sum.
n=1
3.1.27. Establish the identity
oo
oo
O ^r—'\
1
E (2n-l)
0
l
2 =
1
1
4^n?'
3.1.28. Prove that
°° 1
(a)
£^
n=l
oo
(b)
^ ^
.
=
7T2
4
= ^0'
oo
<«>
¥'
.
B-«"5r+T " I-
n=0
3.1.29. For the sequence {a n } defined recursively by
a\ = 2,
a n + i = an-
an + l
for n > 1,
oo
n=l
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69
3.1. Summation of Series
3.1.30. Let {an} be defined as follows:
a\ > 0,
a n + i = In
ean — 1
for n > 1,
and set bn = a\ • a^ •... • an. Find ]T 6 n .
n=l
3.1.31. Let {a n } be defined by setting
ai = 1,
an+i =
\/2
ai + a 2 -f ••• 4- a n
for
n > 1.
oo
Determine the sum of the series ^ a n .
n=l
3.1.32. Find the sum of the following series:
(a)
E(-l)-1^
n=l
n(n + l ) '
™=i
°° /
1
1
1
\
3.1.33. Calculate
D-i)"-b.(i + i).
n=l
x
'
3.1.34. Compute
3.1.35. Determine the sum of the series
S(=-H))-
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Problems. 3: Series of Real Numbers
70
3.1.36. Suppose that a function / is differentiable on (0, oo), that its
derivative / ' is monotonic on a subinterval (a, H-oc), and lim fix) =
x—++oo
0. Prove that the limit
lirn^ Q / ( l ) + /(2) + /(3) + ... + f(n - 1) + ± / ( n ) - £
f(x)dx^J
exists. Consider also the special cases of the functions f(x) = ^ and
f(x) = lnx.
3.1.37. Determine the sum of the series
,lnn
E(-D- n
71=1
3.1.38. Find
f / , 2 n + l
>
^
n In
V
x
n=l
2n-l
3.1.39. Given an integer k > 2, show that the series
y>/
1
2s ^ \[((n
n --l )1f )k
c +4-1
l
+
1
1
+
+
(n - 1 )k+42 9 """" n * - l
(ra-l)A;
nA:
converges for only one value of x. Find this value and the sum of the
series.
3.1.40. For the sequence {an} defined by setting
a 0 = 2,
a n + i = an H
3 + (-l)n
,
compute
n=0
3.1.41. Prove that the sum of the series
oo
n=l
-
oc
n=l
1
v
'
is irrational.
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3.1. Summation of Series
71
3.1.42. Let {en} be a sequence where en is either 1 or — 1. Show
oo
that the sum of the series Y ^
is an irrational number.
U
n=l
'
3.1.43. Show that for any positive integer k the sum of the series
(-i)n
(n\)k
"
^
v
n=l
'
is irrational.
3.1.44. Suppose that {n^} is a monotonically increasing sequence
of positive integers such that
hm
fc^oo niTl2 • ... • Tlk—l
= -foo.
oo
Prove that Y] — is irrational.
2=1
3.1.45. Prove that, if {n^} is a sequence of positive integers such
that
lim
= +oo
/c-+oo mn2 • ... • njfe_i
and
lim
^ ^
> 1,
nk-\
oo
then V —
is irrational.
n
.—' i
1= 1
3.1.46. Suppose that {n^} is a monotonically increasing sequence
oo
of positive integers for which lim ^v/n^ = oo. Prove that J ] ^- is
k-^oc
k=1
k
irrational.
CO
3.1.47. Let Y" 2^, where pn,Qn £ N, be a convergent series and let
—- w
n=l
Pn
qn - 1
Pn+1
qn+i - 1
Pn
qn
Denote by A the set of all n for which the above inequality is sharp.
oo
Prove that Yl ~
1S
irrational if and only if the set A is infinite.
n=l
3.1.48. Prove that for any strictly increasing sequence {n^} ofposoo
itive integers the sum of the series ^T ^-j is irrational.
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Problems. 3: Series of Real Numbers
72
3.2. Series of Nonnegative Terms
3.2.1. Determine whether the following series converge or diverge:
oo
oo
(a) £ ( v ^ T T - ^ T T ) ,
n=l \
^(2n-3)!!
...
(d)
w £(2^jM'
n=2
00
(e)
£
n=l
v
'
/
V
i \
1-cos-
n /
,
'
A /
n \
\
n2
•
'
/
n(n+l)
^UTTJ
N
n=l
°°
(f)
'
£(^-l)»,
n=l
OO
(g)
2 i i
(b) E ( ^ ~ ^ T
n=l
.
•
^(^S-l), a>l.
n=l
3.2.2. Test the following series for convergence:
(a)
g l l n f l + iV
v
n=l
W
(b) g 1 ln=±l,
'
n=2
(d)
Ln2_lnn'
n=l
OO
(e)
v
2^(l n n )lnn
n=2
v
y
-.
2 ^
n=2
H n n V1 n l n n "
(Inn)
3.2.3. Let Y2 ani Yl bn be series of positive terms satisfying
n=l
n=l
&n+l
. ^n+1
^
r
< —— for n > n 0 .
oo
oo
Prove that if Yl bn converges, then Y2 an ^so converges.
n=l
n~l
3.2.4. Test these series for convergence:
(a)
D^T'
n=l
(b)
ZLT^An=l
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3.2. Series of Nonnegative Terms
73
3.2.5. Determine for which values of a the given series converges.
(a)
oo
oo
(b) £ ( ^ - l ) a ,
£ ( ^ - 1 ) * , a>l,
n=l
00
1\
//
n+1
(<> E f ( i + s )
n=l
N
\
n=l
\
a
i \a
°° /
-«) - «) £ ( * — - J
x
n=l
/
'
/
oo
3.2.6. Prove that, if a series ^ an with positive terms converges,
n=l
then
OO
y ^ (a a n - 1),
where
a > 1,
n=l
also converges.
3.2.7. Investigate the behavior (convergence or divergence) of the
following series:
/
OO
1\
°°
(a)
^-lnfcos-j,
n=l
V n^
(c)
/
^
y % **""+*,
n=l
a,6,c,deR,
n2n
°^
n=l
(b)
T
x—TT7
r^—r~ > a, 6 > 0.
(n -h a) n + fc (n + b)n+a
v
/
\
/
oo
3.2.8. Suppose a series ^ a n of nonnegative terms converges. Prove
71=1
OO
that ]P y/anan+i
also converges. Show that the converse is not true.
71=1
If, however, the sequence {an} is monotonically decreasing, then the
converse statement does hold.
oo
3.2.9. Assume that a positive-term series ^ an diverges. Study
the behavior of the following series:
(a)
w
oo
Sift'
^irfe'
71=1
OO
n=l
(b)
(d)
71=1
oo
^rrk'
Ei^f71=1
OO
n=l
n
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Problems. 3: Series of Real Numbers
74
3.2.10. Assume that a positive term series ^ an diverges. Denote
n=l
by {Sn} the sequence of its partial sums. Prove that
oo
—
n=l
and
oo
diverges
—
converges.
n=lSn
3.2.11. Show that under the assumptions of the preceding proposition the series
oo
E
converges for each (3 > 0.
a
n
q qP
n=2 °n>->n-l
3.2.12. Prove that under the assumptions of 3.2.10 the series
oo
E
n=l
a
n
n
converges if a > 1 and diverges if a < 1.
oo
3.2.13. Prove that, if a positive term series 7^2
converges and
1 = 1 an
rn =
Yl ft/e, n G N, denotes the sequence of its remainders, then
k=n+l
oo
(a)
V^ ——
diverges,
n=2r"-l
converges.
3.2.14. Prove that under the assumptions of the foregoing problem
oo
converges if a < 1 and diverges if a > 1.
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75
3.2. Series of Nonnegative Terms
3.2.15. Show that under the assumptions of Problem 3.2.13 the series
53 a n+i In rn converges.
n=l
oc
3.2.16. Let Yl an be a series of positive terms. Suppose that
n=l
lim n In —— = #.
n->oo
Prove that
an+i
oo
53 an converges if g > 1 and diverges if g < 1 (the
n=l
cases # = +oo and # = — oo are included). Show that if g = 1, then
the test is inconclusive.
3.2.17. Study the behavior of the following series:
oo
oo
1
n=l
(d)
00
oo
1
n=l
-.
E ^
a>0
n—1
..
n=l
00
..
(e) E ^ n ^ '
>
n=2
a>0
-
3.2.18. Discuss convergence of the series
00
£V + * + -+*,
a >0.
n=l
3.2.19. Use the result of Problem 3.2.16 to prove the limit form of
the Test of Raabe.
Let an > 0, n G N, and let
lim n [ —
n->oo
\a
n +
i
1 1 — r.
00
Prove that 53 a™ converges if r > 1 and diverges if r < 1.
n=l
3.2.20. Let {a n } be defined recursively by setting
Q>1 = «2 = 1,
« n + l = &n H
ofln-l
f°r
™ > 2.
oo
Study convergence of the series 53 ~ •
71=1
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Problems. 3: Series of Real Numbers
76
3.2.21. Let a\ and a be positive. Define the recursive sequence
{an} by putting
a n + i = ane~a"
for
n=l,2,....
oo
Determine for which values of a and f3 the series J2 an converges.
n=l
3.2.22. Determine for which values of a the series
n!
^ ( a + l ) (( ao - + 2 ) . . . . - ( a + n)
converges.
3.2.23. Let a be an arbitrary positive number and let {bn} be a
positive sequence converging to b. Study the convergence of the series
E
n=1
nlar
(a + 6i)(2a 4-fe2)• - • (na + 6 n )'
3.2.24. Prove that, if a sequence {a n } of positive numbers satisfies
Qn+l __ i _ 1
7n
n
ttrj
n In n'
OO
where 7 n > T > 1, then ^ a n converges. On the other hand, if
71=1
Qn+1
^n
i _ 2.
OTI
n
n In n '
oo
where 7 n < T < 1, then ^ a n diverges. (This is the so-called Test
of Bertrand.)
71=1
3.2.25. Use the tests of Bertrand and Raabe to derive the following
Criterion of Gauss.
If {an} is a sequence of positive numbers satisfying
Qn+1 _ ,
an
PL
n
An
nx'
oo
where A > 1 and {$ n } is a bounded sequence, then J2 an converges
when a > 1 and diverges when a < 1.
n=l
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3.2. Series of Nonnegative Terms
77
3.2.26. Discuss the convergence of the series
y , a(a + 1) • ... • (a + n - 1) /?(/? + 1) •••••(/? + n - 1)
^
n!
7(7 + 1) •... . ( 7 + n - 1)'
where a,/3 and 7 are positive constants.
3.2.27. Determine for which values of p
'(2n-l)!!\p
^-f V (2n)!!
converges.
3.2.28. Prove the following condensation test of Cauchy.
Let {a n } be a monotonically decreasing sequence of nonnegative
00
numbers. Prove that the series Yl an converges if and only if the
n=l
00
series ]T 2 n a2^ converges.
n=l
3.2.29. Test the following series for convergence:
00
00
1
^n(mn)a'
n=2
v
'
'
^
n=3
-
n • Inn • l n l n n '
3.2.30. Prove the following Theorem of Schlomilch (a generalization
of the Cauchy theorem, see Problem 3.2.28).
If {gk} is a strictly increasing sequence of positive integers such
that for some c > 0 and for all k 6 N, gk+i — gk < c(gk — 9k~i) and
if a positive sequence {an} strictly decreases, then
^
a n < 00
if and only if
^ ( # f c + i - gk)a9k < 00.
3.2.31. Let {a n } be a monotonically decreasing sequence of positive
numbers. Prove that the series Yl an converges if and only if the
following series converges:
71=1
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Problems. 3: Series of Real Numbers
78
(a)
^3na3n,
(b)
n=l
^nan2,
(c)
n=l
^n2an3.
n=l
(d) Apply the above tests to study convergence of the series
in Problem 3.2.17.
3.2.32. Suppose that {an} is a positive sequence. Prove that
lim (an)^^
n—>oo
oo
< C
implies the convergence of J2 an.
71=1
3.2.33. Suppose that {an} is a positive sequence. Show that
lim (na™)^ 1 ^ < n—->oo
oo
C
implies the convergence of ^2 an.
71=1
3.2.34. Let {a n } be a monotonically decreasing sequence of positive
numbers such that
2 CiOn
— <9 < 1.
oc
Show that ^ an converges.
71=1
3.2.35. Let {an} be a monotonically decreasing sequence of nonnegoo
ative numbers. Prove that if ]P an converges, then
71=1
n
lim nan = 0.
-+°°
Show that this condition is not sufficient for the convergence of the
series.
3.2.36. Give an example of a convergent series with positive terms
for which the condition lim nan = 0 does not hold.
71—>OC
OO
3.2.37. Let J2 an be a convergent positive series. Give necessary
71=1
and sufficient conditions for the existence of a positive sequence {bn}
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3.2. Series of Nonnegative Terms
79
such that the series
oo
oo
and
6
Yl "
Yl T
n=l
n—1
both converge.
3.2.38. Does there exist a positive sequence {an} such that the series
OO
OO
Y] an
^—
and
n=\
both converge.
3.2.39. Show that
oo
-.
V " ——
^ n2an
l
n=l
1
V^ _
1
"* Q n + 1
n=l
diverges for any positive sequence
{an}.
3.2.40. Let {an} and {bn} be monotonically decreasing to zero and
oo
oo
such that the series ^2 an and J2 bn diverge. What can be said
n=l
n=l
oo
about the convergence of ]T) cn, where cn — min{a n ,6 n }?
n=l
3.2.41. Let {an} be a monotonically decreasing sequence of nonnegoo
ative numbers such that Y2 ^f diverges. Assume that
n=l
bn = min <J an , ,__,
ln(n
I.
oo
Prove that J2 ^T a * s o diverges.
n=l
3.2.42. Let {a n } be a bounded, positive and monotonically increasing sequence. Prove that
OC
/
v
converges.
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Problems. 3: Series of Real Numbers
80
3.2.43. Let {an} be an increasing positive sequence diverging to
infinity. Prove that
OO
/
v
diverges.
3.2.44. Let {an} be a positive monotonically increasing sequence.
Show that for any a > 0
OO
„=i
a
converges.
^ <
3.2.45. Show that for any positive and divergent series Y2 an there
n=l
exists a sequence {cn} monotonically decreasing to zero such that
OO
]T ancn diverges.
n=l
oo
3.2.46. Show that for any positive and convergent series ]T an
n=l
there exists a sequence {cn} monotonically increasing to infinity such
OO
that Yl anCn converges.
n=l
oo
3.2.47. Let Y2 an be a convergent series with positive terms. Let
n=l
oo
{rn} denote the sequence of its remainders. Prove that if Y^ vn
n=l
converges, then lim nan = 0.
n—>oc
3.2.48. Let {an} be a positive sequence diverging to infinity. What
can be said about the convergence of the following series:
oo
1
(a) y±,
n=l ""
oo
oo
^
oo
(b) Y4-,
n=l
^
(C) y-,4-'
I
-—.
n=l " n
"«
I
_?
In n
3.2.49. Study convergence of ^ a n , where
n=l
ai = 1,
a n + i = cosa n
for
nG
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3.2. Series of Nonnegative Terms
81
3.2.50. Let p be an arbitrarily fixed nonnegative number. Study the
OO
convergence of Y2 a n , where
n=l
an+i = n_psinan
ai = 1,
for
n G N.
3.2.51. Let {a n } be a sequence of consecutive positive solutions of
oo
the equation tanx = x. Study the convergence of Y2 ^-.
n=l
an
3.2.52. Let {a n } be a sequence of consecutive positive solutions of
oo
the equation tan y/x — x. Study the convergence of Yl ~ •
n=l ° n
3.2.53. Let a\ be an arbitrary positive number and let a n +i =
oo
ln(l + an) for n > 1. Study the convergence of Yl an71=1
3.2.54. Assume that {an}
oo
is a positive monotonically decreasing
a
sequence such that Y2 n diverges. Show that
71=1
Ql + Q 3 + ... +a2n-l
y
rwoo
_ -
a 2 4" a4 + ... + &2n
3.2.55. Let 5^ = 1 + ^ + ... + ^ and let kn denote the least of all
positive integers k for which Sk >n. Find
,.
&n+i
lim — — .
n—>oo kn
3.2.56. Let A be the set of all positive integers such that their decimal representations do not contain zero.
(a) Show that
^2 n
neA
conver
ges.
(b) Determine all a such that Y2 ^
n€A
converges.
oo
3.2.57. Let Y2 an be a series of positive terms and let
71=1
lim —2=- = p.
n-^oo Inn
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Problems. 3: Series of Real Numbers
82
Prove that if g > 1, then the series converges, and if g < 1, then the
series diverges (here g may be equal to +oo or — oo.)
Give examples showing that in the case of g = 1 the criterion is
indecisive.
3.2.58. Show that the Raabe test (see Problem 3.2.19) and the test
given in Problem 3.2.16 are equivalent. Moreover, show that the
criterion in the preceding problem is stronger than each of the above
mentioned tests.
oo
3.2.59. Study the convergence of ^ an whose terms are given by
71=1
Ol
Vi,
2 - y 2 + V 2 + ... + y/2,
N
n>2.
(n — 1)—roots
3.2.60. Let {an} be a sequence monotonically decreasing to zero.
Show that if the sequence with terms
(oi - an) + (a 2 - an) + ... + (a n _i - an)
oo
is bounded, then Yl an must converge.
n=l
3.2.61. Find a series whose terms an satisfy the following conditions:
fll = 2 '
«n = « n + l + « n + 2 + •••
for
72 =
1,2,3,....
oo
3.2.62. Suppose that the terms of a convergent series Yl an whose
n=l
sum is S satisfy two conditions:
^i > a2 > a 3 > ...
and
0 < an < an+\ + a n + 2 + ...,
n £ N.
Show that it is possible to represent any number 5 in the half-closed
00
interval (0, S] by a finite sum of terms of the series Yl an
00
infinite subseries Y2 ank, where {ank}
n=l
is a subsequence of
or
by an
{an}.
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83
3.2. Series of Nonnegative Terms
3.2.63. Assume that ]T) an is a series of positive and monotonically
71=1
decreasing terms. Prove that if each number in (0,5), where S denotes the sum of the series, can be represented by a finite sum of
oo
terms of {an} or by an infinite subseries J2 anki where {ank}
fc=i
is a
subsequence of {a n }, then the following inequality holds:
&n < a n + i + «n+2 + ••• for each
n G N.
oo
3.2.64. Let Y2 an be a divergent series of positive terms and let
n=l
lim |p- = 0, where S n = a\ + a2 + ... + an. Prove that
..
lim
n-^oo
a i 5 f 1 + a 2 5 ^ 1 + ... + a n 5 - 1
*
^—
.
— — 1.
ln6n
3.2.65. Using the preceding problem, show that
n-^oo
Inn
3.2.66. Let Y2 an be a convergent series of positive terms. What
n=l
can be said about the convergence of
Ql + a 2 + - + Q n 9
n=l
3.2.67. Prove that if {an} is a positive sequence such that ^ ^ a^ >
n
2n
oo
fc=n+l
n=l
fc=i
]T} afc for n € N, then ]T a n < 2eai.
3.2.68. Prove the following Inequality of Carleman:
If {an} is a positive sequence, then
n=l
n=l
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Problems. 3: Series of Real Numbers
84
provided that ^ an converges.
n=l
3.2.69. Show that if {an}
positive integer A;,
is a positive sequence, then, for every
oo
oo
•
n=l
N
..
n=l
, i \ n
'
3.2.70. Let {an} be a sequence of positive numbers. Prove that the
convergence of J2 ^~ implies the convergence of
71=1
n
oo
/
]CI
n2an
• n
(]C
x
n=l \
ak
-2^
x
fc=l
3.2.71. Let {an} be a monotonically increasing sequence of positive
oo
numbers such that ]T ^- diverges. Show that
an
n=l
y
-
oo
^
1
n a n - (n - l ) a n _ i
is also divergent.
3.2.72. Let {pn} be a sequence of all consecutive prime numbers.
oo
Study convergence of ^
n=l
—.
3.2.73. Study convergence of
oo
.,
I
,
{n-
l)pn-i'
y
~2npn-
where pn denotes the nth prime number.
3.2.74. Evaluate
oo
lim
-^
'
k=2
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3.2. Series of Nonnegative Terms
85
3.2.75. Let {an} be a sequence satisfying the following conditions:
0 < an < 1 for all
and
n e N
a\ ^ 0.
Let
Sn = cti + ... -f an
and
Tn = Si + ... + Sn.
oo
Determine for which values a > 0 the series £3 ^ - converges.
n
n=l
3.2.76. Let A: be an arbitrary positive integer. Assume that {an}
is a monotonically increasing sequence of positive numbers such that
^2 -^- converges. Prove that the series
n=l
E
m an
,
and
n=l
^-^ In n
>
n=l
are either both convergent or both divergent.
3.2.77. Assume that / : N —• (0, oo) is a decreasing function and
(f : N —» N is an increasing function such that (f(n) > n for all
n € N. Verify the following inequalities:
¥>(1)-1
ip(n)-l
n-1
fc=i fc=i fc=i
(2)
V>(n)
£
n
/(*)>E^w)^(*)-v(*-1))-
fc=yj(l)-l fc=2
3.2.78. Prove that under the assumptions of the foregoing problem,
if there exists q such that for all n e N the inequality
/(y(n))(y>(n+l)-y>(n))
7W
oc
holds, then ]T f(n)
n=l
,
-g<1
converges. On the other hand, if
/(y(n))(<p(n)-y?(n-1))
TT-T
> 1,
n €N ,
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Problems. 3: Series of Real Numbers
86
then J2 f(n)
diverges.
71=1
3.2.79. Derive from the preceding problem the following test for convergence and divergence of positive series.
oc
The series ^ an whose terms are positive and monotonically
n=l
decreasing is convergent when
hm
n->oc an
= g< Z
and divergent when
hm
=g>
-.
3.2.80. Derive from Problem 3.2.78 the following test for convergence
and divergence of positive series (compare with Problem 3.2.34).
A positive series ]P an whose terms are monotonically decreasn=l
ing is convergent provided that
hm
n—+oc an
and divergent provided that
lim ^
= g<1
> 2.
3.2.81. Using Problem 3.2.77, prove the criteria given in 3.2.31.
3.2.82. Prove the following Test of Kummer,
Let {an} be a positive-valued sequence.
(1) If there are a sequence {bn} of positive numbers and a positive
constant c such that
bn—
Gn+l
bn+i > c for all
n £ N,
oo
then J2 an converges.
n=l
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3.2. Series of Nonnegative Terms
87
(2) If there is a positive sequence {bn} such that £] ^- diverges
n=l
and
^-^--671+1 <0
for all
n
neN,
oc
then ^ a n diverges.
n=l
3.2.83. Show that the tests of d'Alembert (the ratio test), Raabe
(3.2.19) and Bertrand (3.2.24) are special cases of the Kummer test
(3.2.82).
3.2.84. Prove the following converse of the Kummer test.
Let {an} be a positive sequence.
oo
(1) If J2 an converges, then there exist a positive sequence {bn}
n=l
and a positive constant c such that
0n
«n+l
On+1 > C.
oo
(2) If ^2 an diverges, then there exists a positive sequence
n=l
{bn}
oo
such that Yl ir diverges and
n=l
n
K—
Gn-fl
bn+i < 0.
3.2.85. Prove the following tests for convergence and divergence of
positive series.
(a) Let A: be a positive integer and let lim ^ ^ = g. If g < 1, then
oo
oo
n=l
n=l
Y^ an converges, and if g > 1, then ^ an diverges.
(b) Let H e a positive integer and let
lim n f -^
1J = g. If
oc
oo
n—l
n=l
g > k, then ]T a n converges, and if g < k, then ]T an diverges.
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Problems. 3: Series of Real Numbers
88
3.2.86. Let {an} and {(pn} be sequences of positive numbers. Asoo
sume that (fn = 0 ( ^ ) • Prove that the convergence of J2 an im~
oo
plies the convergence of Yl
n=2
a
n=2
n~ipn-
3.3. The Integral Test
3.3.1. Prove the following integral test
Assume that / is a positive and decreasing function on the inoo
terval [l,oo). Then the series Y2 f(n)
n
sequence {/n}> In — J f{x)dx,
l
converges if and only if the
71=1
is bounded.
3.3.2. Let / be a positive and differentiable function on (0, oo) such
that / ' decreases to zero. Show that the series
OO
OO
.//
\
either both converge or both diverge.
3.3.3. Let / be a positive and decreasing function on [1, oo). Set
„N
N
SN = y2f(n)
and
IN = /
n=l
f(x)dx.
Jl
Show that the sequence {SN — IN} is monotonically decreasing and
its limit belongs to the interval [0, / ( l ) ] .
3.3.4. Show that the limits of the sequences
(a)
(b)
! + + . . . + _ _inn,
2
n
1
1
fn 1
1 +
^ + - + — - /
— dx>
0<a<l,
both belong to the interval (0,1).
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3.3. The Integral Test
89
3.3.5. Apply the integral test to study convergence of the series given
in 3.2.29.
oo
3.3.6. Let YJ an be a positive divergent series and let 5 n = ai +
n=l
ai + ... + an > 1 for n > 1. Verify the following claims:
oo
(a)
^£Sk
diverges
'
oo
(b)
Y^
n
^
converges.
_ 1 ^n l n &n
3.3.7. Let / be a positive and decreasing function on [1, oo). Assume
that a function p is strictly increasing, differentiable and such that
ip{x) > x for x > 1. Prove that, if there exists q < 1 such that
^
— # ^ or sufficiently large x, then
f(x)
Prove also that, if ^
series diverges.
/(l)
X
^ /(n)
n=l
converges.
- 1 f° r sufficiently large x, then the
3.3.8. Let / , g be positive continuously differentiable functions on
(0, oo). Moreover, suppose that / is decreasing.
(a) Show that, if lim (-g(x)^verges.
x—•oo ^
- g'(x)\
> 0, then £ fin)
n=l
n
(b) Show that, if the sequence with terms J -h\dx
l
and for sufficiently large x, —g(x)j^j—gf(x)
con
"
is unbounded
g{x)
< 0, then ]T / ( n )
n=l
diverges.
3.3.9. Let / be a positive continuously differentiable function on
(0, oo). Prove that
(a) if
lim (-^jffi)
x—>oo ^
f'( )
'
> 1, then £ f(n) converges,
n=l
°°
(b) if - Xf(x) < 1 for sufficiently large x, then ^ /(n) diverges.
n=l
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Problems. 3: Series of Real Numbers
90
3.3.10. Let / be a positive continuously differentiate function on
(0, oo). Prove that
(a) if lim
X—>00
- ^ J x l n x > 1, then J2 f(n)
[-JT^T
^
converges,
71=1
'
(b) if ( — yr^Y — ~: J x l n x < 1 for sufficiently large x, then Y
^
diverges.
n=l
'
f(n)
3.3.11. Prove the following converse of the theorem stated in 3.3.8.
Let / be a positive decreasing and continuously differentiate
function on (0, oo).
oo
(a) If Yl / ( n ) converges, then there exists a positive continuously
71=1
differentiate function g on (0, oo) such that
^(-•<* ) 7§|-H > f t
OO
(b) If ^ f(n)
diverges, then there exists a positive continuously
71=1
different iable function g on (0, oo) such that the sequence with
terms
r I
/h 9{x)-dx, n = 1,2,...,
is unbounded and for sufficiently large x,
-«*)£$-•(.><<).
3.3.12. For 7 > 0, study convergence of the series
00
^
2-, (lnn)( lnn ) 7 '
71 = 2
V
'
3.3.13. Study convergence of the series
00
y^
1
_L
.
^—' n 1+ nrhr^ Inn
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3.3. The Integral Test
91
3.3.14. Let {An} be a positive monotonically increasing sequence
and let / be a positive and increasing function satisfying the condition
,00
j
Jx1 tf{t)
Show that
£(1-TM^T<OO.
A n +1/ /(A n )
3.3.15. Prove the following generalization of the integral test.
Let {An} be a sequence strictly increasing to infinity and let /
be a positive continuous and decreasing function on [Ai, oo).
(a) If there exists M > 0 such that A n + i - An > M for n G N
oo
and if the improper integral J f(t)dt
converges, then the series
Ai
oo
J2 f(^n) a lso converges.
n=l
(b) If there exists M > 0 such that A n + i - An < M for n G N
oo
and if the improper integral J f(t)dt
diverges, then the series
Ai
oo
J2 /(^n) also diverges.
71=1
3.3.16. Suppose that / : (0, oo) —» R is a positive and differentiable
oo
function with positive derivative. Prove that J2 jh) converges if
n=l
and only if J2 rHn)
does.
n2
n=l
3.3.17. Define lni x = lux and InkX = ln(hifc_ia;) for k > 1 and
sufficiently large x. For n G N, let (p(n) be the unique positive integer
such that 1 < ln^( n ) n < e. Does the series
oo
y
-
^
n(lni n)(ln 2 n) • ... • (ln^ (n) n)
converge or diverge?
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Problems. 3: Series of Real Numbers
92
3.4. Absolute Convergence. Theorem of Leibniz
3.4.1. For the indicated values of a, decide whether the series are
absolutely convergent, conditionally convergent, or divergent:
n+ 1
n=l
(b)
„(lnn)»
£(-1)"^-,
ae
n=2
(c)
V(-l)"sin-,
aG
71
n=l
W
e
/ x
()
v^
1
S^nU
n=l
00
/a2 —4a-8\n
+ ea-ls) '
x
x
_x , n ^
^K\{-8,2},
n
E^->a^0'
V^
ft
n=l
3.4.2. For c 6 l , study convergence and absolute convergence of the
series
00
nn-\
n l
na ~
-f Inn
where na is an index depending on a such that nan
n > n0.
1
-f In n 7^ 0 for
00
3.4.3. Suppose that a series ^ a n with nonzero terms converges.
71=1
Study the convergence of the series
n=l
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3.4. Absolute Convergence. Theorem of Leibniz
93
3.4.4. Does the condition lim ^ n = 1 imply that the convergence
oo
n^oo °
oc
a
of ^2 n is equivalent to the convergence of ^2 bn?
n=l
n=l
oo
3.4.5. Assume that a series ]T) an converges conditionally and set
n=l
pn =
|an|
2
Qn
, qn =
|Qn|
2
° n . Show that both
£ Pn and X) g n
n=l
diverge.
n=l
oo
3.4.6. Assume that a series ]T] a n converges conditionally.
Let
n=l
oo
{P n } and {Q n } be the sequences of partial sums of Yl Pn and
n=l
oo
^ <?n, defined in the foregoing problem, respectively. Show that
lim - ^ = 1.
n-+oo
g
n
3.4.7. Study convergence and absolute convergence of the series
y^ (-1)1*1
n=l
3.4.8. For a G l , decide whether the series
£
(-l)[>/n]
n=l
converges absolutely, converges conditionally, or diverges.
3.4.9. Decide whether the series
-
(_l)[lnn]
^
n
71=1
is absolutely convergent, conditionally convergent, or divergent.
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Problems. 3: Series of Real Numbers
94
3.4.10 i. Let
+ 1
/
for
22k
<n<22k+\
2 f c
+2,
( - i for 22fc+l < n < 2
where fc = 0,1,2,.. .. Discuss the convergence of the series
£•77, ~
=
oo
OO
(a)
5
n=l
£n
(b) V
n
n=2
3.4.11. Study the convergence of the series
oo
r~
-\
n=2
3.4.12. Investigate the behavior (absolute convergence, conditional
convergence) of the following series:
(a)
oo
£(_!)«( ^ _ i r ,
71=1
OO
(b)
^(-i)™^-!),
0 > i )
71=1
(c)
OO
£(_!)«( ^ _ i ) ,
n=l
( ir i+
s - (( =)*"-)•
3.4.13. For a, 6 > 0, discuss convergence of the following series:
n=i
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3.4. Absolute Convergence. Theorem of Leibniz
3.4.14. Let Yl (~l)n~lan
95
be an alternating series which satisfies
71=1
the conditions of the Leibniz test, that is, 0 < an+i < an for all
n and lim an = 0. Denote by rn the nth remainder of the series,
n—•oc
oc
rn — Yli (~l) fc_la fc- Show that rn has the same sign as the term
/c = 7 l + l
( - l ) n a n + i and \rn\ < a n + 1 .
3.4.15. Suppose that a sequence {an} tends to zero. Show that the
series
oc
y^an
oo
y^(q n -f a n +i)
and
n=l
n=l
either both converge or both diverge.
3.4.16. For a sequence {an} convergent to zero and for a, 6, c such
that a + 6 + c ^ 0 , prove that the series
oo
^
oo
an
and
n=l
^ ( a a n + 6 a n + i + ca n + 2 )
n=l
either both converge or both diverge.
3.4.17. Let {an}
be a sequence with lim an = a ^ 0 and with
n—>oo
nonzero terms. Prove that the series
V7an+i-an)
and
VM
I
either both are absolutely convergent or both do not converge absolutely.
oc
3.4.18. Show that, if a sequence {nan} and a series Yl n(an — an+i)
71=1
OO
both converge, then Y2 an also does.
71=1
3.4.19. For a sequence {an} monotonically decreasing to zero, study
the convergence of the series
oo
£(-D^rn + l l
71=1
a
+ a 2 + ... +CLn
n
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Problems. 3: Series of Real Numbers
96
3.4.20. Decide for which values of a the series
oo
(—1) n! sin asm - •... • s i n 71 = 1
converges absolutely and for which it diverges.
3.4.21. For positive a, b and c, study convergence of the series
3.4.22. Discuss convergence of the following series:
oc
oo
(b) j> i n n r-
(a) j>osnr,
n=l
n=l
3.4.23. Let {an} be a positive sequence. Prove that
(a) if lim n f—2^
n—•oo
- 1] > 0, then ]T (~l) n «n converges,
^
(b) if n ( - 2 1
^
lim n [ - ^
'
n=l
l j < 0, then ]T) (-l)nan
n=l
'
diverges (in particular, if
1) < 0, then the series diverges).
3.4.24. Assume that for a positive sequence {an} there exist a e R,
£ > 0 and a bounded sequence {/3n} such that
an+i
n
n1+£'
oo
Prove that the series Y, ( — ^) Ua n converges for a > 0 and diverges
for a < 0.
71=1
3.4.25. Discuss the convergence of the series
n=l
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3.4. Absolute Convergence. Theorem of Leibniz
97
3.4.26. Assume that the series ]T an converges and {pn} is a posn=l
itive sequence which increases to +oo. Show that
j.
aiPi + Q2P2 + ... + anpn
n—oo
=
pn
3.4.27. Let {an} be a positive sequence decreasing to zero. Prove
00
that, if the series ^ anbn converges, then
71=1
lim an(bi +b2 + ... + 6n) = 0.
3.4.28. Let a be a given positive number. Prove that, if the series
00
J2 %£ converges, then
n=i
n
r
hm
ai + a 2 + ... + a n
= 0.
3.4.29. Let {kn} be a strictly increasing sequence of natural num00
00
n=l
n=l
bers. Then the series J2 akn is called a subseries of the series ^
an.
Show that, if all the subseries of a series converge, then the series is
absolutely convergent.
3.4.30. Let fc, I be integers such that k > 1, / > 2. Must the converoc
gent series £^ an be absolutely convergent if all its subseries of the
form
n=l
00
n=l
are convergent?
00
3.4.31. Give an example of a convergent series ^ an such that
71=1
OO
J2 O'n diverges.
n=l
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Problems. 3: Series of Real Numbers
98
3.4.32. Does there exist a convergent series Y2 an s u c n t n a t all the
71=1
OO
series of the form ^ an, where k is an integer greater than or equal
to 2, diverge?
71=1
3.4.33. Let {an} be a monotonically decreasing and positive seoo
a
quence such that the series ^
n diverges. Suppose that the series
71=1
OO
^2 ^n^n? where en is —1 or 1, converges. Prove that
71=1
lim£l+£2
+
-+g"<0< K
£ l + e 2 +
-
+ e
".
3.4.34. Assume that {an} is a positive monotonically decreasing
OO
sequence and that the series ^ snan^ where en is —1 or 1, converges.
72=1
Show that
lim (ei + e2 + ... + en)an = 0.
(See 3.2.35.)
n—>oo
3.4.35. Suppose that the series
oc
J2 bn converges and {pn} is a
71=1
monotonically increasing sequence for which
lim pn = -f oo and
n—>oo
oo
E — = +oo. Show that
n=l
r
lim
Plh + P2&2 + ... 4" PnK
. n . 77— Pl&l + P2&2 + - + PnK
< 0 < lim
.
3.4.36. In the harmonic series Yl n ^
71=1
u sa
^ a c n the s ^S n "+"> P
n
times, consecutively, then the sign "—", q times, consecutively, then
"+", p times, consecutively, etc. Show that the new series converges
if and only if p = q.
3.4.37. Prove the following generalization of the Toeplitz theorem
(see 2.3.1 and 2.3.36).
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3.5. The Dirichlet and Abel Tests
99
Let {cn,k : n , f c e N } be an array of real numbers. Then for any
convergent sequence {an} the transformed sequence {bn} given by
oo
n
K — 2jcn?fcafc,
^ 1»
is convergent to the same limit if and only if the following three conditions are satisfied:
(i)
cn k —> 0
for each k G N,
oo
(ii)
^ C n , f c = l,
(hi)
there exists C > 0 such that for all positive integers n
J2\Cn,k\<C'
k=l
3.5. The Dirichlet and Abel Tests
3.5.1. Using the Dirichlet and Abel tests, study convergence of the
following series:
(a)
£(-1)"
sin 2 n
71=1
oo
n=l
(c)
^
>
2
x
1
/
V"+V
7T
sin^
—
, .. , sin 2f '
n=l
n
n
2
2
f- ln n
Z
V
— 7 5 — COS
n = 92
(d)
n
\
,
a>0.
4
oo sin (n + - )
3.5.2. Does the series V —-^— converge?
In Inn
n=2
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Problems. 3: Series of Real Numbers
100
3.5.3. For a G l , study convergence of the series
£ sin(na) sin(n a)
2
(a)
71=1
OO
(b)
E
sin(na) cos(n 2 a)
71=1
3.5.4. Show that the series
E
cos n sin(na)
71=1
converges for each a G R.
£2, sin(na)
3.5.5. Determine whether the series jT, — - — , a G i , converges
n=i
n
absolutely.
3.5.6. Show that for a G R and n G N,
E
.
it=i
sin(a/c)
k
<2v^.
3.5.7. Prove that the series
£(-!)"arctan n
71=1
y/n
converges.
3.5.8. For x > 1, study convergence of the series
OO
n/i
71=1
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3.5. The Dirichlet and Abel Tests
101
3.5.9. Prove the following lemma of Kronecker.
oo
Let Y2 an be a convergent series and let {bn} be a monotoning
cally increasing sequence such that lim bn = -foo. Then
n—»oo
oo
k=n
k
/
1 \
n
\
n
k=i
/
where o(bn) means that
lim ^ 4 - ^ = 0 .
3.5.10. Assume that the series Y2 ncn converges. Show that for
71=1
OO
every n e N the series Y2 (k -f l)cn+fe also converges. Moreover,
k=0
oo
show that if tn = Y2 (& + l)cn+fc, then lim tn = 0.
n
k=o
^°°
oo
3.5.11. Assume that the partial sums of the series Y2 an form a
oo
n=l
bounded sequence. Prove that if the series Yl fin — &n+i| converges
n=l
and
oo
a so
lim 6 n = 0, then for every natural k the series Yl anbn
n—oo
^
n = 1
converges.
oo
3.5.12. Prove that if ^ (6n—6n+i) converges absolutely and Y2
n—l
n=l
oo
a
"n
converges, then the series Y2 anbn also converges.
71=1
OO
3.5.13. Using Abel's test, show that the convergence of Yl an
im
~
71=1
OO
plies the convergence of the series Y2 anXn for \x\ < 1.
71=1
3.5.14. For a given sequence {an} show that if the Dirichlet series
oo
n=l
nx
converges at x = XQ, then it converges at every x > XQ.
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Problems. 3: Series of Real Numbers
102
3.5.15. Prove that the convergence of the Dirichlet series YL ^
n=i
implies the convergence of the series
OO
i
E
,
n=l
v
z
n
™n ,
M
v
—' xix + l)...(:r + n)
7
x
* * 0,-1,-2,....
'
3.5.16. Prove that if the series Y2 an%n converges for |x| < 1, then
71=1
a
Yl n j z ^ r also converges.
n=l
3.5.17. Must the convergent series ]T) a n be absolutely convergent
n=l
if all its subseries of the form
^Gfc/n,
& > l , i > 2,
n=l
converge ?
3.6. Cauchy Product of Infinite Series
3.6.1. Prove the following theorem of Mertens.
OO
If at least one of the two convergent series Y2 an
n=0
an
OO
d
Yl ^n
n=0
converges absolutely, then their Cauchy product (that is, the series
OO
Yl cn, where c n = aobn + a i 6 n - i + ... + an^o ) converges. Moreover,
71 = 0
OO
OO
OO
71=0
71=0
71=0
if Yl an = A and ^ 6n = B, then J ] c n = AB.
3.6.2. Find the sum of the series
OO
(a)
^nxn_1,
\x\ < 1,
71=1
OO
(b)
V]cn,
n=0
71
where
c n = ^2,xkyn~k,
\x\ < 1, jj/j < 1,
fc=0
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3.6. Cauchy Product of Infinite Series
oo
(c)
5>'
n
where c =
103
^
" E f c ( f c + 1)(n_fc + 1) r
n=l
k—1
3.6.3. Form the Cauchy product of the given series and calculate its
sum.
v(a)
'
(b)
(c)
^ 2n
y ^ —r
*-^ n\
1
V^ -—-,
^ 2nn\
and
n=0
oo
n=0
1
E(-D"n
n=l
oo
^ ( n + l)x
oo
and
n
1
£^>
n=l
oo
^ ( - l ) n ( n + l)x n .
and
n=0
n=0
3.6.4. Assume that the series Yl an is convergent and set An =
71=0
oo
ao+ai + ...+a n . Prove that for \x\ < 1 the series J^ Anxn
converges
71=0
and
^2 anXU = (1-X)^2 An*n>
71=0
71=0
OO
3.6.5. Find the Cauchy product of the series ^ ( _ 1 ) n r a p > x G M,
with itself.
Hint. Use the equality £
fc=0
n=0
(£) 2 = Cn)-
3.6.6. For a > 0 and |x| < 1, verify the claim
1
1 x
1-3 x2
+
2oT2 + 2~^4a + 4 + ' "
( a
X
/
1
1-3 o
(1+2X+2^2 + -
+
+
1 - 3 •... • (2n - 1) xn
2 - 4 - . . . • (2n) a + 2n
1-3-... - ( 2 n - l )
2-4...-(2n) '
- V l I a + 1 X l (Q + 1 )( Q + 3 ) X 2
a\
a+2
(o + 2)(a + 4)
(o+l)-...-(o + 2 n - l )
+
"
(o + 2 ) - . . . - ( o + 2n)
n
+
+
'*'
"
^
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Problems. 3: Series of Real Numbers
104
3.6.7. Prove the following theorem of Abel.
oo
oo
If the Cauchy product ^ cn of the two convergent series J2 an —
n=0
OO
n=0
A and ^ bn = B converges to C, then C = AB.
n=0
3.6.8. Show that the series
r
^
H
v
r
y
i
2 /
n +1 V
x
n=l
i
2
ii +
i\
nj
'
OO
is the Cauchy product of the series £ ( - l ) 7 1 " 1 1 with itself, and find
71=1
its sum.
3.6.9. Study the convergence of the Cauchy product of the series
oo
£ ( - l ) - i - i = with itself.
n=l
3.6.10. Prove that if at least one of two positive series is divergent,
then their Cauchy product diverges.
3.6.11. Must the Cauchy product of two divergent series be divergent?
3.6.12. Prove that the Cauchy product of two convergent series
oo
oo
^2 Q>n and £] bn converges if and only if
n=0
n=0
n
lim Y] ak(K + bn-i + ... + 6n-fc+i) = 0.
n—>oo * — '
fc=l
3.6.13. Suppose that {an} and {bn} are positive sequences monotonically decreasing to zero. Show that the Cauchy product of the
oo
oo
71=0
71=0
series ^ ( _ l ) n « n and J2 (~l) n &n converges if and only if
lim an(b0 + 6i -f ... + bn) = 0
and
lim bn(a0 4- ai + ... + a n ) = 0.
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3.7. Rearrangement of Series. Double Series
105
3.6.14. Show that the Cauchy product of
oo
ytDl
and
yU£.,
a,/3>o,
n=l
n=l
converges if and only if a + /? > 1.
3.6.15. Assume that positive sequences {an} and {bn} are monotonically decreasing to zero. Prove that the convergence of the series
oo
^2 anbn is a sufficient condition for convergence of the Cauchy prod-
n=0
oo
oo
71^
n=0
uct of the series ^ (—l) n a n and ^2 (~l)n^n? and that the COnveroo
gence of ]P (a n 6 n )
1+Q
for every a > 0 is a necessary condition for
n=0
the convergence of this Cauchy product.
3.7. Rearrangement of Series. Double Series
3.7.1. Let {ra/c} be a strictly increasing sequence of positive integers, and put
b\ = ai + a 2 + ... + a m i ,
00
Show that if the series ^
62 = a m i + i + a m i +2 -f ... + am2> ••• •
00
a
n converges, then ^ bn also converges
71=1
and both series have the same sum.
71=1
3.7.2. Consider the series
1_1
i _ i _ i
!_
" 2 ~ 4 + 3 " 6 " 8 + 5 """"'
which is obtained by rearranging the terms of the series ^2 —
n=l
in such a way that each positive term is followed by two negative
terms. Find the sum of this series.
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Problems. 3: Series of Real Numbers
106
3.7.3. Let us rearrange the terms of ^
*—£— so that blocks of
n=l
a positive terms alternate with blocks of /? negative terms, that is,
+
1
3
+
"'
+
+
1
1 1
_J_
1
1
2a-l~2
4 "' 2/3 + 2a + l + 2a + 3
1
1
1__
_J_
4 a - 1 ~ 2(3 + 2 ~ 2/3 + 4 ~ '" " 4/3 + '*''
+
"'
Find the sum of the rearranged series.
3.7.4. Show that
i_i_i_i
~2~4~6~8
+
i__^_j
L__L
I_
16 + 5 ~ "'" " "
3~10~12~14
3.7.5. Find a rearrangement of the series 2^ —n— which doubles
n=l
its sum.
3.7.6. Rearrange the terms of 2J —
to obtain a divergent se-
n=l
ries.
3.7.7. Study convergence of the series
1
V5
1
\/2
1
1
\/5
1
>/7
\/4
obtained by taking alternately two positive terms and one negative
term of the series V *—4=—.
«
i
n=l
^
3.7.8. Prove that any rearrangement of an absolutely convergent series is convergent and has the same sum.
3.7.9. Assume that a function / : (0,+oo) —• (0, oo), decreasing to
zero as x —• oo, is such that the sequence {nf(n)} increases to -f oo.
oo
Let S denote the sum of the series ^ ( - l ) n _ 1 / ( n ) - Given /, find a
n=l
rearrangement of this series convergent to S + /.
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3.7. Rearrangement of Series. Double Series
107
3.7.10. Assume that a function / : (0, +oo) —> (0, oo), decreasing to
zero as x —> oo, satisfies lim nf(n) — g, g € (0, +oo). Let 5 denote
71—•OO
OO
the sum of the series Y ( _ l ) n _ 1 / ( n ) - Given Z, find a rearrangement
n=l
of this series convergent to S + 1.
oo
3.7.11. Rearrange the terms of J2 (~~l) n _ 1 ^> P ^ (0> 1 ) ' t o increase
71=1
its sum by /.
3.7.12. For a > 0, using the result in 3.7.10, find a rearrangement
oo
of Y (~^)n~ln
wnose sum
n=l
is
m
2 + | In a.
3.7.13. Is it possible to accelerate by rearrangement the divergence of
a divergent series with positive and monotonically decreasing terms?
oo
3.7.14. Assume that the series ]T an with positive terms diverges
and that
71=1
lim an — 0. Show that it is possible to slow down its
71—>00
divergence arbitrarily by rearrangement; that is: for any sequence
{Qn} satisfying
0 < Qi < Q2 < ... < Qn < ...,
oo
there is a rearrangement Y
fc=i
a
nk
sucn
lim Qn = +oo,
n—>oo
tnat
Oni + 0>n2 + ••• + « n m < Qm
for
771 E N.
3.7.15. Let {r n } and {sn} be two strictly increasing sequences of
positive integers without common terms. Assume also that every
positive integer appears in one of the two sequences. Then the two
oo
subseries Y
71=1
OO
Y
a
rn
an<
oo
l S
a
sn
are
called complementary subseries of
71=1
Q>n- We say that the rearrangement shifts the two complementary
71=1
subseries relative to each other if for all positive integers m and n
such that m <n the term aTjn precedes aTn and aSrn precedes aSn.
oo
Prove that the terms of a conditionally convergent series Y
71=1
a
n
can
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Problems. 3: Series of Real Numbers
108
be rearranged, by shifting the two complementary subseries of all its
positive and all its negative terms, to give a conditionally convergent
series whose sum is an arbitrarily preassigned number.
oo
3.7.16. Let Y2 ank be a rearrangement of a conditionally converoo
gent series Y2 an. Prove that if {n^ — k} is a bounded sequence,
71=1
OO
OO
a
then YJ nk = Yl an- What happens if the sequence {n^ — A;} is
n=l
k=l
unbounded?
oo
3.7.17. Let Yl ank be a rearrangement of a conditionally convergent
oo
k=i
oo
oo
series Yl an. Prove that J2 ank = Yl an if a n d only if there exists
n=l
n=l
k=l
a positive integer N such that each set {n^ : 1 < k < m} is a union
of at most N disjoint blocks of successive positive integers.
3.7.18. With an infinite matrix {a^/e}, i = 1,2,..., k = 1,2,..., of
oo
Y2 ai,k- We say that the
real numbers we associate a double series
i,k=l
double series converges to S € M if, given £ > 0, there exists no £ N
such that
\Smin - S\ < e for m,n> n 0 ,
where
771
*^7n,n
=
/
j
i=l
Then we write
^
&i,k-
fc=l
oo
S =
We say that
71
/
lim Sm,n = Y ^ a i.k'
771,71—>00
oo
*—'
i,fc=l
J ] a ^ converges absolutely if
i,/c=l
oo
Y2 \°>i,k\ converges.
i,k=l
Note that the terms of an infinite matrix (ai)/e)i,fc=i,2,... can be oroo
dered into a sequence {c n }, and then the corresponding series Y2 cn
oo
is called the ordering of Yl
i,fc=l
71=1
a
i,k into a single series. Prove that if
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3.7. Rearrangement of Series. Double Series
109
one of the orderings of a double series converges absolutely, then the
double series converges (absolutely) to the same sum.
oo
3.7.19. Prove that if a double series
^
a ^ converges absolutely,
oo
then any of its orderings ^ cn converges and
n=l
oo
oo
ai k =
z2Cn-
22 >
i,k=l
n=l
3.7.20. Show that any absolutely convergent double series is convergent.
o o / o o
3.7.21. We say the iterated series ^ I J2 aiyk 1 is absolutely coni = i \k=i
oo
vergent if ^
i=l
/ oo
\
\k=l
/
n ] |ai?fc|
'
/
oo
converges; similarly for J ]
k=l
/ oo
\
\i=l
J
^ ai)fc I •
Prove that an absolutely convergent iterated series is convergent.
oo
3.7.22. Prove that if the double series ^
a^fc converges absolutely,
i,k=l
then both the iterated series
oo
/ oo
\
i=l
\k=l
/
aik
Y\J2 )
oo
and
/ oo
J2[J2a^
H,k
fc=l
\i=l
converge absolutely and
oo
i,fc=l
i=l
/ oo
\
oo
\k=l
/
k=l
/ oo
\i=l
/
3.7.23. Prove that if one of the four series
oo
Yl
i,k=l
o o / o o
K*I>
\
o o / o o
Y Yl i *.*i ' S 1 E iai=l
\fc=l
a
/
fc=l
y ^ ( | a n , l | + l a n - l , 2 | + | f l n - 2 , 3 | + ••• +
n=l
\i=l
lal,n|)
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Problems. 3: Series of Real Numbers
110
converges, then all the series
oo
o o / o o
i,fc=l
oo
i=l
\
\fc=l
o o / o o
/
fc=l
2 ^ ( O n , l + O n - 1 , 2 + O n - 2 , 3 + ••• +
n=l
N
\i=l
/
«l,n)
converge to the same sum.
3.7.24. Calculate
£—' n\k\(n + k + l)'
3.7.25. Find
n,fc=l
3.7.26. Show that
2
y>
n!fc!
= n
^ f n + fc + 2)! ~ 6 '
n,k=0
3.7.27. For 0 < x < 1, consider the infinite matrix
/
x
\ x(l-x)
ht(l-x)2
V
—x2
-x2(l-x2)
-x2(l-x2)2
—x3
x2
x2(l-x2)
x2(l-x2)2
-x3(l-x3)
-x3(l-x3)2
x3
x3(l-x3)
x3(l-x3)2
-X
...
...
/
Prove that only one of the iterated series associated with this matrix
converges (not absolutely).
3.7.28. Study convergence of the following double series:
oo
(a)
b
()
] T xiyk,
i,k=0
oo
J2 ^P>
|x|, \y\ < 1,
where
<*,/3>o,
i,k=l
oo
fc)
..
where
^
y ^ ———,
where
p > 0.
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3.7. Rearrangement of Series. Double Series
111
3.7.29. Find the sums of the following double series:
oo
(a)
^
V
(b)
,
u
oo
,
where p > - 1 ,
1
Jt,w
OO
^
3.7.30. Given an infinite matrix (&i,fc)i,fc=i,2,...> prove that there is
00
^2 ai,k
only one double series
&m,n
— /
2,fc=l
/
v
i=l
J
sucn
that
Q>i,k — ^771,n?
^ 5 ^ — J-j A ••• •
k=l
3.7.31. Taking
^ = (~ir+fc(^ + ^)>
<,fc = l,2,...,
in the preceding problem, study convergence of the corresponding
double series
00
^
a^.
i,fc=l
00
3.7.32. Show that if |x| < 1, then the double series
xlk con-
^
verges absolutely. Using this fact, prove that
OO
i,k=l
0
k=l
0
^
.
0
0
n=l
OO
n=l
n
2
00
n=l
where 6{n) denotes the number of all natural divisors of n.
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Problems. 3: Series of Real Numbers
112
3.7.33. Show that if \x\ < 1, then the double series
verges absolutely. Moreover, prove that
OO
OO
i,k=l
,
£.
E
ixlk
i,k=l
con-
OO
k=l
n=l
where cr(n) denotes the sum of all natural divisors of n.
3.7.34. Let C(P) = E ^ > P > ^ ^ e the Riemann zeta function.
n=l
Set
^ = E ^ = C(P)-I,
P>I.
n=2
Show that
OO
(a)
OO
J > = 1,
..
p
(b)
5 > l ) S P = 2.
p=2
p=2
3.7.35. Prove the following theorem of Goldbach.
li A = {km
: m,fc = 2,3,...}, then E
nGA
^ r j = *•
3.7.36. Let C denote the Riemann zeta function. Prove that for any
integer n > 2,
C(2)C(2n - 2) + C(4)C(2n - 4) + ... + C(2n - 2)<(2) = (n + 0
C(2n).
3.7.37. Using the result in the foregoing problem, find the sums of
the series
E -n^
6
and
y^-g.
^ ^ n8
n—\
n~\
3.8. Infinite Products
3.8.1. Find the value of:
w
n ('-?). a.) n = ^ .
oo
•
n=2
x
1
\
oo
'
n=2
3
1
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3.8.
Infinite Products
113
OO
c
x^2m(^+kn),
l[cos^,
()
meN,
keZ,
71=1
(d)
OO
II
cosh
OO
y
n=l
oo
N
n=l
xeR
|r>
e
( ) Il(1 +
>
..
OO
' '
)'lxl<1'
n
n—\
oo
JL
n=\
a;2n
n=0
v
v
OO
~ o
n=l
n
3.8.2. Study the convergence of the following infinite products:
n=2
x
n—9
X
y
n=l
x
7
'
3.8.3. Assume that a n > 0, n G N. Prove that the infinite product
oo
oo
Y[ (1 + a n ) converges if and only if the series J2 an converges.
n=l
n=l
3.8.4. Suppose that an > 0 and an ^ 1 for n G N. Show that the
oo
infinite product Yl 0-~an)
n=l
converges.
oo
converges if and only if the series £) an
7i=l
3.8.5. Set
1
1
0 2 n - l = —7= + ~»
Show that the product
1
tt
2n
=
7=>
n
^
€ N.
f| (1 + a n ) converges although the series
71=1
£ a n diverges.
71=1
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Problems. 3: Series of Real Numbers
114
3.8.6. Study convergence of the products:
(a)
TTcos-,
n
(b)
x±
TTnsin-,
*
n
A
n=l
n=l
n—\
^
n=l
'
(e)
^
'
(f) n n ^ -
oo
n^
n=l
n=l
3.8.7. Assume that the series ^ a n converges. Prove that the inn=l
n
=l
oo
oo
a
finite product Yl 0- + n) converges if and only if the series ]T) a\
n=l
n=l
oo
a
does. Prove also that if the series Yl n diverges, then the infinite
n=l
oo
a
product Yi (1 + n) diverges to zero.
n=l
3.8.8. Assume that the sequence {an} decreases monotonically to
oo
zero. Show that the product J\ 0- + (~ ^)n°"n) converges if and only
71=1
OO
if the series ^
n=l
a
n converges.
OO
3.8.9. Prove that the product
J
v
n=l ^
'
f1 + (-l)n+1-^J
n
oo
diverges al-
though the series XM~l) n + 1 77^ converges.
71=1
3.8.10. Show that if the series
OO
•
a
-
v
a
]^( Ti~2 ^)
71=1
^
OO
and
Slanl3
71=1
'
both converge, then the product Yl (1 + an) also converges.
71=1
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3.8. Infinite Products
115
3.8.11. Does the convergence of the product Yl (1 + an) imply the
2=1
convergence of the series Y an and Y
n=l
n=l
Hint. Consider the product
1
2a ) V
where \ < a
+
2a
+
22a J (l
an?
3<* ) \
+
3"
+
32"
<\.
3.8.12. Prove the following generalization of the result in 3.8.10. For
k > 2, if both series
£ («. 4a" + - + t2k-*)
n=l ^
and
'
SK|fe+1
n=l
OO
converge, then the product Yl (1 -f a n ) also converges.
3.8.13. Prove that the convergence of Yl (1 + an) and of Y a 2
n=l
implies the convergence of Y
n=l
a
3.8.14. Show that if the products
00
converge, then both series Yl an
n=l
n=l
n-
an
Yl U + a n)
n=l
oo
an
d
II U
n=l
— a
n)
d X^ an converge.
n=l
3.8.15. Assume that the sequence {an} decreases monotonically to
1. Does the product
1
1
d\ - — • as • — • as •...
&2
0*4
always converge?
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Problems. 3: Series of Real Numbers
116
oo
oo
3.8.16. Assume that the products Yl an and
\\ bn with positive
71=1
factors both converge. Study the convergence of
(b)
JJ(fln + M»
(a)
(c)
71=1
IK
71=1
71=1
OO
OO
(d) n ^
J | Q>nbn,
71=1
71=1
3.8.17. Show that for xn G (0, f ) , n G N, the products
n
oo
oo
,
and
cosx n
T T sinx n
II
n=l
Xn
n=l
oo
converge if and only if the series J2 xn converges.
71=1
OO
3.8.18. Let YJ an be a convergent series with positive terms and
71=1
let Sn denote its nth partial sum. Show that
1+
«n( £r - £ OO
/
7l=Z
x
\
OO
'
71=1
3.8.19. Show that if the infinite product
J~J (1 + a n ), an > — 1,
71=1
converges to P , then the series
£
^ ( l + ai)(l + a 2 ) - . . . - ( l + an)
also converges. Moreover, if S is its sum, then 5 = 1 - -p.
3.8.20. Suppose that the infinite product
oo
17(1 + On),
where
an > 0, n G N,
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3.8.
Infinite Products
117
diverges. Prove that
oo
V
=2
= 1.
^ ( i + < * l ) ( l +<*2) • . . . • ( ! +On)
3.8.21. Show that
x
E —(l + x)(l7— + x T7) - . . . - (7-l + i " )r = 1
for
2
n=1
x > 1.
OO
3.8.22. Let a n ^ 0 for n G N. Prove that the infinite product n an
n=l
converges if and only if the following Cauchy criterion is satisfied. For
every e > 0 there is an integer no such that
1^71^71+1 * ••• * 0>n+k — l | < £
for n > no and k £ N.
3.8.23. For |x| < 1, verify the following claim:
00
i
1
n
n( +^ )=^—-—.
n=i
n(i-s2n-"1)
n=l
3.8.24. The product Yl ( l + a n ) is said to be absolutely convergent if
71=1
OO
OO
Yl (l + |a n |) converges. Show that the product Yl ( l + a n ) converges
n=l
00
n=l
absolutely if and only if the series Yl an converges absolutely.
n=l
00
3.8.25. Show that every absolutely convergent product Yl (1 + an)
is convergent.
71=1
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Problems. 3: Series of Real Numbers
118
3.8.26. Prove that if the product Yl (1 + an) converges absolutely,
71=1
then
n=l
n=l
ni,ri2 = l
ni<ri2
oo
"r ••• ~r
y ^
cini(xn2...(xnk
-f- ....
ni,n2,...,nfc=:l
ni<n2<...<nk
3.8.27. Assume that the product Yl (1 + a n) converges absolutely.
n=l
oo
Show that the product
a
Yl (1 + n^) converges absolutely for each
n=l
x G R and it can be expanded in an absolutely convergent series.
That is,
oo
oo
n=l
fc=l
JJ(l + anx) = l + 5^i4fca;fc,
where
Ak —
/
anian2...an
v
ni,n 2 ,...,nfc = l
nn\<ri2<
i<n2<...<nfc
3.8.28. Establish the equality
9
n(n+l)
2
3.8.29. Verify the identity
OO
OO
n2
n , i + ^ . I , _ 1 + : g _ ? _ J r _ _ K J i . , I,I <i.
3.8.30. Assume that the series Yl an converges absolutely. Prove
that if x ^ 0, then
OO
n=l
OO
•
-
\
5
n(i+a^)(i+^)=B0+i: «(^+^).
n=l
n=l
^
^
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3.8. Infinite P r o d u c t s
119
where Bn = An + AxAn+1
+ A2An+2
oo
+ ..., n = 0,1,2,..., and
oo
J J ( l + ana;) = i4o + 53^ib« fc (see 3.8.27).
n=l
fc=l
3.8.31. For \q\ < 1 and x ^ O , establish the identity
n( i -^)n( i +^"M(i+v) =i+ s« (xB+^)-
n=l
n=l
^
'
n=l
^
'
3.8.32. For \q\ < 1, verify the following claims:
(a)
(b)
(c)
oo
oo
n=l
oo
n—1
oo
n=l
oo
n=l
oo
rj(i_^)JJ(1_92n-i)2
oo
= 1+
2^(_ir^)
n=\
oo
n(i-^)ii(i+^i)2=i+2E^2>
n=\
oo
rj(! - «2n) II( 1 +^ 2 ") 2 = ! + £<?" 2+n -
n=l
n=l
n=l
3.8.33. For x > 0, define the sequence {a n } by setting
i
«1 =
n-l
T—1 ,
1+x
«n
7
n T-T x — k
x + n 1* x + k
vn>
=
n>l
-
fc=i
Show that the series Yl an converges and find its sum.
n=l
3.8.34. Prove that if the infinite product f| (1 + can) converges for
71=1
two different values of c G i \ {0}, then it converges for each c.
3.8.35. Prove that if the series
5>n(* 2 - fc2 )
OO
71
n=l
fc=0
converges at x = XQ, XQ £ Z, then it converges for any value of x.
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Problems. 3: Series of Real Numbers
120
3.8.36. Let {pn}
greater than 1.
be the sequence of consecutive prime numbers
(a) Prove the following Euler product formula:
-1
1 1 1N
n=l
oo
nx I
F n /
Z-*i n.x
n=l
oo
(b) Prove that the series ^
71=1
— diverges (compare with 3.2.72).
3.8.37. Using DeMoivre's law, establish the identities
(a)
sinx =
x n ( l - ^ 2 J ,
n=l
^
'
3.8.38. Applying the result in the foregoing problem prove the Wallis
formula
3.8.39. Study convergence of the products
(a)
(b)
oo
Il(1 + ^ ) e " " '
n=l
U^f¥->
n=l
n
x 1
>~ >
x> 1
~oo
3.8.40. Prove that the infinite product fj (1 + an) converges abson=l
lutely if and only if any rearrangement of its factors does not change
its value.
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3.8. Infinite Products
121
3.8.41. Find the value of the product
H)H)-("±)
"O-D-O-SJVTX1^)-"
obtained by rearranging the factors of Yi (14n=2 ^
„
J in such a way
'
that blocks of a factors greater than 1 alternate with blocks of /?
factors smaller than 1.
3.8.42. Prove that the convergent but not absolutely convergent infinite product
Yl (1 + an)> «n > -1? can be rearranged to give a
71 = 1
product whose value is an arbitrarily preassigned positive number, or
to give a product that diverges to zero or to infinity. (Compare with
3.7.15).
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Solutions
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Chapter 1
http://dx.doi.org/10.1090/stml/004/04
Real N u m b e r s
1.1. Supremum and Infimum of Sets of Real
Numbers. Continued Fractions
1.1.1. Set A = {x G Q : x > 0, x2 < 2} and s = sup A. We may
assume that s > 1. We will now show that for any positive integer n,
(-;)' s 2 £ (' + s)'-
<»
Since s — ^ is not an upper bound of A, there exists x* G A such
that s — ^ < x*. Hence
( . - I ) ' <(*?< 2.
Assume that ( s + ^ ) < 2. If s were rational, then s + ^ G A and
5 + ^ > s, which would contradict the fact that s = sup A. If 5 were
irrational, then w = Hn+ H _j—i_ i s a rational number such that
'
n-\-l
n+1
2
s < w < s + ^ . Hence w < (s + ^-) < 2 and w G A, a contradiction.
So, we have proved that (s + ^) > 2. By the left-hand side of (1),
52
~~ 7T < s2 ~ it + ^ - 2 ' w n i c n S i v e s ^ T < £• Letting n —• 00,
we get s2 - 2 < 0.
As above, the right-hand side of (1) gives ^ ^ > — ^, which
implies s 2 - 2 > 0. Therefore s2 = 2.
125
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Solutions. 1: Real Numbers
126
1.1.2. Suppose A is bounded below and set a = inf A. Then
(1)
x > a for all x G A,
(2)
for any e > 0 there is x* G A such that x* < a + e.
Multiplying the inequalities in (1) and (2) by — 1, we get
(1') x< -a
for all xe
(-A),
(2') for any e > 0 there is x* G (—A) such that x* > — a — e.
Hence —a = sup(—A). If A is not bounded below, then —A is not
bounded above and therefore sup (—A) = —inf A = -foo. The other
equality can be established similarly.
1.1.3. Suppose A and B are bounded above, and put a = sup A
and b — supB. Then a is an upper bound of A and b is an upper
bound of B. Thus, a + b is an upper bound of A -f B . Moreover, for
any e > 0 there are x* G A and y* G B such that x* > a — | and
y* > 6 - f . Therefore, x*+y* > a + b-e. Since z* = x* + y* G A + B ,
the equality a + b = sup (A + B) is proved. If A or B is unbounded
above, then A + B is also unbounded above, and by the definition of
the supremun sup(A + B) = sup A + s u p B = +oo.
The second equality is an immediate consequence of the first one
and of the foregoing problem. Indeed, we have
sup(A — B) = sup(A + (—B)) = sup A + s u p ( - B ) = sup A - inf B.
Similar arguments can be applied to prove the equalities
inf(A + B) = i n f A + infB,
inf (A - B) = inf A - sup B.
1.1.4. Suppose that both sets are bounded above, and put a = sup A
and b = supB. Since elements of A and B are positive numbers,
xy < ab for any x G A and y G B . We will now prove that ab is
the least upper bound of A • B . Let e > 0 be arbitrarily fixed. There
exist x* G A and y* G B such that x* > a-s and y* > b-e. Thus
x*y* > ab- e(a + b-e). Since e(a4- b - e) can be made arbitrarily
small if e is small enough, we see that any number less than ab
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1.1. Supremum and Infimum. Continued Fractions
127
cannot be an upper bound of A • B. Therefore ab — sup (A • B ) . If
A or B is not bounded above, then A • B is not either. Therefore
sup( A • B) = sup A • sup B = +00.
The task is now to prove sup (-^-) = T—-^ if a' — inf A > 0.
Then, for any x e A , the inequality x > a' is equivalent to ^ < •£?.
So, -T is an upper bound of -£-. Moreover, for any e > 0 there is
x* G A such that x* < a' + e. Hence
1
1 _ 1
£
x*
a' + s
a'
ar(a'-\-£)'
Since Q /(J + £ \ can be made arbitrarily small, ^ is the least upper
bound of ^-. We now turn to the case a' = 0. Then the set ^ is
unbounded (indeed, for any e > 0 there is x* G j^ for which x* > ^).
Therefore, sup -^ = +00.
Assume now that A , B are bounded sets of real numbers (positive or nonpositive) and put a = sup A, b = supB, a' = inf A and
b' = inf B. If a' and b' are nonnegative, the desired equality follows
from the above. If a' < 0 and a, 6' > 0, then xy < a& for any
x G A and y G B . Take e > 0 so small that a — e > 0. Then there is
a positive number x* in A for which x* > a — e. Moreover, there is
y* G B such that y* > b — e. Hence
x*y* > x*(b - e) > (a - e)(b - e) = ab - e(a + b - e).
So, in this case we have sup (A • B) = ab.
We now consider the case where a', b' < 0 and a, b > 0. Then,
for any x G A and y G B , we have
xy < max{a&, a'b'}.
Assume first that max{a&, a'b'} = a'b'. By the definition of the
greatest lower bound, for sufficiently small e > 0, there exist x* G A
and t / * G B for which x* < a' + e < 0 and y* < b' +e < 0. This gives
x V > x*{b' + e)> {a' + e)(6' + s) = a'b' + e{a' + &' + e).
Notice that a' -f &' 4- £ is negative. Therefore a'6' is the least upper
bound of A - B . In the case where max{a&, a'b'} = ab similar reasoning yields sup(A • B) = ab. All other cases can be proved analogously.
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Solutions. 1: Real Numbers
128
1.1.5. Suppose first that A and B are bounded above. Put a =
sup A and b = sup B. Of course we may assume that a < b. Then, for
any X G A U B , X < 6. Moreover, for any e > 0 there is x* 6 B such
that x* > b - e. It is obvious that x* belongs to A U B . Therefore,
the first equality is valid. If A or B is not bounded above, then
A U B is not either. So, sup(A U B) = -f oo, and we assume that
max{+oo, c} = max{+oo, H-oo} = +oo for any real c. The proof of
the second equality is similar.
1.1.6. We have
Ai =
U
A2 =
( 3
3
3
3
"|
f3Ar-1
3fc-2
3k-3
,
„}
\ 3 f c T l , ~ 6 i r ' - 2 ( 3 f c ^ l ) ; f c €N } -
Therefore inf Ai = — ^ , sup Ai = 5 and inf A2 = —\ , sup A2 = 1.
1.1.7. sup A = | , inf A = 0.2, s u p B = | , inf B = 0.
1.1.8. One can show by induction that for n > 11, 2 n > (n + l ) 3 .
Hence
0< (n±a! < fe±il! =
1
for
2n
(n+1)3
n-fl
Therefore 0 is the greatest lower bound of our set.
„> n .
It is also easy to show that 2 n > (n + 1) 2 for n > 6. Hence
^ ^ - < 1 for n > 6. The numbers 2, f, f|, | | (greater than 1) also
belong to our set. Thus the least upper bound of the set is | .
1.1.9. It follows from the foregoing problem that the greatest lower
bound of this set is equal to 0. By the inequality mentioned in the
preceding solution, 2 n m > (nm -f l ) 2 for nm > 6. Since nm + 1 >
n -h m, for n, m G N, we have
jn + m)2
2nm
(n + m) 2
(nm + l ) 2
(n + m) 2
(n + m) 2
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1.1. Supremum and Infimum. Continued Fractions
129
For nm < 6 we get the following elements of our set: 1,2, | , f§, | § .
Hence its least upper bound is | .
1.1.10.
(a) It is obvious that 2 is an upper bound for the set A. We will,
show there are no smaller upper bounds. Indeed, if e > 0 is arbitrarily fixed, then for any positive integer n* > [|] , we obtain
n*
> 2 - e. The greatest lower bound of A is 0, because
7 > 0 for m , n G N . Given e > 0, there is n such that A < e.
(b) Clearly, 0 < y/n - [yfn] < 1. Taking n = fc2, k G N, we see that
0 G B . Thus inf B = 0. To show that supB = 1, observe first
that [y/n2 -f 2n ] = n for each positive integer n. Suppose now
0 < e < 1. A simple calculation shows that the inequality
\/n2 + 2n - [\/n 2 + 2nl =
2
>l-e
is satisfied for any integer n > ^ 2 g ; .
1.1.11.
(a) sup{x G l : a ; 2 + x + l > 0 } = -f oo,
(b) inf {z = x + x _ 1 : x > 0} = 2,
(c) inf{z = 2* + 2 i : x > 0} = 4.
The first two equalities are easily verifiable. To show the third
one, observe that 9^ > \fab for a, b > 0. Therefore,
with equality if and only if x = 1. Thus (c) is proved.
1.1.12.
(a) Using the inequality ^ - > yfab for a, 6 > 0, we get
m
471^,
— + — >4
n
m
with equality for m = 2n. Therefore, inf A = 4. Taking m = 1,
one can see that the set A is not bounded above. This means
that sup A = +oo.
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Solutions. 1: Real Numbers
130
(b) Similarly, we get
1
ran
1
"I " 4m 2 + n 2 - 4 '
with equalities for m = — 2n and m = 2n, respectively. Consequently, inf B = —\ and s u p B = \.
(c) We have inf C = 0 and s u p C = 1. Indeed, 0 < ^ ^ < 1, and
for any e > 0 there exist positive integers ni and m\ such that
ni + 1
< £
and
rai
+ 1
> 1 — e.
(d) inf D = - 1 and supD = 1.
(e) One can take m = n to see that the set is not bounded above.
Hence supE = +oo. On the other hand, for any ra,n € N we
have ,™n, > h with equality for m = n = 1. Therefore inf E =
1
3"
1.1.13. Setting s = a\ + a>2 -f ... + a n , we get
_ afc+i _ Qfc+2
Ofc
afc
s ~ a,k + a fc+ i + a/c+2
As a result,
!<Y
n
a
<n_2.
-h
£-[ a/b + afc+i + Gfc+2
Now, our task is to show that inf Y] —.
a
* —
= 1 and that
fc=l
n
, „ a*
sup Y)
= n - 2. To this end we take a,k = tk , t > 0.
k—1
Then
£J a
fc
k=
t + t2 + t3
+ a fc+ i + ak+2
tn~2
tn-l
tn
+ - + .„ o • ... , • .- + .-. , • .-. • . +
,
^
i
r~2
,
2
n l
^ 1+ t+t ^ t~
2
+ t"- + 1
(
+
r-1
n l
t~
+t + l
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1.1. Supremum and Infimum. Continued Fractions
Letting t —• 0 + , we see that sup Y] —.
0
'
afc
. — = n - 2, and next
^ fc=l
-—' afc.-fafc.j_i +a,fc4-2
letting £ —• +oo, we conclude that inf Y\ „ ,.,
to
'
7^—'
131
afc
'
, „ — = 1.
ak+ak+i+ak+2
1.1.14. Fix n G N and consider the n + 1 real numbers
0, a — [a], 2a - [2a],..., na - [na\.
Each of them belongs to the interval [0,1). Since the n intervals
In' ^7r)' i = ^, * > •*•'n ~ 1 > cover [0,1), there must be one which contains at least two of these points, say nia — [n\d\ and n^oc — [n^a]
with 0 < n\ < ri2 < n. So,
1722a; — [^2^] — ™i<* + [n-iQ:]| < — .
n
Now, it is enough to take qn = ri2 — n\ and pn = [n2&] — [n\a\.
It follows from the above argument that qn < n; that is, the second
inequality also holds.
1.1.15. We will show that in any interval (p, q) there is at least one
element of A. Put 0 < e = q—p. It follows from the preceding problem
that there are pn and qn such that
a
Pn < 1
2 '
qn
Since a is irrational, lim qn = +oo. Therefore
n—*oo
\qnOt-Pn\
< —
<6
qn
for almost all n. Now set a = \qna — pn\. Then at least one of the
numbers ma, m G Z, belongs to the interval (p,q)\ that is, mqna —
rapn or —mqna -f 77_pn lies in this interval.
1.1.16. Let t G [—1,1]. Then there is an x such that t = cos x. By the
result in the foregoing problem, there exist integer sequences {mn}
and {kn} such that x — lim (fcn27r + ran). This and the continuity
n—•oo
of the cosine function imply that
t = cos-c = cos( lim (kn27r -f ran)) = lim cosra n = lim cos | r a J .
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Solutions. 1: Real Numbers
132
Hence each number of [-1,1] is a limit point of the set {cos n : n G N}.
The desired result is proved.
1.1.17. It is obvious that, if there is n for which xn is an integer,
then x is rational. Assume now that x = | with p G Z and q G N. If
x—[x] ^ 0, then 2 — £ = ^, where Z is a positive integer smaller than
q. Thus, the denominator of x\ = | is smaller than the denominator
of rr. This means that the denominators of xi,X2,... are successively
strictly decreasing and cannot constitute an infinite sequence.
1.1.18. We will proceed by induction. It is easily verifiable that
Rk = ^-
Qk
for k = 0,1,2.
Assume that for an arbitrarily chosen ra > 2,
Pm
Rr,
Pm-ia>m+Pm-2
Qm
qm-lQ>m
+ <7m-2
Note now that if we replace a m in Rm by a m H
the convergent i? m +i. Therefore
_ Pm-l {am + ^ J
R>m+1 —
/
•
— , then we get
+Pm-2
r
qm-l ( «m + ^ ^ 7 J + Qm-2
(Pm-lQm + P m - 2 K + 1 + P m - 1 _
Pm+1
(^m-l^m + g m _2)<2 m + i -f qm-l
qm+1
1.1.19. Denote
A*; = p/c-i^fc - ^fc-iPA:
for k = 1,2,..., n.
Then, for A: > 1,
Afc = Pk-i(qk-ia>k + qk-2) - qk-i{Pk-iQ>k +Pk-2)
= ~{pk-2qk-i
- qk-2Pk~\) = -Afc_i.
Since Ai = poqi — goPi = «o^i — (aoQ>i 4-1) = —1, we obtain A& =
(-l) f c . This implies that p*; and qk are relatively prime.
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1.1. Supremum and Infimum. Continued Fractions
133
1.1.20. As in the solution of 1.1.18, we have, for n > 1,
n—l
yn
Rn
Qn
Qn-l^n
+
Qn-2
Analogously
QnXn+l
"f
for n = l,2,....
Qn-1
Hence
p
_
Pn^n+l + Pn-1
Pn
9n«n+l+Qn-1
Qn
Pn-lQn
(~l)
-Qn-lPn
( t f n ^ n + l + Qn-l)Qn
n
(tfn^n+l + 9 n - l ) 9 n '
where the last equality follows from the result in 1.1.19. Therefore
— Rn}
> 0
for even n,
< 0
for odd n.
Thus x lies between two consecutive convergents.
1.1.21. We first prove that if a is a positive irrational, then the set
{n — ma : n,m € N} is dense in R+. To this end take an interval
(a, 6), 0 < a < b. We will show that this interval contains at least
one element of our set. Put e = b — a > 0. By the preceding problem
there exists a convergent Rn such that
(1)
0<Rn-a<
—.
Qi
Indeed, take an odd n and observe that
( < ? n £ n + l + Qn-l)Qn
Since
>
Qn -
lim qn = -foe, for sufficiently large n we have -±- < e. This
and (1) imply that 0 < pn — aqn < ^- < e for sufficiently large
n. Therefore there is n 0 G N such that no(pn - aqn) e (a, b). Now
let t E [—1,1]. There is a positive x such that t = sinz. It follows
from the above considerations that there exist sequences of positive
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Solutions. 1: Real Numbers
134
integers {mn} and {kn} for which x =
continuity of the sine function,
lim (mn — 27rkn). By the
n—^oo
t = sinx = sin( lim (ra n — 27rkn)) = lim sinra n .
n—>oo
n—KX>
So, we have proved that any number of the interval [—1,1] is a limit
point of the set {sinn : n € N}.
1.1.22. Let pn and qn be the integers defined in 1.1.20. Since xn+\ =
°n+l + - ^
> On+1 > We g e t ( g n ^ n + l + t f n - l ) ^ > (tfnGn+l " h ^ n - l ^ n
=
Qn-hiQn- Therefore, by 1.1.20,
\x — Rn\ <
QnQn+l
Since <?n+i = g n «n+i + #n-i > qn^n+i > qn, the desired inequality
follows. We will show now that the sequence {qn} contains infinitely
many odd numbers. Indeed, it follows from the result in 1.1.19 that
qn and g n + i cannot be both even.
1.1.23. It is enough to apply the formula given in 1.1.19.
1.1.24. Observe first that the sequence {qn} is strictly increasing
and qn > n. Moreover, by Problem 1.1.20,
\x~ Rn\ =
{qnxn+i +qn-i)qn
'
This and the inequality x n +i < a n + i -f 1 imply that
\x-Rn\
>
1
1
(g n (a n + i + 1) + qn-i)qn
(qfn+i + qn)qn
Since a n +2 > 1? we have
\X - Rn+1\ <
-
Wn+lfln+2 + qn)qn+l
<
—.
Wn+1 + <ZnJ<Zn
These inequalities yield the desired result.
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1.1. Supremum and Infimum. Continued Fractions
1.1.25. Let \x — J| < \x — Rn\ < \x — Rn-i\Rn and Rn-\ (see Problem 1.1.20),
< \Rn-l
Rn-l
S
135
Since x lies between
— Rn\-
Therefore, by the result in 1.1.23,
\rqn-i - sPn-i\
sqn-i
1
qn-iqn
1 <
^ —-—
L
u o v _ —-—
Moreover, we have
because |ro n _i - s » n _ i | > 1.
Sqn — 1
Qn — lQn
'
'
Hence s > qn.
1.1.26. Following the algorithm given in 1.1.20, we get
a0 = [ V2] = 1,
xx = -^—^
= V2 + 1.
Therefore, a\ = [xi] = 2. Similarly,
x
2 = 7-7=———= x\
(v / 2 + l ) - 2
and
a2 = a\ = 2.
By induction,
/s
,
1|
1|
1|
1|
1|
1|
Likewise,
\/5-l
= — + — + — + ...,
2
II
II
II
1.1.27. Since k < yjk2 + k < fc + 1, a0 = [v7^2 4-fc]= k. As a result,
xi = ^fc2+fc+fc. Consequently, 2 < xi < 2 + £ and ax = 2. Moreover,
x2 =
j
v/fc2+fc-fc ~~
= k + y P + fc.
Thus 2A; < x 2 < 2/c -I- 1 and a2 = 2k. In much the same way we
obtain 03 = 2. Now, by induction,
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Solutions. 1: Real Numbers
136
1.1.28. Since 0 < x < 1, we have a0 = 0 and x\ = 1/x. Therefore,
a\ = n implies [1/x] = n. Hence, 1/x — 1 < n < 1/x, which gives
l / ( n + l ) < x < 1/n.
1.2. Some Elementary Inequalities
1.2.1. We will use induction. For n = 1 the inequality is obvious.
Take an arbitrary positive integer n and assume that
(1 + oi) • (1 + a2) -... • (1 + On) > 1 + a\ + a 2 + ... + a n .
Then
(1 + ai)(l + a 2 ) •... • (1 + a n ) ( l + a n + i )
> (1 + ai + a2 + ... + a n ) ( l + a n + i )
= 1 + ai -f a 2 + ... + an + a n + i -f an+i {a\ + a 2 + ... + a n )
> 1 + ai + a 2 + ... -|- an + a n +iThus the claim is established.
1.2.2. Induction will be used. For n — 1 our statement is clear. We
suppose now that the claim holds for an arbitrarily chosen n. Without
loss of generality we can assume that numbers ai,..., an+i satisfying
the condition ax • a 2 • ... • a n + i = 1 are enumerated in such a way
that a\ < a 2 < ... < an < an+\. Then a\ < l a n d a n + i > 1. Since
a 2 • as -... • an • (a n +i • ai) = 1, by our induction assumption, we have
a2 + a>z + ... + an + (fln+i • «i) > ^- Hence
ai H- a 2 + ... + a n + a n + i > n + a n + i + ai — a n + i • a\
= n + a n + i • (1 - ai) + ai - 1 + 1
= n + 1 + (a n +i - 1)(1 - ai) > n + 1.
1.2.3. The inequalities follow from the statement proved in Problem
1.2.2. Indeed, replacing there the numbers a7 by n/n aj n , we get
An> Gn. The inequality Gn > Hn follows from the already proved
inequality An > Gn provided one replaces aj by its reciprocal ~.
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1.2. Some Elementary Inequalities
137
1.2.4. Using the arithmetic-geometric mean inequality, we have
y/(l + nx) • 1 • ... • 1 <l + x
(n factors).
1.2.5.
(a) Apply the arithmetic-harmonic mean inequality.
(b) Use the arithmetic-harmonic mean inequality.
(c) The left-hand inequality can be shown as in (a) and (b). To prove
the right one, let us observe that
1
1
1 1
1
2n
2
+ «
7: + - + T- + -z
r < «
7+ «
~ < 773n + l
3n-f2
5n
5n + 1 3 n + l
3n + 2
3
(d) By the arithmetic-geometric mean inequality,
2 3 4
n+1
n/
- + - + - + ... +
> nyVTl.
1
Hence
z
6
n
l + l + l + - + l + - + ... + l + - > n
1
6
n
\/n+l
and
i + J + ^ + ... + ->n(^TFT-i).
2 3
n
To prove the other inequality we use the arithmetic-geometric
mean inequality and get
1 2
3
n
n
+
7
7
+
T
+
+
7
>
2 3 4
n+ 1
Vn+1
This implies
2
3
n
V
A/nTT
n+1/
1.2.6. By the inequality Gn < An we get
a w
Vl-x---x^<
1 +
+ g:
"' +
2n
l
2n
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Solutions. 1: Real Numbers
138
1.2.7. The right-hand side of the inequality is a direct consequence
of Gn < An. The other one can be proved by induction. This is clear
for n = 1. Now we will prove that the inequality holds for n + 1
provided it does for n. To this end we show that ( a i a n + i ) n + 1 <
(ai •... • an • a n + i ) 2 , whenever {a\an)n < {a\ • ... • an)2. We have
x n+1
fa
( a i a n + i ) n + 1 < ai • an(ai • ... • a n ) 2 - ' - ^ ± i
Hence it is enough to show that
an+l
tt
n+l
^
2
a
n
Note that the last inequality can be rewritten as
(
\ n—1
T
1
+ —T7
Tw
oi + ( n - l ) d /
^di
+ in-
l)d,
where a n = a\ + (n — l)d, which is easy to prove by induction.
1.2.8. It is an immediate consequence of the foregoing result.
1.2.9. One can apply the arithmetic-harmonic mean inequality.
1.2.10.
(a) By the arithmetic-harmonic mean inequality we have
"iv1
and consequently
n~
n
-i
9
y->-.
fc = l
Similarly, the inequality
n
n
^
l
\ -—1
V——I
£{ — «*/
1
n
<-V(«-a
fc )
~nt{
implies
-.2
;£} s - afc
a(n - 1)
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139
1.2. Some Elementary Inequalities
From this and from the equalities
n
n
fc=1*-°*
= s >s
ti ~ak
n
n
1
— n and >
ti
i
= s > ak— — ra
ak
ti
the desired result follows.
(b) See the solution of part (a).
(c) This follows by the same method as in (a).
1.2.11. Use the inequality ^g± > v /5£.
1.2.12. We have
n
n
/ n
\ ^
n
1
2
n
n
fc,j=l
1.2.13. This inequality is equivalent to the following:
n
n
which in turn is a direct consequence of the obvious inequality a^aj -f
6fc6,-<(o2 + ^)i(o? + ^ ) i .
1.2.14. The claim follows from the Cauchy inequality.
1.2.15.
(a) By the Cauchy inequality,
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Solutions. 1: Real Numbers
140
(b) By (a),
n
n
n
n
Ofc
fc=l
a k
k=\
k
fc=l
k=\
fc=l fc=l fc=l
n
2
fc=l
n
> n -n^afe =n^(l-afc).
fc=i
fc=l
(c) By our assumption, log a ai + logaa2 + ... + log a a n = 1- This and
the Cauchy inequality (Problem 1.2.12) give the desired result.
1.2.16. The inequality is equivalent to
0 < - 4 a y^flfcfrfc
+4j>2+aa5;«.
fc=l
k=i
fc=l
which holds for each real a, because
n
n
fc=l
/c=l
A = 16 (£>&*) -16£>£]T&2<0.
SJfc=l
1.2.17. Applying the Cauchy inequality, we obtain
n
/
n
n
\
2
n
Kk=l
fc=l fc=l
k=l
1.2.18.
(a) By the Cauchy inequality,
/
<n
\
^
/
r
X>fc) =(E^%) <E^E?-
\jfe=i
fc=i fe=i
U=i
(b) Likewise,
n
Sjfe=l
SJfe=l
<E
fc3a
fc=i
n
*E p '
*:=i
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1.2. Some Elementary Inequalities
141
1.2.19. The Cauchy inequality gives
7c
fc=l
/
\fc=l
/
fc=l
fc=l
1.2.20. By the Cauchy inequality,
n
n
n
fc=l
k=l
fc=l
n
/
\
\/c=l
Hence ^2 a\ — n» w ^ n equality for a^ = ^ , k = l,2,...,n. Therefc=i
fore, the least value we are looking for is ^.
1.2.21. In much the same way as in the solution of the last problem
we get
(
n
\
2
/ n
l \
2
n
n
1
Thus,
n
1
^
with equality for a t = ^
E-
fc=i
^ ^-
pt
.So the least value in question
1.2.22. It follows from the solution of Problem 1.2.20 that
/
n
\fc=l
\
/
2
n
fc=l
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Solutions. 1: Real Numbers
142
Hence
(E a f c )
\fe=l
=
/
(( a i+ a 2)+E a f e ) ^( n _ 1 )f( a i+ a 2) 2 +E a fe
V
fc=3
/
\
fc=3
)
= ( n - l ) | ^ a | + 2a 1 o 2 j .
1.2.23.
(a) By the Cauchy inequality,
(X>*+w 2 ) = ( X > * + 2 a ^ + h D )
\/c=l
/
\fc=l
\k=l
J
fc=l
(
n
\ 2
/
n
\k=l
J
fc=l
\ 2
(b) By (a),
( E* -
\
n
2
/
2
fc=l
n
\
^
/
n
\
^
- \fc=l
E*2- / - \fc=l
EK-M2 / •
/
This and the inequality established in Problem 1.2.17 yield
(
Similarly,
n
\
fc=i
/
(E H
n
\
&
fc=l
/
2
/
n
\fc=i
2*
/
n
- \k=lE 4
\
^
/
\ "2
/
n
fc=i
n
^E
I^-M
k=l
and the desired result is proved.
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1.2. Some Elementary Inequalities
143
1.2.24. Since £ Pk^k = 1, we have 1 = £ Pk°>k = X) (Pk-a)ak
fc=i fc=i fc=i
n a
a J2 k for any real a. Now, by the Cauchy inequality,
fc=i
1< J>-a) 2 + a2
\/c=i
£a 2 +
/
\fc=i
+
£>
\/c=i
y
Hence
£ « 2 + ( X > ) >(E(pfc-a)2 + a2)_1.
fc=l
\fc=l
/
\fe=l
'
n
Putting a = ^ Y ]T) p/c, we obtain the greatest lower bound. Therefore,
fc=l
\k=l
/
(„ , , \ v^ „2
where the equality is attained for
(n + l)pk
=
(n +
-Y<Pk
fc=i
7
fe=i
\~*'
l)±pl-(±Pk)
Vfc=i
/
1.2.25. We will proceed by induction. For n — 1 we get the equality
ai&i = a\b\. Moreover, if the inequality holds for n, then
- {n + 1)^2akbk
Y^ak^h
fc=i
k=i
n
/c=l
fc=i
n
n
fc=l
fc=l
n
= /^(fen+i - M( a fc - a n +i) < 0.
fc=i
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Solutions. 1: Real Numbers
144
1.2.26. We will use induction on p. For p = 1 the equality a\ = a\
holds. Assuming the inequality to hold for p, we will prove it for
p+ 1. Obviously, without loss of generality, we may suppose that the
numbers a^ are enumerated in such a way that a\ < a,2 < ... < a n .
Now, by our induction assumption and the result in the foregoing
problem,
^
re-
k=i
it
fc=i
- . 1 1 /
,v
fc=i
1.2.27. We have
(1 +c)a2 + ( 1 + - V
= a2 + b2 + (yfca - -^=b\
+ 2ab > (a + b)2.
1.2.28. Clearly, y V + b2 + VQ 2 + c2 > |6| + |c| > |6 + c|. Hence
|62 - c 2 | < \b - c\ (\/a 2 + b2 + Va 2 -f c2) , which is equivalent to the
desired inequality.
1.2.29.
(a) For any real numbers a, 6, c we have a2 + 62 -f c2 > a& + 6c -f ca.
Thus 62c2 -f a2c2 + a 2 6 2 > afcc(a + 6 + c), which is equivalent to
our claim.
(b) The desired result follows from the inequality a2 -f b2 -f c2 >
ab + be -f ca in much the same way as in (a).
(c) This is a consequence of the arithmetic-harmonic mean inequality.
(d) We have
62-a2
b+ a
6 +a
— — = —— (6 - a) = —— ((6 + c - (c + a ) ) .
c+a
c-f-a
c+ a
Putting u — a + 6, v = 6 -f c and 2 = c - f a , w e obtain
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1.2. Some Elementary Inequalities
145
b2 — a?
(? -b2
a2 - c 2
u,
, v,
z,
N
x
1
—+ —
= -iV
- Z)
-U)} + -(U
- V)
K
J + ~{Z
K
}
c+a
a+ b
6+ c
z
u
vK
_ u2v2 -f v2z2 + z2u2 — (u2vz + v2uz -f z2uv)
uvz
_u2(v2
+ z2)+v2(u2
>o.
+ z2)+ z2(u2 + v2)- 2(u2vz + v2uz + z2uv)
2uvz
(e) For a = b the inequality is clear. Assume now that 0 < b < a.
Then
a-b
2y/a
(y/a — y/b)(y/a + y/b)
2Va
n
v
rr
a-b
lyfb
and thus
4o
4a
1.2.30. Let rn=^.
Then
m(6i + ... + 6n) = £ (6i + 62 + ... + bn) = ^
< ^ 6 1 + ^ 6 2 + ... + ^ 6
b\
o2
bn
n
+ ^ 6 2 + ... + ^ 6
n
= a i + ... + a n < M(6i + ... + bn).
1.2.31. T h e inequalities follow from the result in the foregoing problem and from the monotonicity of the tangent function on (0,7r/2).
1.2.32. Apply the inequality given in 1.2.30 with a* = lnc^ and bi =
Ki,
2
=
1,Z,
...jTl.
1.2.33. Note that
Y^M-W?^M--T^JM
and use the inequality proved in 1.2.30.
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Solutions. 1: Real Numbers
146
1.2.34. By the arithmetic-harmonic mean inequality (see, e.g. 1.2.3),
(x - a\) + (x - a 2 ) + ... + (x - an)
n
—
x—a\
h —
h... H
x—a,2
— ~
x—an
n
— nX ~ ( Ql + a 2 + — + ^ n )
n
The desired result follows easily.
1.2.35. Observe that
1 + ci + c 2 + ... + cn = (1 + l ) n = 2 n ,
and apply the Cauchy inequality (Problem 1.2.12) with a^ = 1 and
&& = y/ck> & = 1,2, ...,n.
1.2.36. Since
n(:)-n(;) -d 2"-'-£(;)•
the claim follows at once from the arithmetic-geometric mean inequality (Problem 1.2.3).
1.2.37. By the arithmetic-geometric mean inequality (see 1.2.3),
AVAk^<{p-l)Al
P
+ Al-\ fc = l,2,...,n,
where A0 = 0. It follows that
K - -^iAVak
+ ~((p-
= AP
*~ ^ T ^
mi+AU)
1 {kAk
~{k~
l)Ak l]
-
= ^ i «* - iMj-, - MO •
Now, adding these inequalities we get our claim.
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1.2. S o m e E l e m e n t a r y I n e q u a l i t i e s
147
1.2.38. Assume t h a t a{ = m a x { a i , a 2 , . . . , a n } . Then
n—1
2—1
i—1
n—1
n—1
/ J a ^ a fc+i = J ^ CLkdk+i + 2 ^ ak°>k+i < «2 2 ^ afc -f a^ 2 ^ afc+i
fc=l fc=i fe=l
a2
/a
\2
a2
= a 2 (o - a*) = — - (^- - di) < —.
1.2.39. One can apply t h e result in 1.2.2.
1.2.40. T h e left inequality follows from 1.2.1.
(a) Observe t h a t
i- 4
i
1 + ak = 1 - ak* < 1 - ak
Hence
n
v
y n
i
-1
n( +°*)< ( n ^ - ^ ) ) •
fe=l
Nte=l
'
Since a i + a 2 -f ... + a n < 1, applying once again the result in
1.2.1, we get
n
/
n
k=l
^
fc=l
ak
f](l + afc) < (l-J2 )
xv
- 11
—
'
'
(b) Use the same reasoning as in (a).
1 . 2 . 4 1 . Apply the inequality given in 1.2.15 (b), replacing ak
1 - ak.
by
1.2.42. Since 0 < ak < 1 for k = 1,2, ...,n, the inequality
n
fc=i
n
fe=i
n
fc=i
1
k
holds for n > 2. Now, applying the inequality from 1.2.15 (b) with
Ofc replaced by yq^j-, A; = 1, 2,..., n, we get
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Solutions. 1: Real Numbers
148
Multiplying both sides of this inequality by Yi a^ and using (1), we
fc=i
get our result.
1.2.43.
(a) By the arithmetic-geometric mean inequality (Problem 1.2.3),
f[(1 + ak)
it=i
(n + l ) n
_ 2ai -f a 2 + ... + fln o>i + 2a2 + ^3 + ... + an
~~
n+ 1
n+1
ai + fl2 + ••• + 2a n
"+1
> n^
(b) The proof of this part runs as in (a).
1.2.44. Observe first that if T j ^ - = n- 1, then J2 ^ k—1
= 1. To
k=l
get our result it is enough to apply the inequality given in 1.2.43 (b)
with a*; replaced by j ^ - .
1.2.45. [M. S. Klamkin, Amer. Math. Monthly 82(1975), 741-742]
We may assume that ai,a2, ...,a n are enumerated in such a way
that a\ = min{ai,a2, ...,a n } and a2 = max{ai,a2, . . . , a n } , and let
An = 1/n be the arithmetic mean of ai,..., a n . Define a new sequence
{a*.} by setting a^ = A n , a'2 = ai + a 2 - A n , a^ = a* for 3 < z < n.
We will show that
It follows from the definition of the sequence {a'k} that (1) is equivalent to
(l + ai)(l + a 2 ) > (l + A n ) ( l + . o i + o 2 - A n )
(1 - ai)(l - a 2 ) - (1 - 4 n ) ( l - ai - a 2 + An)'
which in turn is equivalent to
(An - ai)(An - a2) < 0.
The last inequality is an immediate consequence of our assumptions.
Now, we repeat the above procedure for the sequence {afk} to get the
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1.2. Some Elementary Inequalities
149
sequence {a'£}. At least two terms of the sequence {a'fc'} are equal
to an An. Moreover, the sequence satisfies an inequality of type (1).
If we repeat this procedure at most n — 1 times, we get the constant
sequence with each term equal to An. In view of (1),
n
l + Qfc > r r 1 + ^n
fe=i
*
=
/n + 1
v
fe=i
1.2.46. Let afcL = max{ai,a 2 , . . . , a n } . There is a fraction on the left
side of the inequality whose numerator is equal to a^. The denominator of this fraction has two terms. Let us denote the greater one by
afc2. Now, take the fraction whose numerator is a^2 and denote by
dk3 the greater of the two terms of its denominator, etc. Note that
(1)
—;
,
>7T
2-1,2,....
It follows from the above construction that there exists an / such
that dkl+1 = Q>ki- Next, observe that the numbers a^ and a>ki+l appear in our inequality as numerators of either two neighbor fractions
or two fractions which are separated by only one term (we assume
here that the last and the first are neighbor quotients). Moreover,
dki+1 appears as a numerator of a fraction that is to the right of the
fraction with the numerator a/Ci. To pass from the fraction with the
numerator a/Cl to the fraction with the numerator a/cz+1, / steps are
needed, where / > ~. Hence, by (1) and the arithmetic-geometric
mean inequality,
°*» + 2a ^ + ...
+ ^>iv/i>-.
'"'
2a
~ V2 ~ 4
2afc2
fc3
kl
l
1.2.47. We have
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Solutions. 1: Real Numbers
150
which implies the desired inequality.
1.2.48. By the arithmetic-geometric mean inequality,
ai
a\+fa
\ (
< -
a2
a2+&2
a\
h ...H
n \a\ + 6i
a
.nl
fa
V «i+6i
an
6i
n
ttn+6n
-
a n -f 6n
1
ai 4- &i
a>2-\-b2
h ...H
Q>n+bn
bn
-—
\
= 1.
an+bn)
1.2.49. [V. Ptak, Amer. Math. Monthly 102(1995), 820-821] First,
observe that if we replace each a,k by ca^ with c > 0, neither the left
side nor the right side of the inequality is changed. Therefore, we can
assume that G = 1. Then an = —. Observe now that if a\ < x < —,
then a: + ^ < ai -f ^ . Hence
n
n
1
1
Now, to obtain our claim we may apply the arithmetic-geometric
mean inequality.
1.2.50. Let us arrange all the positive divisors of n into pairs (fc, I) in
such a way that kl = n. By the arithmetic-geometric mean inequality,
^ > y/kl. Adding these inequalities, we get
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Chapter 2
http://dx.doi.org/10.1090/stml/004/05
Sequences of Real
Numbers
2.1. Monotonic Sequences
2.1.1.
(a) Let {an} be an increasing sequence bounded from above. Then
sup{a n : n G N} = A < oo. Thus for any n G N, an < A. Since
for any e > 0 the number A — e is not an upper bound of the
set {an : n G N}, there is anQ such that ano > A — e. By the
monotonicity of the sequence, A > an > A — e for any n > n$.
Hence lim an = A.
n-+oo
Assume now that {an} is not bounded above. Then for any M
there is ano such that ano > M. Again, by the monotonicity of
the sequence, an > M for n > no, and therefore lim an = +00.
n—*oo
(b) See the solution of (a).
2.1.2. We have
<
Sn—1
for
n > 2.
$n
Indeed, by 1.2.19,
(1)
4 < Sn+lSn-l151
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Solutions. 2: Sequences of Real Numbers
152
We will show that {xn} is an increasing sequence. The inequality
/
p
\ 2
xi < #2 follows from ( JZ afc )
\k=i
J
p
< P X) afc ( s e e t n e solution of
k=i
1.2.20). Assume now that xn-i < xn> Then
(2)
Sn-l<Sn^.
Hence, by (1) and (2),
> -W-T^r = * n .
2.1.3. We have a n + i = ^ a n < a n , n > 1. Therefore {an} is a
strictly decreasing sequence. Since it is bounded below (e.g. by 0),
lim an = g exists. The number g satisfies the equation g = \g.
n—•oo
Consequently, g = 0.
2.1.4. Let bn = an - ^ t r - We have 6 n + i — bn = an+i ~an -f- ^ > 0.
Hence the sequence {bn} converges, and so does { a n } .
2.1.5.
(a) We will show that the sequence {an} is monotonically decreasing
and bounded below. Indeed,
v n + l ( y n + 1 -h y n )
Moreover, by the inequality given in the hint (one can prove it
by induction), an > 2{y/n + 1 — \fn — 1) > —2.
(b) The proof follows by the same method as in (a).
2.1.6. We first show by induction that | < an < 2 for n e N and
that the sequence {an} is strictly increasing. These two facts imply
the convergence of {an}. Let g = lim an. Because an = y/3an-i — 2
we get g = y/3g — 2, and consequently, g = 2.
2.1.7. One can establish by induction that an > 2c. Of course, a\ <
ei2. Moreover, if an > a n _i, then
«n+l = K i ~ C)2 > (tt n _i - C)2 = On-
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2.1. Monotonic Sequences
153
The last inequality follows from the monotonicity of the function
f(x) =x2 on R+.
2.1.8. By the arithmetic-geometric mean inequality and by our assumptions we get
qn + ( l - a n + i )
r—r
r 1
2
> V a n ( l - an+ij > 2*
Hence an — a n +i > 0. Therefore the sequence {an} converges to a
g. Since an(l — a n +i) > | , we get g(l - g) > \. The last inequality
is equivalent to (2g — l ) 2 < 0, which gives g = \.
2.1.9. Obviously, 0 < an < 3 for n > 1. Moreover, a 2 + 1 - a 2 =
—a2 + a n + 6 > 0 for 0 < an < 3. Thus the sequence is monotonically
increasing and bounded above, so it converges. By the definition of
the sequence, lim an = 3.
n—>oo
2.1.10. We see at once that 0 < an < 1 for n > 1. To prove the
monotonicity of the sequence we will need the following form of the
principle of induction:
W(ri) is true for all natural numbers n, if the following two conditions hold:
(i) W(l) is true.
(ii) The truth of W(k) for 1 < k < n implies W(n 4- 1) is also
true.
Assume now that a n _i > a n _2
and
an > an-\.
Then
a n + i - an = - ( a n - a n _i + a^_x - o?n_2) > 0.
Therefore the sequence is convergent. Let g denote its limit. Then
g = | ( 1 + g + g3). Consequently,
^ = 1 or
g=
- l + >/5
or
g=
-l-v7^
.
Observe that all the terms of the sequence are nonnegative and less
than ^ ± V I . Thus lim an = =±±&.
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Solutions. 2: Sequences of Real Numbers
154
2.1.11. We have a n + i = anj£f3
solution of 2.1.3) we get g — 0.
< a>n, n > 1. Therefore (see the
2.1.12. Since a n + i = « n | ^ f < ^ n , n > 1, the sequence is monotonically decreasing. It is bounded below by zero, so it converges.
2.1.13.
(a) Clearly, {an} is monotonically increasing. We will show that it
is also bounded above. Indeed,
1 1
1 , 1 1
1
2 + ^7
an = 1 + 2^7
- — + 2--—
3 2 + ... + n- 2^ < 1 + 1-2
3 + ... + ( n - l ) n
(b) Obviously, {an} is monotonically increasing. Moreover,
1
1
!
1
,
1
1
1
Hence it follows from (a) that the sequence is bounded above.
2.1.14. For n > 1, we have
a n + i — an =
.
:H .
y/n(n + l) ^2n(2n
= H .
: < 0.
+ 1) y/(2n + l)(2n + 2)
Hence the sequence is convergent, as it is monotonically decreasing
and bounded below.
2.1.15. From the arithmetic-geometric mean inequality we get
a n +i > {Upn
V
1
-^zT = ^
n > 1.
An
Therefore
a
n
&
a£ — a
a n + i - an =
+ ——r
r r r < 0)
™ > 2,
which shows that the sequence converges and lim a n = tfa.
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2.1. Monotonic Sequences
155
2.1.16. Clearly, 0 < an < 2 for n > 1. Moreover,
a
n+i ~ a n = \/^n - y/^n-i
> 0 provided
a n > a n _i.
Hence the sequence converges to a # which satisfies the equation
9 = V2 + y/9R e m a r k . Using Cardan's formula for real roots of cubic polynomials
one can show that
1 / 3 / 1 ^
. .
f^TZ.
.
3/I.
9 = o I V o (79 + 3 ^ 2 4 9 ) + V 2 (79 ~ 3 ^ 2 4 9 ) "
X
)•
2.1.17. Note that a n +i = 2 (2 - ^ 3 ) , n > 1. Now one can
establish by induction that 0 < an < 2, n > 1. Moreover,
(qw + l ) ( q w - 2 )
a n 4- 3
Hence the sequence converges and lim an = 2.
n—>-oo
2.1.18. One can show by induction that the sequence {a n } is strictly
increasing. If it were bounded above then there would exist a number
g such that g — lim an. We would also have g2 - 2g -f c = 0.
n—>-oo
This equation has a real solution provided c < 1. So, assume that
0 < c < 1. Then the sequence {a n } is bounded above by 1 — y/1 — c,
and lim a n = 1 — y/Y^~c.
n—>-oo
For c > 1, the sequence is strictly increasing and it does not
converge. So, it diverges to +00.
2.1.19. Since
an+l = o ^ l - 2 ^ - ^ j ,
n>l,
we get
if a n > y/a,
then
a n + i < an,
if an < y/a,
then
a n + i > an,
if a n = y/a,
then
a n +i = \/&-
Observe now that
an-^
> y/a
if and only if
(q n - y/a)3 > 0,
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Solutions. 2: Sequences of Real Numbers
156
which in turn is equivalent to an > y/a. Finally,
if 0 < a\ < y/a, then {an} is increasing and bounded above by y/a;
if a\ > y/a, then {an} is decreasing and bounded below by y/a;
if ai = y/a, then {an} is a constant sequence.
In each of the above cases the sequence converges to y/a.
2.1.20. One can show by induction that
(3n-l _
n
X)
_ (3n-l _
n
3
)
(3 - 1) - (3 - 3)ai
a i
for
n — 1,2,3,....
Therefore the sequence is not defined for a\ = ^n+iZl with n £ N.
Moreover, if a\ — 1, then an = 1 for n = 1,2,3,.... For other values
of a\, the sequence converges to 1/3.
2.1.21. We have a n + i = (a n - a) 2 + a n > a n for n > 1. Hence the
sequence is monotonically increasing. Moreover, if it converges, then
lim an = a. Therefore if ai > a, then the given sequence diverges. In
n—KX)
the case where a — 1 < a\ < a, we have also a — 1 < an < a for n > 1.
Thus for such a\ the sequence converges. Finally, if a\ < a - 1, then
a2 > a, and consequently, the sequence diverges.
2.1.22. It is obvious that the sequence may converge either to a or
to b. We will consider the following cases.
1°
Ob.
Then a 2 = ca\lb > c = a\ and by induction a n + i > an. Hence
lim an = -hoo.
n—»oo
2° c = 6.
Obviously, an = b for n = 1, 2,3,....
3° a < c < 6 .
One can establish inductively that the sequence {a n } is monotonically decreasing and bounded below by a. Hence lim an = a.
n—>oc
4° c = o.
Clearly, an = a for n = 1,2,3,....
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2.1. Monotonic Sequences
157
5° 0 < c < a.
Induction is used once again to show that {an} is monotonically increasing and bounded above by a. It then follows that
lim an — a.
n—»oo
2.1.23. Note that a n + 1 = 6 (l - ^ p f ) for
tion
if ai < 2, then an < 2,
if ai > 2, then an > 2,
Moreover,
(a n + 3)(a n
a
n + l ~ 0>n —
n G N. Hence by inducn G N;
n G N.
- 2)
~~Z
a n -f 7
•
Therefore
l ° i f 0 < a i < 2 , then the sequence is increasing and bounded above
by 2, and lim an — 2,
n—KX>
2° if a\ > 2, then the sequence is decreasing and bounded below by
2, and lim an = 2,
n-^oo
3° if ax = 2, then an = 2 for n G N.
2.1.24. Since 0 = ai < a2 and a^ +1 — a^ = an — a n - i , we see by
induction that a n +i > &n for n G N. The sequence is bounded above,
e.g. by \/l + 4c. One can easily establish that lim a n = 1 + v l 1 + 4 c .
n—+oo
7
2.1.25. Since a 2 = v 2 \ / 2 > v ^ = ai and a ^ + 1 - a ^ = 2(o n —a n _i),
one can show inductively that o n +i > a n for all positive integers.
The sequence is bounded above by 2, and lim an — 2.
n—• oo
n
2.1.26. For /c = 1, we get an = 5 , n G N, and therefore {a n }
diverges to -hoc.
For k> 1,
a 2 = y 5\/5 > \/5 = ai and a^ +1 - a£ = 5(a n - a n _ i ) .
It then follows by induction that {an} is strictly increasing. Moreover, an < k~\f5 for n G N. One can easily verify that lim an =
n—KX>
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Solutions. 2: Sequences of Real Numbers
158
2.1.27. We see (by induction) that 1 < an < 2, n e N. The
monotonicity of the sequence follows from the equality
3(a n - a n _i). Hence for 1 < a\ < 2 the sequence is monotonically
increasing and its limit is 2. On the other hand, if a\ = 1 or a\ = 2,
the sequence is constant.
2.1.28.
(a) We have a\ < a2 and an+1—an = an — a n _ i . It follows by induction that the sequence is monotonically increasing and bounded
above by c. Obviously, lim an = c.
n—>oo
(b) Since 62 = y/cy/c > \fc = 6i and bn+l - bn = c(6 n - 6 n _i),
using induction we conclude that the sequence is monotonically
increasing and bounded above by c, which is its limit.
2.1.29. One can establish by induction that 0 < an < 6, and next
prove that the sequence is strictly increasing. Its limit is equal to b.
2.1.30. The sequence is strictly increasing and bounded above, e.g.
by 3. Its limit is ^ ± ^ .
2.1.31. We have a\ < a2 < 03. Moreover, we see that for any n € N,
if
an < a n + i < a n + 2 ,
then
an+2 < an+3.
It then follows from the principle of induction stated in the solution
of Problem 2.1.10 that the sequence {an} is strictly increasing. It is
also bounded above by 4, and lim an = 4.
n—KX>
2.1.32. As in the solution of the foregoing problem, one can show
that the sequence {an} is monotonically decreasing, bounded below
by 4, and lim an = 4.
n—+00
2.1.33. By the arithmetic-geometric mean inequality, an > bn. Thus
an+i =
< an,
n G N.
This means that the sequence {an} is decreasing. On the other hand,
the sequence {bn} is increasing because
bn+i = yjbnan > \J\?n = bn,
nE N.
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2.1. Monotonic Sequences
159
Moreover, b\ < an, bn < a\. Therefore both sequences converge. Let
a = lim an and (5 = lim bn. Passage to the limit as n —•> oc in
n—>oo
an+i =
Qr| bn
^
n—»oo
gives a = ^ ^ or, in other words, a = f3.
2.1.34. Since 2(a£ + 6^) > (a n + 6 n ) 2 , we get an > 6 n , n G N.
Therefore
—r~ S
—7— — am n e J^i,
a n + bn
an + on
which means that the sequence {an} is decreasing.
a n +i —
In much the same way we show that {bn} increases. Moreover,
we see that b\ < a n , bn < ai, and consequently, both sequences
converge.
Let a = lim a n , (3 = lim bn. Letting n —• 00 in 6 n + i =
n—>oo
an bn
n—>oo
t
,
we obtain f3 — ^ ^ , or a — (3.
2.1.35. By the arithmetic-harmonic mean inequality, an > bn- Hence
an+i =
< an,
ne N,
which means that {a n } decreases. On the other hand, {bn} increases
because
2a n 6 n
fe
"+i =
TT" ^ &n, ™ e N.
Moreover, b\ < a n , 6 n < ai, and therefore the sequences converge.
Let a — lim a n , (3 — lim 6 n . Passing to the limit in the equan—>-oo
n—>oo
tion a n + i - ^ ^ y i e l d s a = H2^- T h u s a = &•
Note also that a n + i 6 n + i = a n 6 n , which means that all the terms
of {anbn} are equal to a\b\. It follows that a = /3 = \/ai6i.
2.1.36. We have a n + i = 2(n+\) ( Qn + *) >
G n + 1 ~" ar\
n G
^- Consequently,
- n o n + (n + 2)
2(n+l)
Now applying the inequality nan > n -f 2 for n > 4 (which can be
established by induction), we see that the sequence is monotonically
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Solutions. 2: Sequences of Real Numbers
160
decreasing and hence is convergent. Put a = lim an. Passage to the
n—>oo
limit in the equation a n + i = 2{'n+i) ( Qn ~*~ ^) g i y e s a = 1. •
2.1.37. It follows from the inequality a n + 2 < ^ a n + i + §a n that
an+2 + |«n+i < Q>n+i + f On- Hence the sequence 6 n = a n + 1 + | o n is
decreasing, bounded, and therefore convergent. Let b be its limit. We
will show that {an} converges to a = | 6 . Let e > 0 be arbitrarily
fixed. Then there exists no E N such that | > \bn — b\ for n > no.
Consequently,
2
2
> |a n +i - a| - - | a n - a|
5
6>
for
n > n0.
Thus |a n +i - a| < | | a n — a| + | . We can see by induction that
+- + | + 1 )g
|«n,+*-«|<(f) K - « l + ( ( | )
k
k
*G)*
no^«i+
1
(2\
l_-^fef2
V
/o\fc
, _ V 6 < (3]
l«"0-«l+2-
vfc
Since ( | ) |a„ 0 — o| < | for sufficiently large k, \an — a\ < e for n
large enough.
2.1.38.
,71+1
(a) 6 n = ( l + I ) n + 1 = (l + I ) a n > a n .
(b) By the geometric-arithmetic mean inequality G n +i < An+\
Problem 1.2.3) with a\ = 1, a2 = 03 = ... = a n +i = 1 + - ,
1 + -1
ny
Hence
D'<0
1 \
n/
n
/
\
<1 +
1
\
(see
n+ 1
n
+i
n+1/
(c) By the harmonic-geometric mean inequality Hn+i < G n +i, n > 1
(see Problem 1.2.3) with ai = 1, a2 = a 3 = ... = a n + i = 1 + ^ - ,
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2.1. Monotonic Sequences
161
which in turn gives bn < 6 n _i, n > 1.
To show that both sequences {an} and {bn} converge it is
enough to observe that a\ < an < bn < &i, n G N. Moreover,
lim bn= lim (l + ^-)a n = lim an.
2.1.39.
(a) By the geometric-arithmetic mean inequality G n +i < An+i (see
Problem 1.2.3) with a\ — 1, a2 = a 3 = ... = a n + i = 1 + ^, n G N,
we see that the sequence is strictly increasing.
If 0 < x < 1, then by the preceding problem,
If x > 1, then there exists a positive integer no such that x < no.
Consequently, the monotonicity of the sequence { (l + ^ ) } and
the result stated in the foregoing problem imply
(1 + !)"<( 1 + !!o)" < ( 1 + J!Lr <e r
V
n/
\
n /
\
nony
(b) It is enough to apply the same reasoning as in (a) and observe
that for x < 0, the sequence is bounded above, e.g. by 1.
2.1.40. Applying the geometric-harmonic mean inequality G n +/+i >
Hn+i+i (see Problem 1.2.3) with a\ = 1, a,2 = a^ = ... = a n +j+i =
1 + *, we get
n+i+J(
,x\n+l
VV
nJ
1
,
+
x{n + l)
n 2 + nl + x + n
s(n + Q
(n + l)(n + /)*
This shows that bn > 6n+i> n G N.
2.1.41. By the inequality given in the hint,
a n + i - an =
log
> 0,
n 1
n
i rc + 1 < 0.
„
6 n + i - bn = — — - log
n+1
n
Clearly, a\ < an < bn < 6i, n G N, and consequently, both sequences
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converge
(to the same limit).
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Solutions. 2: Sequences of Real Numbers
162
2.1.42. Monotonicity and boundedness of the sequence {an} are
easily verified. It follows from the equality an+l = an that its limit
is 1. We now show the monotonicity of {cn}. Assume first that x > 1.
Then
cn = 2n(an - 1) = 2 n (a£ + 1 _ 1 } = 2n{an+l
- l)(an+1 +1)
_ on+l/,,
Qn+i + 1
—^
\P>n+l — J-)
7,
> c
n+l-
This means that for x > 1, the sequence {cn} is strictly decreasing.
The same reasoning applies to the case 0 < x < 1. For x = 1,
the sequence is constant. The monotonicity of {d n } can be proved
analogously.
For x > 1, the sequence {c n } converges (because it is monotonically decreasing and bounded below by 0). On the other hand,
for 0 < x < 1, the sequence {dn} is monotonically increasing and
bounded above by 0. Now, it follows from the equality
that both sequences tend to same limit for all positive x different from
1. If x = 1, then c n = dn = 0.
2.2. Limits. Properties of Convergent Sequences
2.2.1.
(a) 1.
(b) 1.
(c) -1.
(d) We have
0 < (>/2 - yf2){y/2 - y/2) • ... • (V2 -
2n+
\/2) < (>/2 - l ) n .
Thus the limit of the sequence is equal to 0.
(e) We
willAmerican
first show
that the sequence
an — ^ converges
toPlease
zero. report unauth
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We have a n + i = a n ^ ^ < an for n > 3. Therefore the
2.2. Limits. Properties of Convergent Sequences
163
sequence is monotonically decreasing. Clearly, it is bounded below by zero. Hence it is a convergent sequence and its limit g
satisfies the equation g — \g. Thus g = 0. We will now find
the limit of our sequence. To this end, set kn = [y/n]. Then
kn < \fn> < kn + 1, which gives
u
2 fc - +1
^ 2v^
'
Therefore the limit of the given sequence is equal to zero.
(f) Let an = ^ - . Then a n + i = an\^$< a n , n € N, which
implies (see the solution of Problem 2.1.3) that g = 0.
(g) Set
1 / 1
x/nVx/l + V^
Then
an =
1
1
V^+v^
wmcn
^s
V / 2fc^I-v/2fc+T
o
2 AT *'
_
'"
a
1
>/2n - 1 + y/2n + 1
consequence of the equality
1.
_
y/2
(h) It follows from the inequalities
(1 + 2 +
'--
x
+ n)
1
1
2
+
+
+
n
^T^^^TT ^T^ - ^T^
and from the squeeze law that the limit is \.
(i) As in (h) we show that this limit is also equal to \.
2.2.2. Set an = j ^ . Then ^
= {^Y
^i- Moreover, we
have lim (n^)
^py — ^pi- Consequently, the sequence {an} is
monotonically decreasing beginning with some value no of the index
n. It is also bounded below, e.g. by zero. Its limit g satisfies the
equality g = —py£- Therefore g = 0.
2.2.3. We have
0 < (n + l)a -na
=na (f 1 + - )
-1
n
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1
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l-ot'
Solutions. 2: Sequences of Real Numbers
164
Thus the limit of the sequence is equal to zero.
2.2.4. Let a = £ , with p e Z and q € N. For n > q the number
n!a7r is a multiple of 7r, which means that the terms of the sequence,
beginning with some value no of the index n, are all equal to zero.
2.2.5. If the limit existed then we would get
0 = lim (sin(n + 2) — sin n) = 2 sin 1 lim cos(n -I-1),
n—>oo
n—+oo
and consequently, lim cos n = 0. Similarly,
71—+00
0 = lim (cos(n -f 2) — cosn) = —2 sin 1 lim sin(n 4-1),
n—>oo
n—»oo
2
2
which is impossible because sin n + cos n = 1, n G N. Therefore the
limit lim sinn does not exist.
n—•oo
2.2.6. See the solution of the foregoing problem.
2.2.7. We have
l i m - ( ( a + - J + [ a + - J + ... + ( a + ^ — - ) |
n->oo n l \
n/
\
n/
\
n J J
2
(n-l
n
(
n
l
)
,
1
+
2
+
...
+
(n
- }1 ) 2 \
v
2
= hm
a +
a
+
^—>
—
9
n-^oo \ n
nl
n6
)
2
1
= a24-a+-.
o
The last equality follows from the fact that l 2 + 2 2 -f ... -f n2 =
n(n+l)(2n+l)
6
2.2.8. We have
an + a 2 + ... + akn - k = (a n - 1) + (a 2 - 1) + ... + (a* - 1).
Moreover,
a1 - 1
Hm _ n — = / for/ = 1,2,...,fc.
n-^oo a n - 1
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Therefore
the limit is
equal to 1 Society.
+ 2 + ...Duplication
+ fc = M^tll.
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2.2. Limits. P r o p e r t i e s of Convergent Sequences
165
2.2.9. Using the equality
1
k(fc + l)(fc + 2)
1
2 k
1 1 1 1
,
/c + 1 + 2 / c + 2
fcGN,
one can show that the limit is equal to \.
2.2.10. Since
k3-l
A;3 + l
we get
(k-l)((k
+ l)2-(k
+ l) + l)
2
{k + l){k - k + 1)
A fc3-l _ 2 n2 + n + 1 _ _
1 1 fc3 + 1 _ 33 ' „2.
n2 + n n-*<x> 3
fc=2
2.2.11.
I.
2.2.12. Since 1 - ^
= ^ g ^ , we obtain
(n + 1) • (n + 2)/
1 n+ 3
3 n + 1 n-oo 3 '
2.2.13. We have
k3 + 6k2 + Ilk + 5 = (k + l)(fc -f 2)(fe + 3) - 1.
Hence
,. ^ / c 3 + 6/c2 + ll/c + 5
hm > n — • o o *—*J
(fc + 3)!
fc=l
™^U
!
(£ + 3)!^ 3*
2.2.14. Observe that
2
2
1-x *
l-x *"
1
1
l-x2fc
for k = l,2,...,n.
Therefore
lim
n
ofc-i
X2
Y-
(
= <
X
1-X
for
|x| < 1,
1
for Id > 1.
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n—*oo fc=l
*—^ 1
Solutions. 2: Sequences of Real Numbers
166
2.2.15. For x ^ 1,
(1 - x){l + x)(l + x 2 ) •... • (1 + x2n) __ 1 - x2n+1
1-x '
1-x
and consequently,
XZ
1 -
1-x
2n+1
Finally,
-oo
lim an = <
1
1-x
I +0 0
for
n = 0,1,..., x ^ 1,
for
n = 0,1,..., x = 1.
for
x < — 1,
for
x = — 1,
for
|x| < 1,
for
x > 1.
2.2.16. For x ^ 1,
rr (A a.
^
A (x2fc + i) 2
^
_ (re + !)(* - l)(x + l)(x 2 + 1) •... • Or2" + 1)
( x - l ) ( x 2 n + 1 + l)
-1
x - 1 x 2 " + 1 + 1'
X+ 1
XZ
Hence
for
|x| < 1,
for
|x| > 1,
0
for
x = —1,
I +oo
for
x = 1.
lim an = {
n—>oo
x+1
"x-1
x+1
X - 1
2.2.17. Let x be different from 1. Then
z
1 + x^ + x '* =
fc
fc
fc
j _X^2-3
2 3 U^33
( 1 + X„33 +
" )(x
-_ 1 n)
3
x *-l
k+1
„3
X3
__ i
k
r* - 1 '
Thus
on+l
3
3fc
2 Society.
3fc
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X"
n
i
x
x
)
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(
+
+
=
Thank You!
x3-l
k=l
2.2. Limits. Properties of Convergent Sequences
167
Let g denote the limit of the sequence. Then
1
1-x3
9= \ +oo
1
-hoc
for
\x\ < 1,
for
|x| > 1,
for
x — — 1,
for
x — 1.
2.2.18. Clearly, k • k\ = [k + 1)! - k\, keN.
hm
n-*oc
1-1! 4- 2-2! + ... + n-n\
-t
—
(
n
Hence
(n-fl)l-l
,
lt
= hm —
—— = 1.
+l)|
n-^oo
(jl + l j !
2.2.19. Note first that the problem is meaningful for x ^ 0. By 2.2.3,
the denominator nx — (n — l ) x tends to zero if 0 < x < 1. Moreover,
if x < 0, then the denominator also tends to zero. For x = 1 it
equals 1. Therefore the sequence diverges to infinity (+oo or -oo) for
x < 1, x ^ O . Now let x > 1 and set k = [x]. Then k > 1 and
fc+i
i - u - i y < i - (i-i
n
It follows from these inequalities that there exist a and /3 such that
a < n (1- (1
n
<(3,
which in turn gives
an
x-l
< nx
1-
1 -
n
<(3n x-l
Hence if x — 1 < 1999, then the sequence diverges to +oo. If x — 1 >
1999, the sequence converges to zero. Now let x = 2000. By the
binomial formula,
,1999
n ! ™ ) n 2000
_ (n _
1
1 )2000
2000'
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Solutions. 2: Sequences of Real Numbers
168
2.2.20. We have
a„. = < n + 1
a
n
if
a > 6,
if
a = b.
Hence lim an = a.
2.2.21. It can be shown by induction that an = (n — l ) 2 . Consequently, lim an = +oo.
ab
2.2.22. We show by induction that an =
Va2+nb2
n-1
2.2.23. One can show that an = 1 — (j)
. Thus lim an = 0.
. Therefore lim a n = 1.
2.2.24. It is easy to verify that an+1 = l + 6+... + 6 n - 1 +bna. Hence
1
an+i = < 1 - 6 +
n+a
f
1
a- 1-6
6n
for
6^1,
for
6=1.
Thus if 6 = 1, a G l , the sequence diverges to -f-oo. If 6 ^ 1 and
a = j3£, the sequence converges to j ^ . In the case a ^ y ^ and
|6| < 1, it also converges to y ^ . In the remaining cases the sequence
is divergent. Namely, if 6 < — 1 and a ^ j ^ , the sequence has
neither finite nor infinite limit. If 6 > 1 and a > y ^ , the sequence
diverges properly to +oo. Finally, if 6 > 1 and a < y ^ , the sequence
diverges properly to — oo.
2.2.25. The formula for the nth term of the Fibonacci sequence can
be proved by induction. We may assume that a > (3. Then a = 1 ^ 5
and (3 = ^ = ^ . Moreover,
afl
< \Jotn - (3n < CM/1 +
Since2000
limAmerican
= 0,
we get limSociety.
tfa^ =
a.
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2.2. Limits. Properties of Convergent Sequences
169
2.2.26. Note first that 6 n + i = Q n+ 3 V From this a n + i - bn+1 =
\{an — bn), which means that the sequence {an -bn} is a geometric
progression with the ratio \ . Hence this sequence converges to zero.
Now it is enough to show that the sequence {an} converges. Assume
first that a < b. Then {an} monotonically increases and an <bn < b.
Therefore it converges. It follows from the above that {bn} also
converges and lim an = lim bn. The same reasoning applies to the
n—>-oo
case a > b.
n—+oo
2.2.27. We have
n
n digits
digits
a + aa + . . . + aa...a — a(l + 11 + ... + 11...1)
= a(10 n " 1 + 2 • 10 n ~ 2 + ... + n • 10°)
= a((l + 10 + ... + 10 n _ 1 ) + (1 + 10 + ... + 10 n " 2 )
+ ... + (l + 10) + l)
= a
/10n-l
lO71-1-!
V 9
9
n
10(10 - 1) - 9n
81
•
Therefore the limit is
102 - 1
9
109
^ .
2.2.28. Note that the sequence with terms >j/n, n > 3, is monotonically decreasing and its limit is 1. Now it is easy to check that
($/n-l)n < f i J
neN.
l)n = 0.
Thus lim (tfn2.2.29. Since
for
lim an = 0, beginning with some value n 0 of the
n—»oo
index n, |a n | n < ( ^ ) n . Consequently, lim a™ = 0.
n—•oo
2.2.30. Let max{ai,a2, ...,a/c} = a/. Dividing the denominator and
numerator by af we show that
lim
n-^oo
Pia x
+ p2a2
+ ... + VkQ>k
±—
^-^— = a/.
p i a y + p 2 ^ 2 + "• + PfcGfc
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Solutions. 2: Sequences of Real Numbers
170
2.2.31.
(a) Let £ > 0 be so small that q + e < 1. Then there exists no £ N
such that
I a n +i
< q + ^ for n > no.
Hence
K l < (<? + e ) n - n o | a n o | ,
n>n0.
This implies lim \an\ — 0, that is, lim an = 0.
n—>-oo
ra—•oo
(b) Let £ > 0 be so small that q — £ > 1. Then, beginning with
some value n\ of the index n, \an\ > (q — £)n~ni\ani\.
Since
lim (q
-hoc, we get lim \an\ = -f oo.
2.2.32.
(a) Take £ > 0 small enough to get q + £ < 1. Then there exists
n 0 G N such that |o n | < (<? + £) n , n > n 0 . Therefore lim a n — 0.
n—•oo
(b) We have |a n | > (q — «s)n for n > n\. If e > 0 is small enough,
then q — £ > 1 and therefore lim (q — £)n = -hoo. So, lim \an\ —
2.2.33. Setting an = naxn,
lim
n—>oo
n—• oo
n—>-oo
+ 00.
«n+i
a
n
we get
,.
(n + 1
lim
n—>oo \
n—>oo
?7,
:r = x,
where
0 < x < 1.
Hence, by Problem 2.2.31, the sequence tends to zero.
2.2.34. Let an denote the nth term of the sequence. Then
0"n+l
m —n
n+ 1
By Problem 2.2.31 the sequence converges to zero.
2.2.35. Assume that \bn\ < M for n G N. Since lim an = 0, for
n—+oo
any £ > 0 there exists n 0 G N such that \an\ < -fj for n > n^.
Hence
|a n 6 n | < £
for n > noCopyright
2000 American
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2.2. Limits. Properties of Convergent Sequences
171
2.2.36. Without loss of generality we can assume that a < b. Suppose first that a < b. Let e > 0 be so small that a + e < b - e. By
the definition of the limit of a sequence, an < a + e < b — e < bn for
n sufficiently large. Hence max{a n , bn} — bn, and consequently,
lim max{a n , bn} = lim bn = b = max{a, 6}.
n—+ 00
n—->oo
If a = 6, then for any e > 0 there exists no such that for n > no,
the inequalities |a n — a\ < e and |6 n — a\ < £ hold. This means that
| max{a n , bn} — a\ < e.
In this way we have proved that
lim max{a n , bn\ — max{a, b).
n—^oo
2.2.37. Since lim an = 0, for any e € (0,1) we have
v7! ~ £ < v7! + « n < v7! + £
for n sufficiently large.
This implies that lim tfl -f a n = 1.
n—»oo
2.2.38. Put x n = f/1 -f a n . It follows from the foregoing problem
that lim # n = 1. Consequently,
r
lim
yrr^-i
an
xn-i
= n->oo
lim Xn
—=r~ I
Xn -
1
lim
— o o ( X n _ i ) ( a £ - i + ... + i)
1
p"
2.2.39. By Problem 1.2.1,
n^l+fll+02 + - + ^ - l l
(1)
^(Vo^)('^)--('+?)-0
'
= Y ( n + a i ) ( n + a 2 ) •... • (n + ap) - n.
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Solutions. 2: Sequences of Real Numbers
172
Moreover, by 1.2.4 we get
^)(-?)--(^H
(2)
V
i + ... + aP
V
QI
~
J2 aiaj
i<j
a
+ ... + q p
p
i<j
np
+ ...+
ai • ... • a p
pnjP - 1
Combining (1) and (2) with the result in the foregoing problem, we
show that the limit is Q l + a 2 + - + Q p .
p
2.2.40. Note that
n+1
VVi2 + n + 1
1
2
\Ai + l
1
+ \/n 2
n-fl
<
:+... + • 2
+ 2
Vn + n + 1 ~ \/n 2 + l"
This and the squeeze law for sequences imply that the limit is 1.
2.2.41. Let a denote the largest of the numbers ai,a2,...,%,. Then
a
. J a ? + o5 + ... + ay
<a.
<
p
^/p
V
By the squeeze law for sequences,
lim
/a? + a£ + ... + a£
n—KX)
a = max{ai, a2, •-., a p } .
2.2.42. Since
1 < A72si
:sin
,1999
n+1
+ cos2
7 1999
n+ 1
< v^2,
it follows that
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2.2. Limits. Properties of Convergent Sequences
173
2.2.43. We will apply the squeeze law for sequences. We have
1 < (1 4 n ( l 4- cosn)) 2 n + n s i n n < (1 4 2n) 2 n + n s i n n .
We will now show that
lim (1 4 2n)2™+"sin^ = 1.
(*)
n—>oo
Indeed,
1 <
(l +
2 n ) 2 - + nsinn
<
(\
+ 2n)"
.
Hence (*) follows from the squeeze law. Thus the limit we are looking
for is 1.
2.2.44. By the harmonic-geometric-arithmetic mean inequality (see
1.2.3), for x > - 1 we have
Now, putting x = ^ , k = 1,2, ...,n, and adding the obtained inequalities, we get
fc=l
^
n2
fc=l
\
/
/c=l
Moreover,
fc _ n(n 4-1)
E 2n
2
An2
1
n->oo 4
and
Z-/ 2 4 -%
^
2n 2 4- A; ~ 2n 2 4- n ^
2(2n 2 4 n) n-oo 4'
Therefore, by (*) and the squeeze law,
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Solutions. 2: Sequences of Real Numbers
174
2.2.45. One can apply reasoning analogous to that used in the solution of the preceding problem. Let x > — 1. By the harmonicgeometric-arithmetic mean inequality,
l + ^ V
= -T-
i
<{/(l + x ) l . K
1 + 3 ?
1 + 1
t
= l + |.
Substituting x = £ j , we get
Moreover,
^Jf_
_ n(n + l)(2n + l)
^ 1
^ 3n 3 "
18n 3
n"^> 9
and
E 3+ 24
n3
V^
=
^
3
l
>
2
3
^ 3n + 2/c ~ 3n + 2n
_ n(n + l)(2n + l)
1
6(3n 3 + 2n 2 )
;
V ^ ju2
2
^
By the above, together with (*) and the squeeze law,
J!?-t(f+3- 1 )-52.2.46. Clearly, lim ?/ak = 1 for k = 1, 2, ...,p. So we find that
r-Y
lim f I V /Ok]
= 1.
2.2.47. For sufficiently large no and for n > no, we have 0 < a +
1 < a + -±- < 1. Thus
i-l
/
.
N
k
y
lim > [ a + - ) = lim
-, V r = ;
•
n—KX)
' " ^ jfc=0
^V
nj
» - » l - a + i
1-a
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2.2. Limits. Properties of Convergent Sequences
175
2.2.48. The equality is obvious for x = 1. Assume now that x > 1.
To calculate the limit, we will apply the squeeze law for sequences.
We have
0<(v/^-l)2=v/^-2v/S+l.
Hence
(2^-i)"<(v / ^r=x 2 .
(*)
Moreover,
yx Vx*
Now, by the Bernoulli inequality we get
(**)
{2tfc-l)n>x2[l+n
•fx
Vx*
xMl-n^-1*5
Also, by the Bernoulli inequality,
x = (^x-l + l)n>l + n ( ^ - l ) > n ( ^ - l ) .
Consequently,
Therefore, by (**),
(***)
(2^-l)
n
>x2
(l
nvr
Combining (*) and (* * *) with the squeeze law, we see that
lim (2^x-l)n
= x2.
2.2.49. As in the solution of the foregoing problem, we may establish
the inequalities
1>
(2^-ir>1_n(^-Da
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Solutions. 2: Sequences of Real Numbers
176
Now, it is enough to show that
hm n——,
= 0.
n—+oc
To this end, note that for n > 3,
n
=
( ^ - l+ l ) " > n ( n - 1 3 ) | ( n - 2 ) ( ^ - l ) 3 .
Hence
0<n(^-l)2<n(
So, lim n( ^/n - l ) 2 = 0.
2
il
?
V(n-l)(n-2)
n—>oo
2.2.50.
(a) We have
|a n+fc - an\ =
arctan(n + 1)
arctan(n + k)
2 ^
h ... H
^ ^
TV (
1
2 I 2n_,_1
1
\
2n+fc /
"'
7T
2n+1'
For arbitrarily fixed e > 0, let n 0 = [log2 f — 1]. Then for any
k £ N and n > no we get |an+fc — an\ < e. Therefore {a n } is a
Cauchy sequence.
(b) One can show by induction that 4 n > n 4 for all n > 5. Hence
1
n
1
|a n +/c ~ a n < 7—TTT?
(n+1)2 +
^ 7—TT^J
(n + 2) 2 + ••• + (n + fc)5
Consequently,
\an+k
— an\
1
1
1
fc-1
1
n +fc—
< -n7( n +-7^
^T + - + (n +
l ) + 7( n + l 7T7
) ( n + 2)
=
1
n
1
1
n+1
1
1
1
n+ 1 n + 2
1
n
n + fc n
"'
+ fc)
1
n + fc
r < - <€
for any fc €N and n > [^].
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2.2. Limits. Properties of Convergent Sequences
177
(c) Note that
1 2n
1
2n
n|
1
2n-l
1
n +1 "
^ 1 1
2n
2
This means that {an} is not a Cauchy sequence,
(d) We have
\an+k
~ &n
( — D^+^-i
n-\-k-2
(-i)i
("l)n
f
+
+
(n + k){n + A ; + l ) ~ ( n - f A ; - l)(n + k) '" (n + l)(n + 2)
1
1
,
,
1
<
(n + fc)(n + fc + l) (n + f c - - l ) ( n +&) '" (n ++ l)(n + 2)
1
1
1
1
1
n + A; n + fe + 1 n + A; — 1 n-f A+
: ...+n + 1 rc + 2
1
1
1
n+1
n + k + 1 < n + 1r < e
for any fc £ N and n > [| — 1].
(e) We have
|On+fc - 0 „ | < M ( | 9 | " + f c + l ^ r + f c - 1 + ... +
= M('M"
+1
(i-i#)^<
i-M
for any k € N and n > no =
M
\q\n+i)
i-M
kr+1 < e
'in *=£"£_!
T^k T
(f) We have
2n
2n - 1
n+ 1
,„ „2 ++ ••• ++
(2n+l) 2 ++ (2n)
' " ( n + 2) 2
2n
2n 2
>n
2 >
(2n + l ) - (3n) 2
9'
&2n
—
&„ —
Therefore {o n } is not a Cauchy sequence.
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Solutions. 2: Sequences of Real Numbers
178
2.2.51. By the given condition,
\<Ln+k ~ an\ = |&n+/c
_
Q-n+k-l
< A ( | a n + f c - l - an+k-2\
fc
1
—
+ Gn+fc-1
<^n+/c-2 + ••• "f ^ n + 1
—
&n|
+ l^n+fc-2 ~ « n + f c - 3 | + ••• + \d>n ~ ^ n - l | )
2
< (A + A^- + ... + A -f A)|a n - a n _ i |
< (Afc + A ^ 1 + ... + A2 + A)A n ~ 2 |a 2 - oi|
vn-1
\0>2 - Oil <
1-A
1 - A a 2 - ai .
Hence for arbitrarily fixed e > 0, for n >
In A
, and for any
k £ N, we have |an+fc - a n | < £•
2.2.52. Since {5 n } is convergent, it is a Cauchy sequence. We will
show that {lncrn} is also a Cauchy sequence. By the inequality in
2.1.41,
In crn+k - In an = In I 1 +
2_
&n+k J
1
<
+ ... +
1
+ ... + In 1 + a
V
n+l
<e
for A; 6 N and for sufficiently large n.
2.2.53. By the result in 1.1.23,
= (#n+fc
—
= (-!)"(
-Rn+fc-l) + ( - R n + f c - l ~ R n + k - 2 ) + ••• +{Rn+l
-l)fe-2
1
(
^
1
Qn+k-iqn+k
+-
Qn+k-2Qn+k-l
~
Rn)
• + ...- <7n+ltfn+2- +9-n ^fo+1/
Hence, by the monotonicity of the sequence {qn} and the fact that
qn>n (see the solution of 1.1.24),
\Rn+k — Rn\ <
1
qn+lQn
nz
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2.2. Limits. Properties of Convergent Sequences
179
2.2.54. Let d denote the common difference of the given progression.
Assume first that d ^ 0. Then
1
Hence
i.
hm
/
1
7 1
1
+
1 \ 1
1
+ ... +
\
1
For d = 0, the arithmetic progression is a constant sequence and
therefore
,.
(
hm
l
1
1
+
n - * o o \aiCL2
\
+ ... +
= H-oc.
Onfln+l/
CL2&Z
2.2.55. Let d denote the common difference of the given progression.
Assume first that d ^ 0. Since
1
y/Ok
_
y/Q<k+l -
+ y/Cik+l
y/Ok
d
we have
lim -= ( — = - ^ — — + - — - ^ — — + ... +
For d = 0, the arithmetic progression is a constant sequence and
therefore the limit is equal to -foo.
2.2.56.
(a) By Problem 2.1.38,
i \
1+ nj
n
<e<
/
i \
1+ \
n
n
+i
Thus
(*)
l<n(v/e-l)<nl(l + ^
" -l) .
Now using the Bernoulli inequality (see 1.2.4) one can show that
n}
nz
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Solutions. 2: Sequences of Real Numbers
180
Hence
n | | l + - Y + n - l ) <l + ~ + \ .
Therefore, by (*) and by the squeeze law,
lim n( \/e — 1) = 1.
n—>oo
(b) For an arbitrarily fixed n,
n(en - 1)
n
Hence in view of (a) we get
i.
lim
eri
n—>oo
2.
n
+ en + ... + en
71
= e - 1.
2.2.57. We have a n +i — a n = — p(an - a n _ i ) . Therefore
an = a + (6 - a) + (a 3 - a2) + ... -f (a n — a n _i)
= a + (6 - a)(l - p + p2 + ... + ( - l ) n p n - 2 ) .
If 6 = a, then {a n } is a constant sequence convergent to a. If a ^ 6,
then the sequence is convergent provided \p\ < 1, and its limit is
2.2.58. Observe that
a n -f 26n
an + bn
c n -f 2
cn + 1
Hence
|cn + i - V2| = ^ - ^ | c
Cn -j- I
n
- V^| < (\/2 - l)|c n - x/2| < \\cn - V2\.
Z
Consequently, by induction,
\cn+1 - y/2\ < ±\cx
- V2\,
which means that the limit of {c n } is y/2.
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2.3.
Toeplitz Transformation and Stolz Theorem
181
2.3. The Toeplitz Transformation, the Stolz
Theorem and their Applications
2.3.1. If all the terms of the sequence {an} are equal to a, then by
(ii),
n
lim bn = a lim ]T cnk = &• Thus it is enough to consider the
n—>oo
n—>oo t . _ i
case where the sequence converges to zero. Then, for any m > 1 and
n> ra,
(*)
l&n " 0| =
/
^Cn,fcQfc
fc=l
- ] C l 0 ™^' l a f c l + X ! l Cn ' fc l' lafcl
/c=l
fc=n
The convergence to zero of {an} implies that for a given e > 0 there
exists n\ such that
for
\an\ < Wg
n >n1.
Of course, the sequence {an} is bounded, say by D > 0. It follows
from (i) that there exists ri2 such that for n > ri2,
n i —1
Next putting m = ni in (*), we get
k=l
k=n\
for n > max{ni,ri2}. Hence lim bn = 0.
n—>oo
2.3.2. Apply the Toeplitz theorem with cn^ = ^ , A; = 1,2,..., n.
2.3.3.
(a) If cUik are nonnegative, then (hi) follows from (ii).
n
(b) By (ii) in Problem 2.3.1, Yl cn,k > \ for sufficiently large n,
fc=i
say n > no- It follows from the divergence of {an} to +oo that,
given M > 0, there exists n\ such that a n > 2M if n > n i .
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Solutions. 2: Sequences of Real Numbers
182
Without loss of generality we can assume that all the terms an
are positive. Set ri2 = max{no,ni}. Then
712
71
n
Ck k
n
Cn kCLk
^ Cn,kO>k = ^ 2 ^ °> + ^ 2 >
fc=l
fc=l
2
- 5 Z Cn,kCLk + M - ^ '
fc=n2-fl
/c=l
and so {bn} diverges to -foo.
2.3.4. This is a special case of 2.3.3. Take cn^ — ^ for k = 1,2, ...,n.
2.3.5. Apply the Toeplitz theorem (2.3.1) with cn,fc =
2(n
~£+1).
2.3.6. Use the harmonic-geometric-arithmetic mean inequality (see
1.2.3), the squeeze principle for sequences and the result in 2.3.2.
2.3.7. Apply the foregoing problem to the sequence { ^ ± L } .
2.3.8. If b ^ 0, we take cn^ = n ~£ +1 and see that condition (i) in
2.3.1 is satisfied. In view of 2.3.2 condition (ii) is also satisfied. In
this case the desired result follows from the Toeplitz theorem. For
6 = 0, setting cn,fc = 1+6nn~*+1 yields
r
lim
ai(l -I- bn) + a 2 (l + 6 n -i) + ... + a n ( l + h)
= a.
Thus, by 2.3.2,
,.
a x 6 n - h a 2 6 n _ i - h ...-fa n 6i
lim
= 0.
2.3.9. We apply the Toeplitz theorem to the sequence {f^} with
Cn,k
= fe1
+
.fc+6n-
2.3.10. One can apply the Toeplitz theorem with cn^ =
b +*\fe
•
2.3.11. For n > 1, we put
and apply the result in the foregoing problem.
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2.3. Toeplitz Transformation and Stolz Theorem
183
2.3.12.
(a) In 2.3.10 we put xn = 1 + -4= 4-... -f -7=, yn = \Ai and show that
the limit is 2.
(b) Set
a2
2
a"
• a + — + ... + — ,
i
• - •
i
2/n =
^ /
7
t
n
n
Beginning with some value of the index n, the sequence {yn} is
strictly increasing. By 2.2.31 (b) we see that lim yn = +oo.
n—>oo
Therefore
n
a22
n^oo a nn+ 1 /\
'" anj\
a 1— 1 '
hm —— a + — + ... + — =
(c) We can apply the Stolz theorem (see 2.3.11) to the sequences
_. (fc + 1)!
(Jfe + n)!
xn = fc! + V 1t ; + ... +
r^,
1!
n!
We have
,.
h m
xn-xn_i
= lim ^
=
"'
™ n ( l - ( l - I )
_
"
lim
fc+1
yn = n fc+1 .
(n + 1) • (n + 2) •... • (n + fc)
I+I—/
-nfc+i
v
f c + 1
"/ = iim
)
v
v
"-*
n_oc l + ( l - I )
+
. .
+
"'
(l-I)^
1
HI'
(d) Set z n = -^= + ... + ^ ,
y n = Vn. Then
^- +
= lim (y/n + V n - 1) ( - 7 = + ,
-,
|
v
J
n-oo^ v
\^H
y/2^^1
x/n^T;
/ 1
/ n
/ n
/n-1
/n—1
= n-^00
hm L/2
-7= + A/
\ / - ^2n— + V 2n - 1
\ / 277n - l 7 - yJ n - 17 + V
= 2(V2-1).
Hence by the Stolz theorem the limit is 2(\/2 - 1).
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Solutions. 2: Sequences of Real Numbers
184
(e) Taking xn = lk + 2k + ••• + nk and yn = nk+1, we see that
yn - yn-i
nk+l - (n - l)k+1
n-+<x> k + 1"
Now it is enough to apply the Stolz theorem.
(f) By the Stolz theorem,
,.
1 + 1 -a + 2- a 2 + . . . + n - an
1
lim
—
=
-.
n-+oo
n • an+i
a—1
(g) One can also apply the Stolz theorem with
x n = (fc + l)(l* + 2* + ... + n * ) - n * + 1
and
yn = {k + l)n*
Then
x n - x n - i _ (A: + l)nk - nk+1 + (n - l) f e + 1
(k + l){nk
Vn-Vn-l
-
(n-
l)fc]
1
n->oo 2 '
2.3.13. Applying the Stolz theorem to
^n = ai + —= + ... + - =n
V2
V
and
yn = V™,
we see that
lim - =
ai + —= + —^ + ... + —=
= 2a.
2.3.14. In the Stolz theorem we set xn = a n + i and yn = n.
2.3.15. Applying the Toeplitz transformation to {an} with cn^
1
2n_ fc+i, we see that
/On
a n _i
ai
\
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2.3. Toeplitz Transformation and Stolz Theorem
185
2.3.16.
(a) Using the Toeplitz transformation to {an} with
1
Cn fc
' ~ (n+l-fc)(n + 2-fc)'
we can show that
,.
(
a
n
x
Q>n-1
Ol
\
(b) As in the proof of (a), we can apply the Toeplitz theorem to {an}
3o (I - l-i)\ n "—fc a n <
with cntk = %
2 ,n->
2n~k
rt
i show that the limit is i o .
2.3.17. Set an = (nwfc). In view of 2.3.7 it is enough to calculate
lim ^ - . We have
n—>-oo
an
(("n+\)fc)
("^
=
(nfc + l)(nfc + 2)-...-(nfc + fc)
( n + l ) ( n f c - n + l ) ( n f c - n + 2)-... -{nk-n
+ k-
1)'
Therefore the limit is equal to (k_\\k-\<
2.3.18. Let {an} be an arithmetic progression with the common
difference d > 0. Set
_ nn(ai •... • an)
(ai + ... + a n ) n "
C"n. —
Then
cn+l
cn
(n + l ) a n + 1
/ Ql+;-+aa i + a 2 + ... + a n + i V/ 1 *;'*'"* 1 /
2an+i
/^2ai + ( n - l ) d ^ n
0 __x
2e~
a,\ + an-i-i \
2ai -f nd / n—>oo
Hence by 2.3.7, the limit equals 2e _ 1 . If d = 0, the limit is 1.
2.3.19. Since bn = 2an + a n _ i , a n = k " " ^ - 1 and a n _i =
bn-i-Qn-2> T h u g a ^ = 2bw-bn-2i+an--2j R e p e a t e d application of this
procedure n — 1 times gives
2 - - 1 6 n - 2 n - 2 6 n _ ! + ... + ( - l ) n ~ 2 2 1 6 2 + ( - l ) 7 1 " 1 ^
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Solutions. 2: Sequences of Real Numbers
186
Thus by 2.3.16 (b), lim an = \b.
2.3.20. Put cn = (ai •... • an)nnx.
£ 2 ± l = (l + ±)
Then
( n -flfa n + 1 —> e*a.
Therefore by 2.3.7, lim nx(ai • a 2 • ... • an)™ = exa.
71—•OO
2.3.21.
(a) We apply the Stolz theorem with xn = 1 + ^ + ... + ^ and
yn = Inn. This gives
%n
%n— 1
Vn-Vn-1
-L
-.
"~m(l + _!_)"
™
because lim l n ( l + - ) n = l, which follows from the inequalities
(l + ^ r < e < ( l + ^ r + 1 (see 2.1.41).
(b) The limit is \ (see the solution of (a)).
2.3.22. We apply the Stolz theorem to
xn = — + —- + ...
1 2
Consequently,
y„-y n -i
H
n
and
in^ + ^ y
j / n = Inn.
— • a.
n—»oo
2.3.23. Use the result in 2.3.7.
(a) 1,
(b) e~2,
(c) e- 2 ,
(d) C3.
(e) We have
lim
n—KX)
for k = 1,
for jfe > 1.
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2.3. Toeplitz Transformation and Stolz Theorem
187
2.3.24. By the Stolz theorem (see 2.3.11),
an+1
= n lim
^ ° ° In (l + - ) n + 1
Urn ^ ^
*-*oo Inn
2.3.25. One can easily verify that
ai=Ai,
a2 = 2A2-
Ai,
an = nAn - (n - l)i4 n _i,
n > 2.
Thus
n
lim &L
n-+oo Inn
= I - ** + ^
+- + ^ - i + * , = A t
n-^oo
Inn
where the last equality follows from the foregoing problem.
2.3.26. [O. Toeplitz, Prace Matematyczno-Fizyczne, 22(1911), 113119] Let {an} be the sequence all of whose terms are equal to 1. Then
n
n
lim an = 1 and bn = ]T cUikak = J2 cn,k- Hence 1 = lim bn =
n->oo
^
n
fc=1
n-+oo
(k)
lim 5^ c n fc. Thus (ii) holds. Now let {an j } b e a sequence whose
fcth term is 1 and whose other terms are all 0. Then lim an
—0
n—Kx>
and 0 = lim bn — lim cnk. Therefore (i) also holds. Suppose that
(iii) is not satisfied.
Then for any C > 0 there exists nc
such
nc
that £ |c n fc| > C. In fact, given C > 0, there are infinitely many
fc=i
c
'
such indices nc- Now let n\ be the least positive integer such that
ni
X) lcni,fc| > 10 2 . We define the first n\ terms of {an} by setting
sgn c n i )fc = sgn a*
and
\ak \ = —.
Then
ni
ni
.
ni = 5^ C n i' f c a / c = ^ T n t a i . * ! > 10*
fc=i fc=i
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b
Solutions. 2: Sequences of Real Numbers
188
By (i), there exists n 0 such that
ni
^jcn^l < 1
for
n > n0.
k=\
Consequently,
/
^n.k&k
< —
for
n > n0.
fe=i
Now we take the least integer ri2 such that ri2 > max{no,ni} and
]C \cn2,k\ > 10 H- 1 -f 10. Define the consecutive terms of {an} by
setting
sgnc n2)/e = sgnafc
and
\ak\ = —^
for
ni + 1 < fc < n 2 .
Then
^2
bn2
C
= 2_^
k-\
n2,k&k
= 2__j Cn2,kO>k +
k—\
J*i
=
-i
C
22 n2,kClk+ J^2
/c=l
^
fc=ni+l
^n2,k^k
J12
X,
lC™2,A:|-
/c=ni-M
It follows from the above that
^ > - ^ + i^( 1 ° 4 + 1 + 1 0 - 1 ) = 1°2We can construct inductively the sequence {an} whose terms with
indices from rifc_i + l through n^ are equal either to + ^ r or — ^ - ;
then the transformed sequence {bn} satisfies
b n f c >10*
for
Jk = 1,2,3,....
Thus the sequence {an} converges to zero whereas the transformed
sequence {bn} has a divergent subsequence {bUk}- This is a contradiction, and so (iii) holds true.
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2.4. Limit Points. Limit Superior and Limit Inferior
189
2.4. Limit Points. Limit Superior and Limit
Inferior
2.4.1.
(a) First, let us show that the given subsequences have a common
limit. Suppose that lim a,2k — a, hm a,2k+i = b and lim a^k —
c. Then
k—>oo
lim a^k = a = c and
k—>oo
fc—>-oo
k—KX>
lim a^k+3 = b = c. Therefore
k—>oo
a = b = c. Now we prove that the sequence {an} also converges
to a. Given any e > 0, there exist positive integers k\ and &2
such that
implies
k > ki
k > /c2 implies
\a,2k — a\ < e,
|a2fc+i
—
o\ < £•
Hence
n > no = max{2/ci, 2/c2 + 1} implies
\an — a\ < e.
(b) No. Consider the sequence {an} denned by an = (—l) n . Then
lim d2k = 1, hm a2/c+i = — 1. But lim an does not exist.
fc—>oo
k-
'
~~
Now take the sequence {an} defined as follows
an =
0
U
if n = 2k, fc = 0,l,2,...,
otherwise.
Then lim a^k = 1 and lim a,2k+i = 1, but lim o,2k does not
fc—»-00
fc—»00
AC—>QO
exist. Of course, the sequence {a n } is divergent.
Finally, consider the third sequence
f 0 if n is a prime number,
^
=
\ l
if n is a composite number.
For this sequence we have lim a^k = 1 and lim a,2k = 1, but
k—>oc
k—>-oo
lim ci2k+i does not exist, because the sequence {02^+1} contains
AC—>-00
a subsequence with prime indexes and a subsequence with composite indexes. (Note that there are infinitely many prime numbers. Otherwise, if pi,p2> •••,Pn are prime, Pi < P2 < ••• < pn,
and no prime greater than pn does exist, then p\ -p2 •... -pn + 1 >
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Solutions. 2: Sequences of Real Numbers
190
pn is also prime, because it has no prime divisors except for itself
and 1. This is a contradiction.)
2.4.2. No. Define the sequence {an} by putting
an =
0
if n is a prime number,
{i
if n is a composite number.
Then every subsequence {a s . n }, s > 1, n > 2, is a constant sequence
and therefore it is convergent. The sequence {an} is divergent (see
the solution of Problem 2.4.1(b)).
2.4.3. Evidently, S p US q U...US s C S. To obtain the inclusion in the
other direction, assume x g S p US q U...US s - Then there exist positive
numbers ep, eqy..., es and positive integers np,nq,... ,ns such that
n> np
implies
\x — aPn \ > ep,
n> nq
implies
\x - aqn \ > eq,
n> ns
implies
\x — aSn\>
es.
Setting e = mm{ep,eq, ...,e s } and m = max{p n p ,g n q ,... , s n s } , we
obtain \x - an\ > e for n > m, This implies that x cannot be a limit
point of the sequence {an}. Therefore
S C SpUSqU...USs.
It follows from the equality S = S p U S q U ... U S s just proved that,
if every subsequence {aPn}, {aqn},..., {aSn} converges to a, then the
sequence {a n } also converges to a.
2.4.4. No. Define the sequence {an} by the following formula:
0
an={1
if
n = 2fe, fc = 0 , l , 2 , . . . ,
otherwise.
Every subsequence
converges to 1, whereas the sequence {an} diverges.
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2.4. Limit Points. Limit Superior and Limit Inferior
191
2.4.5. Assume that the sequence {an} does not converge to a. Then
there exists e > 0 such that for any positive integer k there is n^ > k
satisfying \ank — a\ > e. If we assume that rik is the minimum of
such numbers, then the sequence {n^} is monotonically increasing.
-hoo. Such a sequence {a nfc } does not contain
Moreover, lim n^
k—KX>
erging
to a,
< which contradicts our hypothesis
any subsequence converging to
Therefore {an} converges to a.
2.4.6.
(a) It is obvious that 1 is the only limit point of the sequence. Hence
S is a singleton, S = {1}.
(b) We have a^ = 0> «3fc+i = 1, G3/c+2 = 0. Hence, by Problem
2.4.3, the set S of the limit points of this sequence has two members, S = {0,1}.
(c) We have
a>2k =
22k+2
1
2fc
2 +3
and
a2fc+i
+
j
22/C+1+3*
Hence S = {0,2}.
(d) We have
a>2k =
21n(6fc) + ln(2Jfc)
r~77T^
ln(4ife)
^
a n d
«2fc+i =
ln(2fc + l)
ln(2(2fc+l))"
Therefore S = {1,3}.
(e)
\6fc+l
G6/C+3 = - 1 ,
a6k+4
=
a 6fe+2 = (-0.5) 6 / c + 2 ,
(-0.5)6/c+4,
a 6fc+5 = (0.5) 6fc+5 .
Thus S = {-1,0, ! } •
(f) We have
2
a>7k = 0 ,
^7fc+l = » ,
4
&7k+4 = -j,
1
G7/C+2 =
1
-,
2
a7fc+3 =
7'
G7fc+6 = =•
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Solutions. 2: Sequences of Real Numbers
192
Therefore S = { 0 , ± , f , f } .
2.4.7.
(a) Let a = - , p G Z , # G N , where p and q are co-prime. T h e n
akq = 0
and
fcp +
afc<?+/ = fcp H
where i = 1,2,..., q-1
H-
and r = ^
2p
2p
_ g
V
+ r
9
V
.q.
1 . Thus
(«-!)?
(q - 1)P
}•
(b) We will show that every real number x G [0,1] is a limit point of
the sequence {na — [na]}. By Problem 1.1.20, there exist pn G Z
and <?n G N such that 0 < a — ^~- < 4-. Since lim qn = +oo,
lim (aqn - pn) = 0. Let x G (0,1) and let e > 0 be so small
n—•oo
that 0 < x — £ < £ + £ < ! . Now suppose that n\ is so large that
0 < aqni
- pni <
< £•
Qrn
Then there is n 0 G N satisfying
(1)
n0(aqni
- p n i ) G (x - e, x + e).
(see the solution of Problem 1.1.21). It follows from (1) that
[n0aqni - n0pni} = 0, or equivalently, n 0 p n i = [n0aqni]. Therefore the term rtoaqni — [noaqni] from the range of our sequence
belongs to the interval (x - e,x -j- e), which means that # is a
limit point of the sequence under consideration. Similarly, one
can show that 0 and 1 are also limit points.
(c) Assume first that a is a rational number of the interval (0,1).
Let a — £ where p and q are co-prime and p < q. Then a^kq —
a>2kq+q = 0, and
Ipn
a>2kq+i = sm -1—
for
I = 1, 2,..., q - 1, q + 1,..., 2<? - 1.
Si
Hence
. 2r>7r
. (qf — l)p7r 1
= < 0 , s i n — , sin
,...,sin— >
q
q
J
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2.4. Limit Points. Limit Superior and Limit Inferior
193
If a e Z, then the sequence is constant. Taking a e Q \ Z, we
can write
a = [a] + {a - [a])
and
a - [a] e (0,1).
So, sinn7ro; = (—1)^ sin(a— [a])n7r, and this case can be reduced
to the foregoing special case.
(d) Let t G [—1,1] be an arbitrarily chosen number. Then there
exists x e M+ such that sin a: = —t. We can restrict our consideration to the case a > 0, because the sine is an odd function.
Since a is irrational, there exist sequences of positive integers
{pn} and {qn} such that
— = lim [pn -qn-A •
(See the solution of 1.1.21.) Therefore x = lim (27rpn — a7rqn).
n—•oo
Hence, by continuity and periodicity of the sine function, we get
—t = s'mx=
lim sin(27rpn — a7rqn) = — lim smaTrqn.
n—>oo
n—+oo
It follows from the above consideration that every number in the
interval [—1,1] is a limit point of the sequence.
2.4.8. We will show that in any interval (a, b) there is at least one
term of our sequence. Since lim (\/n + 1 — tyn) — 0, there exists
n-+oo
no € N such that
\/n -f 1 - \/n < b - a,
n > UQ.
Let mo be a positive integer satisfying ^mo > y/no — Q> and let
A = {n € N : yfn— ^mo < a}- The set A is nonempty (e.g., no £ A )
and bounded above. Putting m = max A and n2 = n\ + 1, we get
^fwi — ^/rriQ > a and %fn^ > a + ^/m^ > ^no. Therefore n 2 > no.
Hence ^/ri2 < %fn{ + b — a < ^/ra"o -f a -f 6 — a, or equivalently,
2.4.9. Boundedness of the set of the limit points of a bounded sequence is evident. Let S denote the set of limit points of the sequence
{an}. If S is finite, then it is closed. Assume that S is infinite and
let s be its limit point. Define the sequence {s^} of members of S
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Solutions. 2: Sequences of Real Numbers
194
in the following way: for Si take any member of S different from
s. For 52 choose any member of S different from s and such that
|«2 -s\ < fax - s | , and inductively, \sk+i - s\ < ^\sk - s|, s fc+ i ^ s.
Such a sequence {s^} satisfies the following condition:
Since s& is a limit point of the sequence { a n } , there exists anfc such
that |anfc — Sfc| < rpr^rlsi — s|- Hence
|anfc ~ s\< \ank - Sk\ + \sk -s\<
5 ^ 2 l s i ~~ 5I>
which implies that s is a limit of the subsequence {ank}.
Therefore
seS.
2.4.10. Let S denote the set of limit points of {an}.
(a) The sequence {an} is bounded. By 2.4.6, S = {0, y , | , ~ } .
Therefore lim an = 0 and lim an = | .
(b) We have S = { — 1 , — | , | , l } , which together with the boundedness of the sequence gives lim an = — 1 and lim = 1.
(c) The sequence is unbounded and the set of its limit points is empty.
Therefore
lim an = — cc
and
lim an = -foo.
(d) The sequence is unbounded above because its subsequence a,2k =
(2fc)2/c diverges to infinity. The subsequence with odd indexes
tends to zero. This shows that
lim an = 0 and
lim an = -foo.
(e) The sequence is unbounded because a ^ + i = 4fc -f 2 — • 4-oo
and a4/c+3 = — 4k — 2 —• - c o . Consequently,
k
and lim an = -foo.
-*°°
k—KX>
lim a n = — oo
n-+oc
n—>oo
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2.4. Limit Points. Limit Superior and Limit Inferior
195
(f) It is clear that the sequence is bounded. Moreover,
S=
>/2
V2
l-e
I
It then follows that
lim a n = — e
(g)
V2
—
and
—
lim an = e + 1.
lim an = 1 and lim a n = 2.
„_^
n—>oo
n—»oo
(h) The sequence is not bounded above, as a^k — 23fc k—>oo + 00.
Moreover, S = {—1,1}. So, lim an = — 1 and lim an — +oo.
(i) We will show first that
lim ~-
= +oc. Indeed, applying the
Stolz theorem (see 2.3.11), we get
lim
n—>oo
Inn
n
,.
Inn —ln(n —1)
, /
= lim
—-—- = hm In 1 +
n — U + 1
n—>oo
n—KX>
1
V
71—1
= 0.
This shows that
lim a,2k = hm
ln(2k)-4k
-00.
fc-^oo fc^oo In 2 + ln(2/c)
So, the sequence {an} is not bounded below. Moreover,
lim d2k+\ — h m
, , ,
/c-oo 2 t e + i
/c-,ooln2 + ln(2fc + l)
This gives lim an — — oo and lim a n = 1.
= 1.
2.4.11. It is enough to apply Problem 2.4.7.
(a)
lim an — minS = 0 and lim an = maxS, where
2p
= {o,?-f ?
(b)
(q - l)p
(q - 1)P
9
lim an — 0 and lim an = 1.
n-oo
(c)
2p
n
^°°
lim a n = minS and lim a n = maxS, where S is the set of all
n-+oc
n
^°°
limit points of the sequence described in Problem 2.4.7(c).
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Solutions. 2: Sequences of Real Numbers
196
(d)
lim an = —1 and lim an = 1.
2.4.12.
(a) If the set S of the limit points of {an} is empty, then lim an =
n—»oo
—oo < A Now assume that S is nonempty. Since S is closed (see
Problem 2.4.9), supS = lim an = L G S. It follows from the
n—KX)
definition of a limit point that there exists a subsequence {aUk}
converging to L. Therefore, for any e > 0, there is fco G N such
that
L — e < ank < A
for fc> fco-
Since £ is arbitrary, we get L < A.
(b) If the sequence {an}
is not bounded below, then
lim an ~
n—>oo
—oo < A. So, assume that the sequence {an} is bounded below,
that is, there exists B e R such that an > B for all n G N.
Moreover, by assumption, there is a sequence rife, n^ > fc, such
that anA. < A. Thus, by the Bolzano-Weierstrass theorem (see
2.4.30), the sequence {ank} contains a convergent subsequence.
Let g denote its limit. Then B < g < A. Therefore the set S of
the limit points of the sequence {an} is nonempty and lim an =
n—•oo
infS
<g<A.
(c) It is enough to apply the argument presented in the proof of (a).
(d) It is enough to use analysis similar to that in the proof of (b).
2.4.13.
(a) Let L = lim an. Suppose that (i) is not satisfied, contrary to
n—•oo
what is to be proved. Then there is e > 0 such that for any
k G N there is n > k for which an > L + e. Thus, by Problem
2.4.12 (d), lim an > L + e, which contradicts our hypothesis.
n—>oo
Now suppose that (ii) is not satisfied. Then there are e > 0 and
k G N such that an < L - e for all n > k. By 2.4.12(a), we get
lim an < L — e, which again contradicts our hypothesis. Thus
n—•oo
we have proved that L = lim an implies (i) and (ii).
n—•oo
Now we prove that conditions (i) and (ii) imply lim an = L.
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2.4. Limit Points. Limit Superior and Limit Inferior
197
It follows from (i) that the sequence {an} is bounded above.
On the other hand, it follows from (ii) that the sequence contains a subsequence which is bounded below. According to the
Bolzano-Weierstrass theorem (see 2.4.30), the sequence contains
at least one convergent subsequence. Therefore the set S of all
limit points of {an} is nonempty. We will show that L = supS.
Indeed, if s is an element of S, then by (i), s < L + e. By the
arbitrariness of e we get s < L. Moreover, from condition (ii),
we see that for any e > 0 there is a subsequence of the sequence
{an} converging to s which satisfies the inequality L — e < s. Of
course s G S. In this way the second implication is also proved.
Thus the proof is complete.
(b) This follows by the same method as in (a).
Now we state necessary and sufficient conditions for infinite limit
superior and inferior. The limit superior of {an} is +00 if and
only if the sequence is not bounded above. Therefore
lim an = +00
if and only if for every M G R and for
every k G N there exists n^ > k such that ank > M.
The limit superior of {an} is — 00 if and only if the sequence
is bounded above, say by L, and the set of its limit points is
empty. Therefore there is a finite number of terms of {an} in
every bounded interval [M,L\. Hence an < M for all sufficiently
large n. This implies that
(2)
n
lim an = — 00
-^°°
if and only if for every M £ M there is
k 6 N such that for every n > fc, an < M.
Similar arguments give
lim an = — 00
(3)
(4)
if and only if for every M G R and for
n->oc
every k G N there exists rik> k such that ank < M,
lim an = +00 if and only if for every M G R there is
~*°°
k G N such that for every n > fc, an > M.
n
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Solutions. 2: Sequences of Real Numbers
198
2.4.14. We prove only inequality (a), because the proof of (b) is
analogous. Inequality (a) is obvious in the case of lim bn = +oo
n—»-oo
or lim an — — oo. If lim an = -foo, then, combining condition (4)
n—*oo
n—KX)
given in the solution of Problem 2.4.13 with the inequality an < bn,
we get lim bn = -foo. Similarly, if lim bn = — oo, then combining condition (3) given in the solution of Problem 2.4.13 with the
inequality an < 6 n , we obtain lim an = — oo.
n—>oo
Assume now that both limits are finite and let
lim an — l\
and
ri—>oo
lim bn = fo-
n—Kx>
We wish to show that l\ < fo. Suppose, contrary to our aim, that
I2 < h. Let e > 0 be so small that I2 + e < l\ — e. Then there is c
such that I2 + e < c < l\ — e. By (ii) of Problem 2.4.13(b), we have
bnk < h + £ < c. On the other hand, by (i) we get c < ij — e < an.
Hence, in particular, c < ank, and therefore the inequality bnk < ank
holds for infinitely many n^, contrary to our hypothesis.
2.4.15. Set
lim an — /1,
lim bn = I2,
n _+oo
n_oo
lim an = Li,
n—00
lim bn = L2.
n->oo
We show first that
(1)
lim (an + bn) > lim an + lim 6 n .
n—>c»
n—>oo
n—»oo
Assume that l\ and /2 are finite. Then, by Problem 2.4.13 (b), for
any e > 0 there is fci such that an > l\ — e for n > k\, and there
exists fc2 for which bn > I2 — £ if n > ^2. As a result,
an + K > h + h — %£ for n > max{fci, £2}Combining this with Problem 2.4.12(c), we obtain lim (an -f bn) >
n—»oo
'i + /2 — 2s. Letting £ —> 0 + , we get (1).
If Zi or I2 is —00, then inequality (1) is obvious. Now we show
that if one of the limits l\ or I2 is +00, then lim [an + bn) = +00.
n—>oo
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2.4. Limit Points. Limit Superior and Limit Inferior
199
Assume, for example, that /i = +00. This is equivalent to condition
(4) given in the solution of Problem 2.4.13:
(*)
for every M G I there is k G N such that an > M if n > k
Since I2 ^ —00, the sequence {6 n } is bounded below. So, condition
(*) is satisfied by {an + bn}. In other words, lim (an + bn) = -foo.
n—•oo
Thus inequality (1) is proved.
The proofs of remaining inequalities are similar, and we will give
them only for finite limits. According to Problem 2.4.13, for any
e > 0 there exists a subsequence {ank} such that ank < h + £
and there is no for which bn < L2 + £ when n > no. This implies
that ank + bUk < l\ + L2 4- 2e for sufficiently large k. Therefore, by
Problem 2.4.12(b), we get lim (an + bn) <h+L2 + 2e. Since £ > 0
is arbitrary, we get
(2)
n—-KX)
lim (an + 6n) < lim a n + lim 6 n .
Similarly, for any £ > 0 there is a subsequence {bnk} such that
6nfc > L2 — £ and there exists np for which an > h — £, for n > no.
Hence ank + 6nfc > /1 -f L2 — 2£ for sufficiently large k. Therefore, by
Problem 2.4.12(c), we have lim (an + 6n) > l\ + £2 - 2e. Since £
n—>oo
can be made arbitrarily small, we conclude that
(3)
lim (an + bn) > lim a n + lim 6 n .
Moreover, for any £ > 0 there is &i for which a n < L\ -f £ when
n > fci, and there is &2 for which 6n < L2 + £, for n > &2- Thus
a n + 6n < £1 + £2 + 2e
for
n > max{fci, k2}.
Combining this with Problem 2.4.12(a), we obtain lim [an -f bn) <
n—>oo
L\ + L2 -f 2£. Since £ can be made arbitrarily small, we get
(4)
lim (an + bn) < lim an + lim 6 n .
n—>oo
n—+00 Duplication
n—>oo
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Solutions. 2: Sequences of Real Numbers
200
Now we give examples of sequences {an}
inequalities (l)-(4) are strict. Let
&n =
and {bn} for which the
0
if
n = 4fc,
1
if
n = 4fc + l,
2
if
n = 4k + 2,
I 1
if
n = 4k + 3,
( 2
if
n = 4fc,
1
if
n = 4k + l,
1
if
n = 4k + 2,
<
I 0
if n = 4fc + 3.
In this case the inequalities given in the problem are of the form
0<1<2<3<4.
2.4.16. No. It is enough to consider sequences {a™}, m = 1, 2,3,...,
defined by setting
( 1 for n = m,
for n ^ m.
Then
lim ( < + a„ + ...) - 1 > 0 =
n—>oo
Now let
lim a\ + lim a ; + ....
n—->oo
C = {0
—1
for
n—+00
71 = 171,
for n ^ m.
In this case
lim {a\ + a2n + ...) = - 1 < 0 = lim a\ + lim o£ + ....
2.4.17. Let
lim an = li,
n_oo
lim 6 n = Z2,
^ ^
hm an = L\,
n^oo
lim 6 n = L2n-+oo
We will only show the inequality
(1)
lil2 < Km {anbn) < hL2.
n—KX)
The same reasoning applies to the other cases.
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2.4. Limit Points. Limit Superior and Limit Inferior
201
Assume first that l\ and Z2 are positive. Then, by Problem
2.4.13(b), for any e > 0, there exists no such that
a>n > h — £•> bn > fa - £ for
n > no.
Consequently, anbn > lil2 — e{l\ H- h) + £2 for 5 so small that l\ —
e > 0 and I2 — £ > 0. Therefore, on account of Problem 2.4.12(c),
lim {anbn) > lil2 — e(h + Z2) + £2 • Letting e —> 0 + , we get
n—>oo
(i)
lil2 < lim (a n 6 n ).
n—»oo
If l\ — 0 or l2 = 0, then inequality (i) is obvious. If Zi = +00 and
l2 = +00, then (by condition (4) in the solution of Problem 2.4.13),
for any preassigned positive number M, we can find no such that
an > V M ,
bn > VM
for
n > n0.
Therefore anbn > M, which means that lim {anbn) = 4-cx).
Assume now that one of the limits, say Zi, is infinite and the
second one is finite and positive. Then for any 0 < e < h and any
M > 0 there exists a positive integer no such that for n > no we have
1.
M
/
On > h ~ £>
•
CLu > -j
l2 - £
Hence anbn > M for n > n 0 . Therefore
inequality (i) is proved.
Now our task is to prove that
(ii)
lim (anbn) < hL2.
lim (anbn) = -foe, and
n—>oo
n—>oo
If /1 and L 2 are finite, then on account of Problem 2.4.13, one can
find a subsequence {n/J such that ank < l\ + £ and 6nfc < L2 + £.
This gives
< ZiL 2 +e(Zi + L 2 ) + ^ •
Therefore lim (anbn) < liL2+e(li
n—>oc
+ L2)+£2. Letting e —> 0 + yields
(ii). If /1 = +00 or L2 = +00, then inequality (ii) is apparent.
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Solutions. 2: Sequences of Real Numbers
202
Now we give examples of sequences {an} and {bn} for which all
the inequalities are strict. Let
1
for
n = 4/c,
2
for
n = 4fc + l,
3
for
n = 4k + 2,
12
for
n = Ak + 3,
3
for
n = 4/c,
2
for
n = 4fc + l,
2
for
n = 4k + 2,
11
for
n = 4k + 3.
bn = {
In this case our inequalities are of the form
1<2<3<6<9.
2.4.18. Assume that lim an = lim an = g. Then, by 2.4.13,
(i) for any e > 0 there is fc G N such that an < g + e \in> k\ and
(i') for any e > 0 there is k G N such that # — 5 < an if n > A;.
Thereby g is a limit of the sequence
{an}.
On the other hand, if lim an = #, then (i) and (ii) in Problem
n—>oo
2.4.13(a) and (b) are satisfied with L = g and / = g. Consequently,
lim an = lim an = p.
Assume now that lim an = -foo. Then statements (1) and (4) in
n—>oo
the solution of Problem 2.4.13 are obvious. If lim an = lim an =
H-oo, then condition (4) means that lim an = -foe. Similar argun—>-oo
ments apply to the case lim an
-OO.
2.4.19. By Problem 2.4.15,
lim an -f lim bn < lim (a n + 6 n ) < lim an + lim bn.
n—»oo
n—>oo
n—»oo
n—»oo
n—>oo
On the other hand, on account of the last problem, a = lim an =
n—>-oo
lim a n . Therefore
n—+oo
lim (an + 6n) = a + lim 6 n . The proof of the
n—>-oc
second equality runs as before.
n—>oc
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2.4. Limit Points. Limit Superior and Limit Inferior
203
2.4.20. Using the inequalities given in Problem 2.4.17, we can apply
the same method as in the solution of the preceding exercise.
2.4.21. We will apply Problem 2.4.13. Let lim an = L. Then conn—>oo
ditions (i) and (ii) in 2.4.13 (a) are fulfilled. Multiplying both sides
of the inequalities in (i) and (ii) by —1, we get:
(i) for every e > 0 there is k G N such that for every n > k we
have — L — e < —an\ and
(ii) for every e > 0 and for every k e N there exists n^ > k for
which — aUk < —L + e.
By 2.4.13(b), we obtain
lim (—an) = —L = — lim an.
The proof of the second equality runs as before. In the case of infinite
limits it is enough to apply statements (l)-(4) given in the solution
of Problem 2.4.13.
2.4.22. We will apply Problem 2.4.13. Let lim an = L. Then by
n—>oo
conditions (i) and (ii) in 2.4.13(a), we have
(i) for every e > 0, there exists k G N such that for every n > k
we have an < L + eL2; and
(ii) for every e > 0 and for every k G N there exists n^ > k for
which L — e~- < ank.
Assume first that L ^ O . Then by (i),
JL
an
>
1
2
L + eL
_ 1
~ L
eL2
L{L + eL2)
>
^_
1
£
'
Assume now that 0 < e < ^ . Then by (ii),
1
ank <
L
1
_ 1
- e £ - L +
e^-
_l
L(L-e^)<L+6'
The above conditions imply (by 2.4.13(b))
lim — = — = - = =
.
an
L
lim an
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Solutions. 2: Sequences of Real Numbers
204
Now suppose that
lim an = 0. Given M > 0, by (i) in Problem
n—• oo
2.4.13(a), there exists an integer k such that an < -^ for n > k.
Therefore — > M for n> k, which in turn, by statement (4) given
in the solution of Problem 2.4.13, means that lim •—- = -foo. Finally,
n—xx)
suppose lim an = -foe. Then for any e > 0 and for any k G N there
n—•oo
exists rik> k such that ank > ^ (see statement (1) in the solution of
Problem 2.4.13(a)). The above inequality is equivalent to -^- < e. Of
course — e < -£-. Thus both conditions given in 2.4.13(b) are fulfilled
for the sequence { —} with I = 0, which means that lim -^- = 0.
n—•oo
The proof of the first equality is finished. The proof of the second
equality is similar.
2.4.23. It follows from our hypothesis that 0 < lim an < -hoc. The
n—>oc
equality lim an • lim — = 1 combined with the preceding problem
yields
n—•oo
7i—^00
a n
lim an — ^=—=- = lim an.
li
n ^ omo
n^oc
_L
"i
n—»oo
Therefore, by Problem 2.4.18, the sequence {an} is convergent.
2.4.24. Assume that {an} is a sequence such that for any sequence
{bn} the first equality holds. Take bn — —an. From Problem 2.4.21
it follows that
0 = lim [an + (—an)) — lim an + lim (—an) = lim an — lim a n .
n—>oo
n—*oo
n—>oo
n—^oo
n-^oo
From this, by Problem 2.4.18, we conclude that the sequence
is convergent.
{an}
2.4.25. Assume that {an} is a positive valued sequence such that
for any positive valued sequence {bn} the first equality holds. Take
bn = ^-. Hence, by Problem 2.4.22, we get
1 = lim I an - — 1 = lim an • lim f — J = lim an • -==
lim an
.
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2.4. Limit Points. Limit Superior and Limit Inferior
205
From this it follows that both the limits superior and inferior of {an}
are positive and lim an = lim an. Therefore the sequence {an} is
n
n->oo
convergent (see 2.4.18).
2.4.26. Evidently,
lim v/a^ < lim v/a^. Now we will show that
lim v/a^ < lim ^
n—*oo
^°°
n oc
n-oo
±i
n
n—>oo
. If lim ^
n—•oo
^
n±i
Gn
= -f-oo, then the inequality is
obvious. So, assume that lim ^n±1 — L < +oc. Then for any e > 0
there exists /c such that
^ ± i < L + £ for n > jfc.
Hence
an _
Uk
an
dn-1
an-i
dk+\
0>n-2
&k
(TJjrE\n~k
Consequently,
_k
Since ^fa^(L + e) ~
—> 1,
n—>oo
$/a£(L + e) n < l +
e
for sufficiently large n. Prom what has already been proved, we have
tfa^ < (1 + e){L + e) = L+{L + l)e + s2
for sufficiently large n. Combining this with Problem 2.4.12(a), we
obtain lim v/a^ < L + (L + l)e + e2. Since £ can be made arbitrarily
n—>-oc
small, lim v/a^ < L — n—>-oc
lim ^dii. To prove lim ^±1
< ^lim ^ {/a^
it is enough to apply Problem 2.4.22 and the inequality just proved
to the sequence {^-}2.4.27. We first prove that
that
lim bn < lim an. To this end, assume
n—>oc
n—>oo
lim an — L < +oo. (For L = +oo the above inequality is
n—•oo
clear). Then, given e > 0, there exists fc€N such that an < L + e
for n > fc. Hence
&n
ai + a 2 + ... + dk + a/c+i + ••• + ^n
n
ai+a2 + ... + ak
k(L + e)
<
+ L + e.
n
n
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Solutions. 2: Sequences of Real Numbers
206
g
m c e
Qi+Q2 + ---+Qfc _
n
k(L+e)
n
^ Q
'
n_^QO
a i + Q 2 + ---+Qfc _
n
k(L+e)
n
^
sufficiently large n. It follows from the above consideration that bn <
e + L -f e for sufficiently large n. According to Problem 2.4.12(a), as
e can be made arbitrarily small, we get lim bn < L = lim an. The
n—• c o
proof of the inequality lim an < lim bn is analogous.
n—+00
n—•oo
n—>-oo
2.4.28.
(a) (b) It is enough to apply Problem 2.4.13.
(c) The equality is not true. To see this, it is enough to consider the
sequences defined by setting
a =
0
1
bn =
1
0
" {i
for n = 2k,
for n = 2fc+ 1,
for n = 2fc,
for n = 2fc+l.
Then
0 = lim min{a n ,6 n } 7^ min{ lim a n , lim bn} = 1.
(d) This equality is likewise not true, as can be seen by considering
the sequences defined in (c).
2.4.29. Assume that the sequence {an} has the property that there
are infinitely many n such that
(1)
for every k > n, dk < an.
Let rii be the first such n, n<i the second, etc. Then the sequence
{ank} is a monotonically decreasing subsequence of {a n }. On the
other hand, if the sequence {an} fails to have the above property,
that is, there are only finitely many n satisfying (1), choose an integer mi such that the sequence { a m i + n } does not satisfy (1). Let
rri2 be the first integer greater than mi for which a m 2 > a m i . Continuing the process, we obtain a subsequence {amn} of {an} which
is monotonically increasing.
2.4.30. By the preceding problem such a sequence contains a monotonic subsequence which is bounded and therefore convergent.
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2.4. Limit Points. Limit Superior and Limit Inferior
2.4.31. Assume first that
207
lim 2n±i = +oo. Then, by 2.4.14(b),
—- ai -h ... + a n + a n + i
hm
= +oo.
Now let
lim
= a < -f oo.
Then, given £ > 0, there exists k such that
(1)
0>n
- < a + e for
n> k.
In other words,
an
(2)
1
>
an+i
for
a +e
n> k.
Hence, for sufficiently large n, we have
ai -f ... + a n + a n + i ^ a* + ... + a n -f a n + i
>
=
—
a
a k
tt/c+1
_J_
_L_
(
a n
~
2
dri
n-2
Qn-1
Qfc+1
Qn-2
&n-l
&n
G/c+2
&n-l
a
^~l
^ n —1
^n
-1
\ n—k
i
a
n-l
1
a
n
Qn+1
1
^n
/
Qn-1
^n
\ H—fc—1
i
a + £/
\a-\-ej
a-he
an
If 0 < a < 1, then the above inequality and Problem 2.4.14(b) yield
lim bn = -f oo.
n—>oo
On the other hand, if a > 1, then by Problems 2.4.14 (b) and
2.4.19 we conclude that
(3)
lim 6 n > a + lim
n-+oo
n-»oo
1^+^ .n-fc+l
7
\
= a +
1
±—
In case a = 1 {e > 0 can be arbitrary) we get
a > 1, then (3) implies
a +e
Q; -J- ^ — 1
lim bn = -j-co. If
n—••oo
lim bn > 1 + a + — i — = 2 + (a - 1) + —^— > 4.
n-^00
a —1
a — 1
4 is an optimal estimate because it is attained for the sequence an =
2 n , n G N.
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Solutions. 2: Sequences of Real Numbers
208
2.5. Miscellaneous P r o b l e m s
2.5.1. Assume first that
lim an = -foo. Put bn = [an]. Then bn <
n—>oo
an < bn + 1. Hence
V
bn + l)
K)
a.
Thus the result in 2.1.38 and the squeeze principle imply
lim ( H
)
n-+oo V
= e.
an
Moreover,
lim I 1 - — ]
= e" 1
because
lim ( 1
n->oo \
an )
I
= lim
n->oo /-j ,
1
r-?rn = e
\
l
.
^•L"t" o n - i y
This implies that
lim 1H
n^oo y
anJ
= e,
if {a n } diverges to
— oo.
2.5.2. One can apply the foregoing problem with an = ^, x ^ 0.
2.5.3. By 2.1.39, 2.1.40 and 2.5.2, (l 4- ^)n < ex < (l + ^ ) Z + n for
Z > x > 0, I G N . Hence for any positive x and for any positive integer
n, ^ < In (l + £) < ~ if / > x. Taking n = 1 we get ln(l + x) < x
for x > 0. Now, set / = [x] + 1. Then we obtain
K)
Infl + i U
nJ
Therefore ln(l 4- z) > ^
"
X '
2+*
for x > 0.
Consider now the function f(x) = ln(l + x) - ^ ^ ,
have
/,(
-)=(,
+
lKx + 2 ) 2 > 0
f r
°
x > 0. We
*>0'
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2.5. Miscellaneous Problems
209
Hence
2x
/ ( x ) = ln(l + x ) - — — > / ( 0 ) = 0 for x > 0.
2 -\- x
2.5.4.
(a) Assume first that a > 1. Set a n = >/a — 1. By the inequality in
2.5.3,
2an
1
_,
.
<
—
lna
=
ln(a
+
1) < a n .
n
an + 2 n
Therefore the squeeze principle implies that
lim n( y/a — 1) =
n—>-oc
lna for a > 1. We see at once that the claim holds if a — 1.
To prove it for 0 < a < 1, it is enough to apply the above with
a
(b) Put an = tfn-1. Then (an + l ) n = n. Hence by 2.5.3, Inn =
nln(a n + 1) < nan. Consequently, lim nan = +oo.
n—•oo
2.5.5. Using differentiation we can show that, for x > — 1, j ^ <
ln(l+x) < x. Since lim a n = 1, an > 0 beginning with some value of
n—>oo
the index n. It follows that 1 ^ ~ 1 1 < lna n = ln(l+(a n —1)) < a n — 1.
Dividing the inequalities by an — 1 and using the squeeze principle
yields the desired result.
2.5.6. By the definition (see 2.1.38), e = lim (l -f \)n . Moreover,
1
_ 1 + 1 + '2!( 1V_ Inf
V . . + i(i-iWi-*'
k\ V
n/ \
n.
+...+j.(l_iv..ri-"n! V
1
n
Hence
(i)
(l + i ) <an.
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Solutions. 2: Sequences of Real Numbers
210
On the other hand,
^H+IHKK) H
+-+F(
1
-
k-1
1
k\ \
n
Passage to the limit as n —> oo gives
(ii)
e>ak.
By (i) and (ii), the limit of the sequence {an} is e. Moreover,
_
1
1
(n-f-1)!
(n + 2)! ""
1
f
1
1
<
+
+
(n + 1)! I
n + 2 (n + 2) 2 + '"
1
n+2
<
(n + l ) ! n + l '
+
1
(n + ra)!
1
1
1
(n + 2)™" J
Keeping n fixed and letting m —> oo, we get
n
~
1
n+ 2
(n+l)!n+l
This and (ii) imply that 0 < e — an < ^ .
2.5.7. We know (see 2.5.2) that ex = lim (l -f - ) n , x G R. For a
fixed x G R, put an = ( l -f § + ^ + ... + ^ ) • We get
•.-(-Hn-K'-o-i)-o
fc — l W x^
n
^-H)-- ^
AT
fc!
By 1.2.1,
\
n)
'" \
n
/
7
^—' n
j=i
fc(fc-l)
2n
for
2<k<n.
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2.5. Miscellaneous Problems
211
Therefore
ifc
an
+
~V n)
\ - ^
k\ - In
^ (k 2nf^(k-2)V
~~5n~~
k=2
Since
lim ~ ^
n_
*oc
U
k=2
K
J
J^_L, = 0 (which follows easily from the Stolz
}
k=2
'
theorem, see 2.3.11), we get lim an = lim (l + f)
n—>oo
n—>oo
— ex.
2.5.8.
(a) By 2.1.38, ^
< In (l + £) < £. So for n > 1 we get
,271 + 1 1
1
1
,
2n
In n
< -n + n + -1 + ... + —
<
In
2n
n- 1
Thus the desired result follows from the continuity of the logarithm function and the squeeze principle.
(b) We have
1
1
1
n + 1 "" 2n + l
y/n(n+l)
1
1
1
< n- +
— - + ... + 2n
--.
n + 1
1
^2n(2n+l)
'"
Therefore the claim follows from (a).
2.5.9. Analysis similar to that in the proof of 2.5.3 gives
x2
x —— < ln(l + x) < x
(*)
for
x > 0.
Set bn = \nan = £ In (l + £ ) . By (*),
ifc
k2
<1
/
n^~2n^ H
+
k\
^J
<
k
^
This and the equalities
V^ i. _
l^k"
n
( n + -1-)
2
'
V^ u2 _
^ " "
n
( n + l)(2n + 1)
6
fc=i fc=i
imply that
lim 6n = ^. Finally, the continuity of the logarithm
n—>-oo
function yields lim an — yfe.
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Solutions. 2: Sequences of Real Numbers
212
2.5.10. One can show by induction that
an = n + n(n - 1) + ... + n(n - 1) •... • 2 + n(n - 1) •... • 2 • 1
n!
n!
n! n!
^(n-1)! + 7
( n - 2^
) ! + ... + 77
1! + 0!
Hence
lim f f f 1 + — J = lim
'ai + 1
oo \
= lim
n-»oo
;—
n!
= lim
n—>oo y
ai
an + 1
an
1 -f — + ... H—r
1!
= e,
n!/
where the last equality follows from 2.5.6.
2.5.11. By 2.5.6,
1
1 6
e = l + - - h ... + — + - ^ - ,
1!
n! nn!
where
0 < 0 n < 1.
Therefore 0 < n\e — [n\e] = ^ < ^ , which proves our claim.
2.5.12. By the arithmetic-geometric mean inequality, the monotonicity of the logarithm function and the inequality proved in 2.5.3, we
get
- l n V ^ < l n i ( ^ + v ^ ) = l n ( ^ ( ^ - l ) + \(Vb-l)
n
I
\l
Z
+ l)
)
<!((tfs-i)+(^-i)).
To get the desired result it is enough to multiply these inequalities
by n and use the result in 2.5.4 (a).
2.5.13. Note first that if lim a™ = a > 0, then
n—+oo
lim an = 1.
n—+oo
Assume now that {an} and {bn} are sequences whose terms are
different from 1. By Problem 2.5.5,
, x
(*)
,.
nlnan
lim —
— = 1.
n-*oo n{an — 1)
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2.5. Miscellaneous Problems
213
The assumption lim a™ = a > 0 and the continuity of the logarithm
n—• oo
function imply that
lim n l n a n = In a. Thus, by (*),
lim n(an - 1) = lim n l n a n = In a.
n—KX>
n—-KX)
Observe that these equalities remain valid if an = 1. Finally,
lim n\n(pan + g&n) = lim n(p(an — 1) + g(6n — 1)) = lna p 6 9 .
n—>oo
n—>oc
2.5.14. We have a n + i — a n = — ^ ( a n - a n - i ) - Consequently,
a n = a + (6 — a) + ... + (a n - a n _i)
= a + (6 - a) (1 - 1 + i - ... + v( - l y) n - 2
*
2!
3!
(n-1)!
Finally, by 2.5.7, lim a n = 6 - (b n—>oo
a)e~l.
2.5.15. Consider the sequence {6 n }, where bn — ^ > a n d a PPty the
same method as in the solution of the foregoing problem to conclude
that an = n\.
2.5.16. As in the solution of 2.5.14, an+\ — an = — ^ ( a n — a n - i ) Thus lim an = 2b — a — 2(6 — a)e~.
n—>oo
2.5.17. (a) We have
_ ~
\ - * fc + 1 — fc ^
1
^ f c ( f c + l)! £ j ( f c + l)(fc + l)!
n
1
^fc&!
n+l ,
fc=Qfc!
-.
n
^ ( f c + 1)!
n
^ ( f c + l)(A+i)!
1
' (n+l)(n+l)r
Therefore, by 2.5.6, lim an = e.
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Solutions. 2: Sequences of Real Numbers
214
(b) By (a) and 2.5.6,
0 < an - e <
[n + l)(n + l)r
R e m a r k . It is interesting to note that this sequence tends to e
faster than the sequence considered in Problem 2.5.6.
2.5.18. It follows from 2.5.6 that e = 1 + ^ + ... + ^ + r n , where
lim n\rn = 0. Moreover,
n—+oo
(*)
——7 <n\rn
n+ 1
< -.
n
So,
lim nsin(27m!e) = lim nsin(27rn!r n )
71—KX>
71—^OO
• sin(27rn!rn)
,
= lim n27rn!r n —-—
= lim n2^n\rn
n—>oo
n—>oo
27TTl\rn
= 27r.
The last equality follows from (*).
2.5.19. We will show that lim (l - ^ ) n = 0. By assumption, for
an arbitrarily chosen M > 0 we have an > M if n is large enough.
Hence
a
.
M
0 < 1 - —n < 1
n
n
Consequently,
•<0-ir)"<0-£)"-
So, by 2.4.12, 2.4.14 and 2.5.2, we get
0
<
lim ( l - ^ Y
V
n J
n"Z^o
< Urn" ( l - ^ )
n->oo V
nJ
n
< e~M.
Letting M —* oo yields
o < lim fi - ^ Y < BS fi - ^ r < o
n-z^o
V
Therefore lim (l - ^ )
n /
n
n-oo V
n /
= 0, as claimed.
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215
2.5. Miscellaneous Problems
2.5.20. We will show that lim (l + M " = +oo. Given M > 0, we
n-»oo
v
n
'
have bn > M for sufficiently large n. So, as in the solution of the
foregoing problem, we get
lirafl + M">l i m fl
n
n^c V
n
J
-"°° V
^V=e«
+
U
)
Since M can be arbitrarily large, we see that lim (l + ^ ) = -f oo.
2.5.21. It is easy to see that the sequence {an} is monotonically
decreasing to zero.
(a) We have -±
V J
a
(b) By (a),
n+i
l
> 1. So, by 2.3.14, lim -±- = 1.
- = -^
l-on
an
J
'
n^OQ
'
lim "<1,"""» - lim I l f c ^ U
n
^
Q O
lim
na n
i^.
n->oo
Inn
n^oo
In n
n-+<x> Inn
Using the Stolz theorem (see 2.3.11), we obtain
..
n(l - nan)
hm —^—j
n-^oo
Inn
n
=
,.
n
~~ ^ ~ 1 J
U
= hm — n + 1 T r r — n-+oo
nlnfl + -J
(^r=^ ~ l)
nan
hm — ^ I—TT- = hm
n^oo l n ^ l _ ^ i . j
n—oo 1 -
an
= 1.
2.5.22. It is easy to see that the sequence {an} is monotonically
decreasing to zero. Moreover, an application of l'Hospital's rule gives
x2 — sin2 x
lim —
x_>o
1
« — — o-
x 2 sin x
3
Therefore
lim
» - ° ° \ an+1
an
Now, by the result in Problem 2.3.14, lim nan = 3.
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Solutions. 2: Sequences of Real Numbers
216
2.5.23. Clearly, the sequence is monotonically increasing. We will
show that it diverges to +oo. We have
« n + l = (
^
a
n
+
T
V
T
J
> ( a n +
ai + . . . + a n /
na
V
)
nJ
> €?n +
-
n
and
Thus {a^} is not a Cauchy sequence. Since it increases, it must
diverge to -foe. Moreover,
(*)
i<^±i<i
J_.
+
By the Stolz theorem
a
n
,.
n
n->oo21nn
(an+l
_
a
n)
,.
= lim - (
™/
2
n-°° 2 v
n ^ o o 2 1 n ( l + ^)
n+1
h7
2x
n}
TTT
I = 1,
2
n-^oo 2 \ o i + a 2 + ... + an
{a\ + a 2 + ... + a n ) /
because
0
< T
To <
-;
(ai + a 2 4-... + a n ) 2
n
and again, by Stolz's theorem,
nan
(n-h l ) a n + i - n a n
lim
= lim
n-+oo a\ + a 2 + ••• + CLn
Gn+1
n-+oc
= lim ( n + 1 — n—— 1 = 1.
n
-+°° \
CLn+l )
The last equality follows from (*). Indeed, we have
An
-
X
+
1<n+l-n—— <
a n +i
1+
nan
nan
Since lim a n = +oo,
n—+oo
1 +
Um
n—»oo 1 -|
n±l
™n_
—
=
i.
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217
2.5. Miscellaneous Problems
2.5.24. By the inequality arctanx < x for x > 0, the sequence is
monotonically decreasing. Moreover, it is bounded below by zero.
Hence it converges, say, to g, which has to satisfy the equation g =
arctany. Thus g = 0.
2.5.25. Note that all the terms of the sequence {an} belong to the
interval (0,1). Denote by xo the unique root of the equation cos a; =
x. If x > xo, then cos(eosx) < x. The function f(x) = cos(cosx) — x
is monotonically decreasing, because /'(x) = sin x sin (cos x) — 1 < 0
for x e R . Thus, for x > xo, cos(cosx) —x < f(xo) = 0. Analogously,
if x < xo, then cos(cosx) > x.
Assume first that a\ > XQ. It follows from the above that as =
cos(cosai) < a\. Since the function y = cos(cosx) is monotonically
increasing in (0, §), we get a 5 < a^. It can be shown by induction that the sequence {d2n-i} is monotonically decreasing. On
the other hand, a,2 = cosai < cosxo = XQ, which implies that
Ci4 = cos (cos 02) > ei2, and consequently, {a2n} is monotonically
increasing.
Similar arguments can be applied to the case where 0 < a\ < xoIf a\ = xo, then all the terms of the sequence {an} are equal to xoIn all these cases the sequences {a2 n -i} and {a,2n} both tend to the
unique root of the equation cos(cosx) = x. It is easily seen that xo
is such a root.
2.5.26. We get, inductively,
an = 1 - ( - l ) n - 1 sin(sin(...sin 1)...),
n > 1.
(n—1) times
Hence
n-1
J2 ( - i ) * " 1 sin(sin(...sin 1)...)
n-1,
n
1 \-^v
- > dk =
%
fc=l
v
(k—1) times
'
•
n *—*
n
k=l
We will show now that
n
E(-l) f c "' 1 sin(sin(...sinl)...)
fc=l
N
(*)
"^
v
(fc-1) times
,.
lim
TO—>-oo
'
= 0.
n
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Solutions. 2: Sequences of Real Numbers
218
If n — 1 is even, then
n-l
^ (—l) k ~ l sin(sin(...sinl)...)
— sinl + sin(sin(...sinl)...)
v
/
v
( n - l ) times
v
k = 1
/
v
( f c -l) times
<
n
< Q_
n
n
Obviously, for odd n - l , (*) also holds. Finally, lim ^ ^ ak = 1.
2.5.27. Clearly, a n G (n7r,n7r+ f ) , n = 1,2,..., and thus lim an =
z
n—>oo
4-oo. Moreover,
7T
1
lim tan(— -f nn — an) = lim
n-+oc
2
1
= lim — = 0 .
n^oo t a n a n
n-*oo a n
By the continuity of the arctan function we get lim (f -\-n7r — an) = 0.
n—>oo
Therefore
lim (a n + i - a n - 7r)
n—>oo
=
n ^ 2 o ("2 "
h n 7 r _ a n
~ (o" + (
n
+
1
)
7 r
-
a
n + l ) j
= 0 .
Consequently, lim (a n +i — a n ) = 7r.
n—>oo
2.5.28. Observe that without loss of generality we can assume that
I oil < f • Indeed, if not, then by assumption |a 2 | < f. Consider
first the case where 0 < a < 1 and 0 < &i < ^ . Then an-\-i —
a s i n a n < an. This means that {an} is monotonically decreasing
and, since it is bounded, it converges. Its limit is equal to zero,
which is the unique root of the equation x = asinx, 0 < a < 1.
Assume now that 1 < a < ^ and 0 < a\ < | . Then the equation
x = asinx has two nonnegative solutions 0 and XQ > 0. If a\ < #o,
then {an} is monotonically increasing and bounded above by x$.
Indeed, a 2 = asinai > «i- Moreover, a 2 = a s i n a i < a s i n x 0 = xo
and, inductively, an < a n +i < #o- Similarly, xo < a\ < | implies
that an > an+i > XQ. Hence lim an = XQ for 1 < a < §. If
n—>oo
— | < a < 0, a\ > 0, then we consider the sequence {bn} defined
by setting &i = a i , 6 n +i = - a s i n 6 n . Obviously, bn = ( - l ) n _ 1 a n . It
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2.5. Miscellaneous Problems
219
follows from the above that in the case where 0 < a\ < | we have
lim an = 0
if
\a\ < 1,
lim an = xo
if
1 < a < —,
n—>oc
n—>oo
lim an
does not exist
n—>oo
2
7T
if
— — < a < — 1.
2
If — | < ai < 0, then one can consider the sequence given by b\ =
—ai, 6 n + i = asin& n , and apply the foregoing. If a\ — 0, then all the
terms of the sequence are also equal to 0.
2.5.29.
(a) Note that an > 0 and a n + i = ln(l + a n ) < a n . Therefore the
sequence converges to a g which satisfies g = ln(l + g), so g = 0.
We now show that lim nan = 2. Using differentiation, one can
n—• oo
prove that (see also 2.5.3)
2iX
x
x
v + x)
y < x - — + —
2 + x < ln(l
2
3
This implies that
1
r*j
1 |
an
1
x > 0.
1<
<
an (l - \an + | a £ )
an
for
an+i
1 I 1
m
an
2
Putting
l
A
0n = — an
we see that
l
an (1 - \an + \a2n)
lim bn = ^.
9 Summing both sides of (*), we get
n—>oo
1
1
1
L
L
L
!
X
!
—
ai + —
a2 + • • • + —
a n + fel -f 62 + • • • + &n <a2— + ••• +a n — +
^n+i
1 1
I n
< — + — + ••• + — + ai
a2
an
I
Consequently,
1
6i + fr2 + - • + bn
(n+l)ai
n+1
1
n
< 7( n + l TV—
+
)ai
2(n+l)'
(n + l)o n + i
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Solutions. 2: Sequences of Real Numbers
220
Hence lim
1
(n+l)an+i
2'
(b) We have
n
n — -—
lim (nan - 2)= lim nan—
n-^oo
Inn
n-^oo
Inn
(1)
To prove that
^
lim
,
an
exists we will use the Stolz theorem
lnn
n->oc
(see 2.3.11). We get
n - - ^
lim — - 5 2 - =
n-^oo
Since lim ^ - we see that
(2)
Inn
lim
lim 2Z±.
n—•oo
lim
n-^oo
ln(1+an)
1 - - 2 - + -2.
^ — ^ .
In ( 1 - f - J
= 1 and lim nln (l + l) = 1,
= um n ( 2 a w + 1 - 2 a
Inn
n—>oo
+a r a a r a + 1 )_
a^
Now it is enough to show that lim
2Qn+1
~ 2 a g + a n Q n + 1 exists. To
this end, we use the inequality (which can be proved by differentiation)
«A/
t*-/
»X/
.
,
v
Jb
Ju
X - y + y - — < ln(l + x ) < X - y + y ,
X > 0.
Thus
g a n - £<4 ~ J a n < 2 a n+i - 2a n + a n a n + i < -a3n + - a * .
This gives
2 a n + i — 2a n + a n a n + i
1
lim
=
3
Combining this with (1) and (2), we see that
2.5,30. Set f(x) = (\)x and F(x) = f(f(x))
that F'(x) < 0 for positive x. We have
F'(x)=(i)
a'
lim
n(n n
^ n ~ 2) = §.
- x. We first show
ln24-l.
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2.5. Miscellaneous Problems
221
Hence
F'{x) < 0 if and only if
4/
(i) x +*
<
In" 4
It is a simple matter to verify that the function on the left-hand side
of the last inequality attains its maximum value of -^j at x = ^ i p .
This implies that F'(x) < 0, which means that F strictly decreases
on (0, +oo). Moreover, F{\) = 0. Therefore F(x) > 0 for 0 < x < \
and F(x) < 0 for x > \. Consequently,
f{f{x))<x
for
x>-.
Since 02 = 1 > | , it follows that a^ = f{f{a,2)) < ^2? and inductively,
we find that {a,2n} strictly decreases. Thus it tends to a g\ such
that /(/Q71)) = g\- The convergence of {a2 n -i} to a #2 satisfying
f(f(92)) — 92 can be established in much the same way. Clearly,
9i=92 = 5.
2.5.31. Observe first that 0 < an < 2 for n > 2. If an > 1, then
a n + i < 1. Set /(x) = 2 1 - x and F(x) = f{f(x)) — x. One can show
that F'(x) < 0 for 0 < x < 2. Therefore
F(x) < F ( l ) = 0
for
1 < x < 2,
F(x) > F ( l ) = 0
for
0 < x < 1.
Next, as in the proof of the foregoing problem, we show that if a\ < 1,
then the sequence {a2n} is monotonically decreasing and the sequence {a2n-i} is monotonically increasing, and both tend to the
same limit 1. Similar considerations apply to the case a\ > 1.
2.5.32. Observe first the all the terms of the sequence are in the
interval (1,2). Since the function F(x) = 2% — x is monotonically
decreasing on this interval, F(x) > F(2) = 0 for x e (1,2). Therefore
the sequence is monotonically increasing and its limit g satisfies g =
22, so g = 2.
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Solutions. 2: Sequences of Real Numbers
222
2.5.33. We apply 2.3.14 to the sequence {an + a n _ i } and obtain
l j m an+an n i _ Q jsjexj. w e consider the sequence 6n = (—l) n a n .
n
n—KX>
Since lim (bn — 6n_2) = 0, we see that
0 = hm
= lim
n—>-oo
.
n—>oo
n
77,
2.5.34. By the Stolz theorem (see 2.3.11),
= lim , , f l n + * — - ^ - -
lim —^
n->oo Inn
lim
TWOO ln(n + 1) — Inn
r
a
i
n-^oo
m
h _|_ l ) n
n+l
= — lim n In
.
n—>oo
an
If lim n f 1 - 2m)
= p
i s finite, then
lim f ^»±i _ i") = o. Now
the desired result follows from the inequalities
1
^~1
,
an4-i
+ ~^T^
i
—
<nln(l+(^-lY\<„(f!2H-l
—
1
•
v ~ ' i
V
~
V &n
i
i
—
'~
/ /
i
\
an
If g = +oo, then the right inequality shows that
—oo, and consequently, lim
n—>oo
lim n l n ^
1
=
. an = +oo. Finally, if g = — oo, then
m n
for any M > 0 there is no such that ^ ^ > —^ + 1 for n > no- Hence
nln^±i>lnf1+^y—>M.
n
/
n—->oo
Since M can be arbitrarily large, we see that
lim
n—>oo
I n - * n-
, m°n = —oo.
2.5.35. By the definition of the sequences,
Gn+i + bn+1 = (ai + 6i)(l - (a„ + 6 n )) + {an + 6 n ).
Set dn = an + 6 n . Then d n + i = d i ( l - d n ) + dn and by induction we
show that dn = 1 - (1 - d i ) n . Likewise,
an = ^-(l-(l-d1D
and
6 n = ^ (1 - (1 -
d1)n).
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223
2.5. Miscellaneous Problems
Since |1 — d\\ < 1, we get
lim an =
n—KX>
— and lim bn =
a\ + b\
—.
d\ + b\
n^oo
2.5.36. Define the sequence {bn} by setting bn = aan. Then bn+i =
bn{2-bn) = - ( 6 n - l ) 2 + l. Hence 6 n + 1 - 1 = - ( 6 n - l ) 2 . Obviously,
the sequence {an} converges if and only if {bn — 1} does or, in other
words, when \bi — 1| = |aai — 1| < 1. Moreover, if ai = | , then
lim an = 0, and if 0 < aa\ < 2, then lim an = a
n—>oo
n—>-oo
2.5.37. This result is contained as a special case in Problem 2.5.38.
2.5.38. One can show that the function / is continuous at (a, a,...,a)
and f(a,a,...,a)
— a. Define the sequence {bn} by setting
h =b2 = ... = bk =min{ai,a 2 ,...,ajfe},
bn = /(&n-l,&n-2,...,&n_fc)
for
71 > fc.
Note that if min{ai,a2, ...,afc} < a, then {6 n } is strictly increasing
and bounded above by a. On the other hand, if min{ai,a2,..., a/e} >
a, then {bn} is strictly decreasing and bounded below by a. Hence, in
both cases, the sequence {bn} converges and lim bn = a. Moreover,
n—>oo
the monotonicity of / with respect to every variable implies an < bn
for n G N. Now define the sequence {cn} by setting
ci = c2 = ... = Cfc = max{ai,a 2 ,...,a/ c },
Cn = /(c n _i,Cn_2,...,C n _fc)
As above, we show that
for
n > fc.
lim cn = a and c n < an for n G N.
n—••oo
Finally, by the squeeze principle, lim an — a.
n—>oo
2.5.39. We have a 3 = a 2 e a 2 _ a i , a 4 = a 3 e a 3 _ a 2 = a 2 e a 3 _ a i and,
inductively, a n + i = a 2 e a ™ -ai for n > 2. Suppose ^ is the limit of the
sequence. Then
(*)
— (/ = e*.
«2
Note that if ^— = e, then the equation (*) has only one solution
g = 1. If ^— > e this equation has two solutions, and if 0 < — < e
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224
Solutions. 2: Sequences of Real Numbers
it has no solutions. Consider first the case where 0 < — < e. Then
the sequence {an} diverges because in this case (*) does not have
any solutions. Moreover, one can show that the sequence {an} is
monotonically increasing and therefore diverges to -hoo.
Consider now the case where — = e. Then a 2 = e a i _ 1 > a\
and, inductively, an+\ > an. Moreover, if a\ < 1, then one can show
by induction that also an < 1. Hence, in such a case, lim an = 1. If
n—>oc
a\ > 1, then {an} is monotonically increasing and diverges to +oo.
Next, consider the case where ^— > e. Then (*) has two solutions,
sa
Y 9ii92, where, e.g. g\ < #2- Assume that a\ < g\. Then
eai
eai
ax > 0
or, in other words, a<i > a\. It follows by induction that {an} is
monotonically increasing and bounded above by #i, which is its limit.
If g\ < a\ < g2, then {an} is monotonically decreasing and bounded
below by #i, which is also its limit. If a\ = g\ or a\ = g2, then the
sequence is constant. Finally, if a\ > g<i, then the sequence increases
to -I-oo.
2.5.40. (This problem and its solution is due to Euler in a more
general case. See also [13]). Applying differentiation, we show that
lnx < | for x > 0. Hence ^ > l n a n = a n _ i l n a , n > 1, and
consequently, an > a n _ i l n a e . Therefore, if a > e*, the sequence
{a n } is monotonically increasing. We will show that in this case,
lim an = +oo. We have an+i — an = aan — an. So for a > e«,
n—+oo
we consider the function g{x) = ax — x. This function attains its
minimum at x0 = ~lj1^
< e. It follows that ax-x > l+l^a)
> 0,
and consequently, a n + i — an > 1 + 1 na
> 0- Since the difference
between two consecutive terms is greater than a positive number, the
sequence diverges to infinity.
Now we will consider the case where 1 < a < e*. We first show
that in this case the equation ax — x — 0 has two positive roots.
The derivative of the function g(x) = ax — x vanishes at the point
xo > 0 such that ax° = y ^ . The function g attains its minimum
value at x 0 , and g(x0) = ax° -x0 = ^ -XQ = 1 ~f n Q j n a < 0, because
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225
2.5. Miscellaneous Problems
if 1 < a < ee, then r^- > e. Since q is a continuous function on
'
a
In a
R, it possesses the intermediate value property. Thus the equation
ax = x has one root in the interval (0, XQ) and the other in (xo, -hoo).
Denote these roots by a and /?, respectively. Note that since g(e) =
ae — e < ye*) — e = 0, the number e is between a and /?.
If x > /?, then ax > aP = j3 and p(x) > 0. This means that
in such a case the sequence {an} is monotonically increasing and
bounded below by /?. Hence lim a n = -hoo.
n—UDO
If a < x < /?, then a < a x < /? and #(#) < 0. Consequently,
the sequence {an} is bounded and monotonically decreasing. Thus
it converges to a.
For either a; = a o r x = (3, we get a constant sequence.
Now if 0 < x < a, then 1 < ax < a and g(x) > 0. Therefore the
sequence {an} increases to a.
Finally, if a = e *, then the number e is the only solution of the
equation ax = x, and the function g attains its minimum value of 0
at e. Thus for 0 < x < e, we get 0 < ax < e and g(x) > g(e) = 0.
This implies that the sequence {an} is monotonically increasing and
its limit is equal to e. On the other hand, if x > e, the sequence
increases to infinity.
We can summarize the results as follows:
( -hoo
-hoo
n
(3
lim an = <
-+°°
\ a
if a > ee and ' x > 0,
if 1 < a < ee and x > (3,
if 1 < a < e« and x = f3,
!
if 1 < a < ee and 0 < x < /?,
-foo
if a = e« and x > e,
e
if a = e^ and 0 < x < e.
2.5.41. The equality can be proved by induction. We have lim an =
2 (compare with the solution of 2.1.16).
n—>oo
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226
Solutions. 2: Sequences of Real Numbers
2.5.42. [20] Note first that
an<)j2 + yf:
2 + ... + V2
<2.
n roots
Observe that if e\ — 0, then all the terms of the sequence {an} are
equal to zero. Assume now that e\ ^ 0. We will show by induction
that the given equality holds. It is evident for n = 1. So suppose that
a n = £ 1 y 2 + £ 2 y 2 + ...+£nV / 2 = 2sin j J J V€i€2.'.£k
Then
/
n+l
\
/
n+l
\
a n+1 - 2 = 2sm ^ g ^ ^ j = -2cos ^ - + - g ^ r r j
(
n+l
\
/
n+l
\
iE^)-^(iE^)-».
which completes the proof of equality. Now, by the continuity of the
sine function,
lim ann = 2sin I — >^ k 1'.'\ .
n-+oo
\ 4 ^
2~
I
2.5.43. One can show by induction that
1
1
n
arctan -2 + ...+ arctan —77
2n 2= arctann + l '
Therefore lim (arctan \ + ... + arctan —5-) = f.
n—>oo
v
z
zn /
4
2.5.44. We have
sin 2 (7r\/n 2 + n) = sin 2 (7ryn 2 + n — ixn) = sin2
—• 1.
2.5.45. One can show by induction that the sequence is monotonically increasing and bounded above, e.g. by 3. Hence it is a convergent sequence whose limit g satisfies g = y/2 + y/S + g and g G (2,3).
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227
2.5. Miscellaneous Problems
2.5.46. [13] We have
3 = Vl + 2 • 4 = V1 + 2>/l6 = Y 1 + 2V 1 +
3
^5
y 1 + 2 + y 1 + 3\/l + 4x/36,
and, inductively,
\\ 1 + 2W 1 + 3y 1 4- . . y i + nV^H- 2)2 - 3.
(i)
Therefore
;3 > \ H 2y 1 + 3y 1 + . . y i + ( n - l ) \ / n T T .
(2)
Now we will use the following (easily verifiable) inequality:
(3)
yTTxa: < >/a>A + z,
x > 0, a > 1.
By (3) with x = n and a = n + 2,
y 1 + ni/(n + 2)2 < \/n + 2\A + n.
Hence
y 1 + (n - l)Y / H-^V / (^ + 2 ) 2 < \/ X + Vn + 2 ( n - 1)>/1 + n
< (n + 2)*^/l + (n-l)>/ra + l,
where the last inequality follows from (3) with a = \/n + 2. In view
of (1), repeating this argument n times gives
(4)
3 < ( n + 2)2 n J 1 + 2y 1 + 3y 1 + . . ^ 1 + (n - l ) V n T I .
Combining (2) with (4) yields
lim \
1 + 2yi+3^/i
+ .../l + ( n - l ) V n + l
= 3.
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Solutions. 2: Sequences of Real Numbers
228
2.5.47. The equation x2 + x — a = 0, a > 0, has two roots a and
/3 such that o: > 0 > /?. Furthermore, we have
l - a = a — an — cuar.
an+1 - a =
_ a - (1 + ot){an - a) - Q(1 + a) _ - ( 1 + a)(q w - a )
Since a + /? = — 1, we see that an+\ — a = /? a ^~ Q . Likewise, a n +i —
/? = a « a z £ . Thus
a n + i - /? _ aan - (3
an+i - a
(3 an- a'
and inductively,
n-l
fli-j9
ai — a
Since
j ^
< 1, we get
lim (%)
n-l
= 0, and consequently,
lim an = (3.
n—•oo
2.5.48. Let a and ft be the roots of x2 + x - a = 0, a > 0. Then
a > 0 > /?. In much the same manner as in the solution of the
foregoing problem we get
an — a
an-(3
=
i a\
\0J
a\ — a
ax-F
Thus lim an = a,
n—>oo
2.5.49. For any positive integer fc, we have
l&n+l+A; — 0>n+l\ =
< j\a>n+k
1
1 + an+/e
1
1 + an
\0>n+k — 0>n\
(l + an+fc)(l + an)
- On|.
Now by induction we get
kn+l+fc - On+l| — ( 7 ) la*:+l - a l | -
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2.5. Miscellaneous Problems
229
Moreover,
|cifc+i - a i | < |afc+i - ak\ + \ak - ak-\\ + ... + |a2 - a\\
<
r|a2
- a i | = - | a 2 - a\\.
1-4
^
Thus {a n } is a Cauchy sequence. Its limit is y/2.
2.5.50. One can proceed as in the solution of the foregoing problem
and show that lim an = 1 + y/2.
n—+oc
2.5.51. Let f(x) = ^ , x > 0, and F(x) = / ( / ( x ) ) . Then F'(x) >
0 for x > 0. It is easily verified that ai < 03 and 04 < a 2 . Moreover, since F is strictly increasing, we see that the sequence {a 2 n }
is strictly decreasing and the sequence {a 2 n +i} is strictly increasing.
The sequence {an} is bounded. Thus both its subsequences {a 2 n }
and { a 2 n + i } converge. One can check that they have the same limit
VTTa-1.
2.5.52. If ai < 0, then a 2 = 1 — ai > 1 and as = a2 - \ > \. By
induction, a n +i = an — ^ t r for n > 2. Consequently,
a
and therefore
n+i = -(K^T
+ ^
+ ... +
\)+a2,
lim an = —ax if ai < 0. Now if ai E (0,2), then
n—+00
a 2 E [0,1) and inductively, we see that a n + i E [0, T>^T], which implies
that in this case lim an = 0. Finally, if ai > 2, then a2 = ai — 1 > 1.
n—>oo
By induction, we get a n +i = an — ^ t r ? a n d consequently, as in the
first case, we show that lim an — ai- 2.
2.5.53.
(a) We have
n-l
n-
j
2a 2
3a3
(n-l)a"-1
n - l + n - 2~ + n - 3~+••• +
1
fa
J-i
+
n - l Vl
2(n-l)a2
n-2
+
3(n - l)a 3
n-3
+
'"
+
(n-l^a"-1^
1
) '
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Solutions. 2: Sequences of Real Numbers
230
Since
J
:=J
~+
:
< J 1 + J - 1 = 3\
we obtain
n-1
ja*
. a + 2 2 a 2 + 3 2 a 3 + ... + ( n - l ) V
n —j
n— 1
.
Now it is enough to observe that, by the result in Problem 2.3.2,
v
hm
a + 2 2 a 2 + 3 2 a 3 + ... + (n - l ) 2 a n " 1
n—>oo
71—1
„
= 0.
(b) Observe that
n
n—1
n
1
n—1
n—1
,
L.
and apply (a).
(c) Apply (b) with
a=\.
3
2.5.54. Since for positive x, x — ^- < sinx < x, we see that
n
n
n + /c
k=l
n
Q
^ 6 ( n + /c)3
k=l
v
n
^SH1n-f/c
'
^ n + /c'
fc=l
n
k=l
3
It is easy to check that lim Y2 6(n+k)3 ~ ®m Moreover, by 2.5.8 (a),
tim
n
J2 ^ T I = 7rln2. Therefore the limit is 7rln2.
n-oofc=1n+*
2.5.55.
(a) Let a n = J"] (1 + -^ J . In view of the inequality (see 2.5.3)
fc=i ^
'
•—^ < ln(l + x) < x for x > 0, we get
—^
- r < l n a n < > —r-.
en6 + kz
fc=i fc=i
z
—' CTTT
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2.5. Miscellaneous Problems
231
Hence, by the equality £ k2 =
n(n+1) (2n+1)
,
6
fc=i
n ( n + l ) ( 2 n + l)
n(n + l ) ( 2 n + l )
1
< man < 6cn 3
6(cn 3 + n 2 )
Therefore lim a n = e 3 c.
n—>-oo
(b) One can show that the inequality ^ j < ln(l + x) < x also holds
if — 1 < x < 0. Therefore, as in the proof of (a), we get
e
lim f j 1
3c
—e
cnc
k=l
.
2.5.56. Since for positive x, x - ~ < sinx < x - \ + fr, we see
that
(i)
(nv/n)"»n
A l i- 6n
n!
3
n-
i.3" —
<
n\
sin
k
riyjri
and
x /r,3n
™
sm •
fc=i
v/n
<
n!
*2
TT / i
(ny^)" ^
\
6n
fe4
3
5!n6
It follows from (1) and from the result in the foregoing problem that
the limit is greater than or equal to e _ i s . We will now show that
lim fr fi - ^13 + *!)
<e
6
n-ooH V
k=l
Indeed,
ln
1
x
6n
I I ( - ^ + ^ 6 ) <k=lE
fc=l
18
5!n /
'
k2
6n 3
+
A;4
5!n6
n ( n + l ) ( 2 n + l)
n(n + l)(2n + l)(3n 2 + 3n - 1)
3
+
36n
30 • 5!n6
Finally, by (2) and the squeeze principle,
_
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Solutions. 2: Sequences of Real Numbers
232
2.5.57. We will first show that an = n^a
-i
2nn
n-l
a
_ ^ _1_ _ ^
"-l^(n)-l.
fc=0 Vfc/
_ ^
k\(n-l-k)\{n-k)
(n-l)!n
v
fc=0
1
-f 1, n > 2. We have
+
;
^fc(n-l-fc)!fc!
fc=0 V fc ^
v
fc=0
fc=o
;
v fc ;
Moreover,
Y^ k
1
=
Y^
^ — 1 — fc 1
=
n—1
^»(V) to ~^~ (V) ~
an
^-^ /c
1
n
" " h~ (V)'
Therefore
k
Efc=o
1
n—1
Vfcy
Finally, by (1), an = a n _i - ^ o n _ i + 1 = ^ o n _ i + 1, which
establishes our first assertion. Prom this we deduce that lim an = 2.
n—•oo
2.5.58. If a = 0, then obviously lim a n = 0. If a > 0, then 0 <
n—»oo
a n < 1 — „ a ' , and consequently,
a < 0. Then
lim an = 0. Assume now that
^ - , - i , - ( . - - i ) ( ( = ) - - i ) . . . . . ( ( ^
T
) - - i ) .
Therefore, if we take a = — 1, we obtain the divergent sequence an =
(—l) n _ 1 . If we take a < — 1, we get
p)
)\\n-p)
J
for 1 < p < n. Moreover, I — I
\an\>(n-a-l)l(-^-)
\p
)\n-p
) ~
— 1 > 2 — 1 = 1. Therefore
a
\ V U —1/
-l)
—> +oc.
/ n-^oo
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233
2.5. Miscellaneous Problems
Likewise, we can see that if — 1 < a < 0, then
|an|<(n-«-l)[(—
2.5.59. We have (2 + y/3)n = £
k=0
- 1 - 0 .
(]J)(V5)fc2n-fc. If we group the
terms with odd and even indices, respectively, we can write
(2 + y/S)n = An + VZBn
and
(2 - y/%)n = An -
V3Bn.
Hence lim (An + \/3B n ) = +oo and lim (An - y/3Bn) = 0. Also,
n—•oo
n—>oo
lim — - — = 1.
n->oo
An
Since the A n are integers and v3Bn < 1, it follows that [V^JBn] =
^4n — 1 for sufficiently large n. Consequently,
{VSBn}
—• 1 or
n—UDO
{A n + v ^ S n } = {\/3J5n} —• 1.
n—KX)
2.5.60. The sequence {Sn} is monotonically increasing. If it were
bounded above it would converge, and then
lim an = lim (S n - S n - i ) = 0-
n—•oo
Suppose now that
n—Kx>
lim Sn = +oo. By our assumption, S n +ia n +i +
n—•oo
an < S n a n + a n _ i , and consequently, 5 n a n -f a n - i < 52a2+ai. Hence
.
, a - i ^ ^2^2 + a i
n
a n < a n + -=—
<
•
Finally, lim an = 0.
n—>oo
2.5.61. By assumption, for any e > 0 there is a positive integer no
such that an < en if n> n0. Hence
al + al + ... + al
n2
=
a 2 + oj + ... 4- a2nQ
n2
Q i + a i + ... + o^0
^
i+- + ^
n2
n g ( a n o + i + ...-fan)
0 +
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Solutions. 2: Sequences of Real Numbers
234
By 2.4.14 and 2.4.19,
-.2 i ^ 2 _j_
_j_ Jl
—- a{
-\-a$ + ... +
an
—- ax + ... + an
hm —
=—5
< e hm
.
n—+00
77r
n—>oo
This obviously implies that
a
a
lim ?+ i+•••+<£
n
=
0.
n—+00
2.5.62. We will apply the Toeplitz theorem (see 2.3.1). Set
An = a i + a 2 + ... + a n , Bn = bx + 6 2 + ... + 6 n , C n = ci + c 2 + ...+c r j
and
an-k+iBk
U>n,k =fliBn + a 2 i? n _i + • •• + &n^i
Now we will show that the positive numbers dn)& satisfy conditions
(i) and (ii) in 2.3.1 (see also 2.3.3 (a)). For fixed fc,
j
/
On.fc <
n
Q
;
n-fc+l
;
n
;
• 0.
ai + a 2 + ... + a n _ fc+ i n-+oo
Clearly, ^ dn,fc = 1. Observe also that
fc=i
Cn _ dlK + a 2 6 n - l + ••• + a n ^ l
,
Cn
Q>lBn + (LiBn-l + ... +
ttn-Bi
—+rf
J5i
— 452
Finally, by the Toeplitz theorem, lim ^
= lim -fe^ = 0.
2.5.63. We know that x - ^ < ln(l H- 2) < x - ^ + ^
n2
+ //
—
' jB n
for x > 0.
Put an — (l + ^) e~ n . Then — ^ < l n a n < — | -f ^ , which implies
lim l n a n = —\. Therefore the limit is equal to e~2.
n—->oo
2.5.64. We have a n + 1 - a n > - 4 , > - ^ r y = ~ ^ i + \ for
n > 1. Let 6n = an — ^ y . Then the sequence {bn} is monotonically
increasing and bounded above; hence it converges. Therefore {an}
also converges.
2.5.65. By the assumption an+i 2 \/2 > a n , we see that
an+12-&
>an2~^rT.
Hence the sequence bn = an2~ 2™-1 is monotonically increasing and
bounded. So, it converges. Obviously, lim bn = lim an.
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2.5. Miscellaneous Problems
235
2.5.66. Let a G (/, L). Suppose, contrary to our claim, that a is not a
limit point of {an}. Then there is a neighborhood of a which contains
only finitely many terms of the sequence. Let e > 0 be so small that
(*) I < a — s < a < a + s < L and an £ [a — £, a + e] for n > n\.
By the assumption, |a n +i — an\ < e for n > n2. By 2.4.13 (b), we
know that there exists ank such that ank < I + e < a for . rc^ >
max{ni,n 2 }. Hence a nfc+i < anfc + |a nfe+ i - ank\ < a + e. Thus by
(*), a nfc +i < a — e:. Therefore, by 2.4.12 (a), L < a — e < L, a
contradiction.
2.5.67. Let a G (/, L). Suppose, contrary to our claim, that a is not a
limit point of {an}. Then there is a neighborhood of a which contains
only finitely many terms of the sequence. Let e > 0 be so small that
(*) l<a — e < a < a + s < L and an £ [a — £,a + e] for n > n\.
By the assumption,
(**)
an - an+i < an < e
for
n>
n2.
It follows from 2.4.13 (a) that there is ank such that ank > L — e > a.
Hence, by (**), we get ank+1 = ank + {ank+i - ank) > a - e. Now,
by (*), a>nk+i > a + s for rife > max{ni,ri2}. Thus by 2.4.12 (c),
Z > a + £ > a > / , a contradiction.
2.5.68. We will use the result proved in the solution of the last problem. By the monotonicity of {a n },
Q
n+1
Q-n
n + l + On+i
n + an~
^
~Qn
.
—
1
( n + l + a n + i ) ( n + o n ) ~" n
Thus, by the result in the preceding problem, the set of all limit points
of the given sequence is the interval [/, L], where
/ = lim
—
n_>oo n + an
and
L = lim
—.
n-^oo n + an
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Solutions. 2: Sequences of Real Numbers
236
2.5.69. Note that
0>2n+l
1
1 + Q2n-1 _ 2
&2T\
2
3
&2n-l
—
^
This implies that the sequence has the two limit points: ^ and | .
2.5.70. We know, by 1.1.14, that for any positive integer n there
exist a positive integer qn and an integer pn such that
Pn
2TT -
< ~2-
Thus |p n | < (27r -f- l)g n , and consequently,
|V / l^Jsinp n | = | v ^ ] s i n ( 2 7 r g n - pn)\ <
f
\pn~\sin-
<
V2F+T
v7^
Since the sequence {qn} is unbounded, it contains a subsequence
divergent to infinity. Therefore zero is a limit point of {an}.
2.5.71. It is enough to show that there is a subsequence {ank} for
which
fc
{ai + a n ff ec++ i ) \x nUk
fn'nkk{a\
> 1
V (nfck + l)o
l)c nfc /
Suppose that the above condition does not hold. Then there exists
no such that
n(ai + a n +i)
< 1 if n > no(n+ l)an
Hence ^n+1 +' ^n +-1 < ^ for n > n 0 . Thus
Qn _ ^ n - l
n
&1
n—1
n '
Q n - l _ &n-2
n-1
a
Ql
n-2
n0 + l
no + 1
n0
n-2'
<
ai
n0 + 1
Summing the above inequalities, we get
<
n
no
a\
1
no 4-1
1
n
——- + ... + -
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237
2.5. Miscellaneous Problems
Therefore, by 2.2.50 (c), lim ^ = -oo, which is impossible because
an > 0.
2.5.72. In much the same way as in the proof of the foregoing problem we show that there is a subsequence {ank} for which
nk{a\ +a n f c + P )\ r i f c
(nk+p)ank
J
>
l
2.5.73. Suppose that our claim does not hold. Then there exists an
no such that for n > no, n ( + ^ n + 1 —11 < 1. The last inequality
can be rewritten as - ^ r < ^ — ^ J r . This, in turn, implies (see the
solution of 2.5.71) that
1
no -f 1
Hence
1
+ ...+ n- <
&nn
n0
<2n
no
n
lim g^ = — oo, which contradicts the fact that {an\
n—•oo
n
is a
positive sequence.
To show that 1 is the best possible constant, take an = n l n n .
Then
,.
/ l + (n + l ) l n ( n + l )
lim n
n
n-+oc \
nmn
1 + (n + 1) ln(n + 1) - n l n n
= lim
——
n-^oo
Inn
=
Um
l + ln(n+l)+ln(l + l ) "
n—oo
Inn
^
2.5.74. Note that an = 1 4- a n - i a n d a\ = \. Clearly, the sequence
strictly increases. We will show by induction that it is bounded above
by ^(1 + y/E). Indeed, if a n _i < ^(1 + \/5), then a2n = 1 + a n _i <
§ + | \ / 5 - Therefore a n < y | -f | \ / 5 = l + ^v 7 ^ and {an} converges
to ±(l + >/5).
2.5.75. [20] Obviously, the sequence {6 n } is strictly increasing. Assume first that a < In 2. Then, by assumption, there is no E N
such that ln(lna n ) < n l n 2 if n > no; or, equivalently, an < e 2 " if
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238
Solutions. 2: Sequences of Real Numbers
n > no. We have
K = y o i + ••• + y ano + ••• + \fa~r,
< yai
+ ... + V a n o _j + ^e2™0 + ... + v ^
< y ai + .. + y a„ 0 _i + e2"0 ^ / l + - + VT.
By the foregoing problem,
bn<
N «i +
..+v~-i+«
2
"
, i :
^-
This means that {bn} is bounded above and convergent. Assume
now that a > In 2. By assumption, given e > 0, there is an no such
that ln(lna n ) > n(a 4- e) for n > n 0 . Setting a 4- e = ln/3, we get
a n > e^ for n > no, where /3 > 2. Thus
= y a i + y a 2 + ... 4- y a ^ 4- ... + V ^
> y ax + ... 4- V a n o _i + e2"-o+i
> e(?y\
In this case, the sequence {6 n } diverges to infinity.
Note, additionally, that if 0 < an < 1, then, although l n l n a n
is not defined, the sequence {bn} is monotonically increasing and
bounded above by l y 5 , and is convergent.
2.5.76. [20] It follows from the assumption that 0 < an < na\. Thus
the sequence {^f} is bounded. Denote by L its limit superior. Then
there is a sequence {m^} of positive integers such that lim -^ = L.
For an arbitrarily fixed n e N, we can write m^ = nbK 4- r^, where
r/c G {0,1,...,n — 1}. Thus, by the assumption, amk < lkQ>n 4- arfe.
Hence
mfc
< —;—;
^n H
•
n/fc 4- rk
mk
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239
2.5. Miscellaneous Problems
Letting k —> oo, we get
w
i < -,
n
which implies
lim — < lrm —.
n—oo n
n _+oo
Tl
As a result the sequence { ^f } converges.
2.5.77. Analysis similar to that in the solution of the preceding problem can be applied.
2.5.78. [20] The sequences {an + 1} and {1 - an} satisfy the assumption of Problem 2.5.76. Hence lim ^ ^ and lim ^ ^ exist
n-^oo
n
n—>oo
n
and are finite.
(a) By the above, since lim 9^L^- = g, lim ^ — g.
n—•oo
n
n—>oo
n
(b) The inequalities follow immediately from (*) in the solution of
2.5.76.
2.5.79. We will show that the sequence {^f} converges to A =
s u p { ^ : n G N}. Let p be an arbitrarily fixed positive integer. Then
n
pln + rn ~ pln + rn '
where rn e {0, l,...,p - 1}. So, by our assumption,
lim ^
> ^.
71—>O0
This, in turn, implies that
lim
9L2L
> lim ^ . Thus the convergence
of the sequence {^f} has been established. Moreover, ^ ^
implies that
>
^
A > lim — > lim — = inf sup —
n^oo
n
P-^OO p
p i>p
I
> ini sup —-— > inf sup — = A.
P men pm
p m rn
2.5.80. We first show the boundedness of the given sequence. Indeed,
if \ < a>n,Q>n+i < a, then \ < a n + 2 = an -fL+i < a. Thus, by the
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Solutions. 2: Sequences of Real Numbers
240
principle of induction stated in the solution of Problem 2.1.10, the
sequence {an} is bounded. Put
I = lim a n ,
L = lim an.
Then for an arbitrarily fixed e > 0 there exist n i , n 2 GN such that
(i)
an < L + e for
n > ni,
(ii)
an> I — e for
n > n2.
By (i), a n + 2 = — - A — > r x 7 , n > n i . Since the positive e can be
arbitrarily small, we get / > -^. In much the same way (ii) implies
that L < j . Thus I = •£- Let {rik} be a sequence of positive integers such that lim a nfc+2 = L. We can assume that the sequences
k—KX>
{a n f c + i}, {ank} and {a nfc _i} converge to /i, / 2 , and / 3 , respectively.
In fact, if this is not the case, we can choose subsequences which do.
By the definition of { a n } ,
2
ii 4- h = y = 2/
and
2
J2 + /3 = 7-,
and since / < li,h,h < L, we get /1 = l2 = I and /2 = ^3 = L.
Hence I = L. This and the equality I = ^ imply that the sequence
{an} converges to 1.
2.5.81. Since 0 < a\ < &i, there exists <p € [0, \) such that a\ —
b\ cos (p. Now, one can show by induction that, for <p ^ 0,
an+l
Therefore
61 sin (p
= ¥^
and
lim an = lim bn =
bn+l =
2^%'
neN
-
fcisini
^. If </> = 0, i.e. ai = 61, then
n—•00
n—->oo
bi sin <p
^
the given sequences {a n } and {bn} are constant.
2.5.82. [181 By assumption, ^ ^ = 1 + £fcn, where £fcn tends to
zero, uniformly with respect to k. Thus
n
(*)
^2
k=l
n
a/c n =
'
n
6/c n +
X^ '
fc=l
Yl ek>nbk>n'
fc=l
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2.5. Miscellaneous Problems
241
Since lim Yl ^k n exists, there is an M > 0 such that
J2 t>k,n
k=l
M, n e N. Moreover, for any e > 0, \£k,n\ < -h
M for k = 1,2, ...,n,
provided n is sufficiently large. Hence
that
n
£
lim ^
>fc=i
n
/ j
£k,nOk,n
< e, which means
k,nbk,n = 0. Thus, by (*)
= lim
lim Y]ak,n
fc=l
Y]bkin.
fc=l
2.5.83. We have
s i n ^ ^
(2fc-l)q
n-»oo
1
uniformly with respect to k.
Thus, by the foregoing problem,
lim
^
. (2fc-l)a
y sin
^ (2fc - l)a
n—^oo ^—^
fc=i
fc=i
= a.
H'
2.5.84. It follows from 2.5.5 that, if the sequence {xn} converges to
zero, then a n,~* —• 1. This implies that
a^ - 1
1
A In a n~>0°
uniformly with respect to k. Now applying Problem 2.5.82 we get
lim >
( a ^ 1 — 1) = lim In a >
n ^ o o ^—' \
k=l
/
n-*oo
kk
11
2
2
—=• = - In a.
^—^ n
k~l
2.5.85. If {xn} is a positive sequence that converges to zero, then,
by Problem 2.5.3, l n ( 1 + a ") —• 1. Applying 2.5.82, we see that
Xn
lim y
n—>oo ^—'
fc=l
n-»oo
In ( 1 + -rJ = lim y
z
\x
n
7
I
-= = -r.
n—»oo ^—' nz
fc=l
2
Thus lim U (l + A ) = e*.
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Solutions. 2: Sequences of Real Numbers
242
2.5.86. One can show that if {xn} is a positive sequence that converges to zero, then
(*)
(l±x»)iz! _ !
Set
c
k,n
=
kq~1
~ >
AC = 1 , 2 ,
...,71.
Then c^ n < max{^ , —•}, and consequently, {ck,n} converges to
zero, uniformly with respect to k. Putting a ^ n = (1 4-Cfc>n)p — 1 and
bk,n = ^CA:jn, and using 2.5.82, we obtain
hm >
14n—>oo ^—' \ y
fC-—J.
\
- 1
n9 /
I
/
= - hm >
V n~*°° ^^
n<?
.
fC—— 1
By the Stolz theorem (see 2.3.11),
hm >
n-^oo^
fc=l
77/?
= hm
-— = hm -—-——
-
n->oo fl? — (n — \)q
n-^oo
nQ
n^1
— iim
B
"*~n»-^(1-^ + ^ 4 , - . . . )
=
—^
M _ l ) y
V
71/
— nq
1
__
^
2.5.87. Set a„ = gffg;;;<ffff. Then
a
_«
6
(i+a-(i+»g)
(i + 5 + ( ! - i ) ) ( i + 2^ + ( * - i ) ) - ( i + »g + ( 5 - i ) ) '
Now let x = - - 1. Then z > 0 and
a
_ a
a
"
=
'(
1
1 +
T%)( 1
+
T%)-(1 + H % ) '
Since
x
x
I^
x
\
I
x
d
'
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2.5. Miscellaneous Problems
243
we get
a•nn ^<
Hence
hx
/
\ •
(iqhr + T^I + - + iTZz)
lim an — 0, because
n—>oo
lim
j H
n-ooll + 2
J + ... H
1+2^
j
1 + n M
= +oo.
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Chapter 3
http://dx.doi.org/10.1090/stml/004/06
Series of Real N u m b e r s
3.1. Summation of Series
3.1.1.
(a) We have a\ = S\ = 2 and an = Sn-Sn-i
oo
we get the series 2 — J2 „,}
n=2
(
n
^
= ~ yi ( n 1 _ 1 ) ? n > 1. So
, whose sum is S = lim Sn = 1.
n-
(b) As in the solution of (a), we get an = ~ ,
*°°
oc
XI W
^
n=l
= 1.
(c) By a similar argument, an = arctann — arctan(n — 1), and consequently, tan an = n s _ 1 n + 1 . Therefore an = arctan n s _ 1 n + 1 anc
E arctan ^
n.= 1
^
= 1.
(d) ai = - 1 , a n = ( - l ) " ^ 5 i )
for n > 1. Moreover,
- 1 + E(- 1 ) n 4 ! L : : ^ = 0±i
n(n-l)
3.1.2.
(a) We have a n = 4j lim S n = 1.
(n^1)5.
Thus 5„ = 1 - T^rrp and 5 =
245
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Solutions. 3: Series of Real Numbers
246
(b) Similarly, an = § (^n-i)*
I (* - (2^)
(c)
a
and 5
- (2n+i) a ) • J t t h e n
follows
that S„ =
= ^ ^ = §•
» = ^ T " ^ - Therefore S n = ^
and S = ^ S
n
= 1.
(d) on = | ( 2 ^ 1 - 2 ^ 1 ) • Thus S = Jirr^ 5„ = \.
(e) a„ = ^fj-^
yn(n+l)
= 4 j - ^ i - j . Consequently, 5 = lim S n = 1.
v
n
vra+l
n—>-oo
3.1.3.
(a) Sn = l n l - In 4 + In 2 + In 4 - In 1 - In 7 + In 3 + In 7 - In 2
- In 10 + ... + Inn + ln(3n - 2) - ln(n - 1) - ln(3n + 1)
n +1
+ ln(n + 1) + ln(3n + 1) - In n - ln(3n + 4) = In
3n + 4'
Hence S = In | .
(b) 5 = In 2.
3.1.4. (a) We have
0>n =
1
n(n + l)...(n + m)
1 /
1
m \ n ( n + l)...(n + m - 1)
(n + l)(n + 2)...(n + m ) / '
Hence Sn = — ^^.....m ~ (n+l)(n+2)...(n+m)J ' ^ ° ' ^
(b) Since aB = l ( i -
s
=
^
!'
k ) , S = i ( l + i + ... + i ) .
(c) We have
n2
2
( n + l ) ( n + 2)(n + 3)(n + 4)
( n + l ) ( n + 2)
1 1 / 1
1
2 V ( n + l ) ( n + 3)
(n + 2)(n + 4)
l
1
1
+ l(
l)(n + 4)
(n + 2)(n + 3 ) , T
1
(n + 3)(n + 4)
4 V(n + 1)(
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247
3.1. Summation of Series
Now, as in (b), after simple calculations we get S — J^.
3.1.5.
(a) For n > 5, we have
^=
sin
7io + s i n 3io + s i n iio + S i l l ^ + S i n ^
(b) Observe that 0 < - ^ - < 1, n € N. Thus 5 = 0.
v
/
— n — Inn
'
3.1.6. Since an = sin ^ T T cos ^ F T = \ (sin ^ r - sin ^ ) , we see
that S = \ sinl.
3.1.7. Note that
1
n!(n + n 2 + l)
1/
n
2 V (n + l)!((n + l)n + 1)
4
n- 1
n\{n(n - 1) 4-1)
+ •
1
(n + 1)! / '
Hence
+I+
1 /
n
2 \^(n+l)!((n+l)ri+l)
&n
S<^)'
and by 2.5.6, we get S = lim Sn = \e.
n—KX>
3.1.8. Note that for n > 1,
n
1
2
(2n+l)-l
3-5-...-(2n+l)
1 /
1
2 V3-5-...-(2n-l)
3 - 5 - . . . - ( 2 n + l)
and a,\ = | . It follows that
5 n
X
3
| * f1
+
2^3
1
\
3-5-...-(2n+l)/'
which implies the desired result.
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Solutions. 3: Series of Real Numbers
248
3.1.9. As in the solution of the last problem, we have
an
an + 1 - 1
(ai + l)(a 2 -f l)...(on + 1)
(ai + l)(o 2 + l)».(a n + 1)
1
1
(ai + l)(o 2 + l)...(a n _i + 1)
(ai + l)(a 2 + l)...(a n + 1)
1
(ai+1)(aa+ 1)...(an+1)-
for n > 1. Hence Sn = 1 3.1.10.
(a) If we take an = n—\ in the last problem, we have g = +oo, and
so the sum of the given series is equal to 1.
(b) Here we take an = 2n — 1, and, as in (a), the sum of the given
series is equal to 1.
(c) We take an = — ^y. Then
lim ((a 2 + l)(a 3 + l)...(an + l))
n—•oo
=
iiTn ( 2 - l ) ( 2 + l ) ( 3 - l ) ( 3 + l) ( n - l ) ( n + l)
n-ioo
22
32
"'
n2
1
2'
Thus, by the result in 3.1.9, the sum of the given series is equal
to 1.
3.1.11. By definition, the sequence {an} increases to infinity. Moreover, we have a 2 - 4 = a^l_1(a^l_1 — 4), and it can be shown by
induction that a 2 - 4 = a 2 • a 2 • ... • Gn-i( a i ~ 4). Hence
lim
^
= , / a 2 - 4.
v
n-^oo ai • a 2 •... • a n _i
(1)
Note also that for n > 1,
1
ai • a 2 •... • an
1 /
an
2 \ a i • a 2 •... • a n _i
^n+i
a\ • a 2 •... • a n
Thus by (1) the sum of the given series is equal to
ai
2ai
2
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249
3.1. Summation of Series
3.1.12. Observe that
1
6
1!
2!
6- 2
(6-2)6 '
2!
3!
(6-2) 6
(6-2)6( 6 + 1) '
2!
6(6 + 1)
n!
6(6 + 1)...(6 + n - 1)
(6 - 2)6(6 + 1)...(6 + n - 2)
(n+l)i
( 6 - 2 ) 6 ( 6 + l)...(6 + n - l ) '
Summing the above equalities, we obtain
s
n
=
_1
6-2
(n+1)!
(6-2)6(6 + l)...(6 + n - l ) '
Hence, by 2.5.87, lim Sn = i - ^ .
n—>oo
° •*
3.1.13. For n = 0,1,..., put
a n (a + n -f 1). Then ^4n_i —
we additionally set J4_I = a
the above equalities from n =
a - AN
= A-i
a n = t i w t f f i n ) » a n d d e f i n e A" =
An = an(b — a — 1), n = 0,1,..., where
and a_i = 1. Summing both sides of
0 to n = N, we get
- vl^ = (6 - a - 1) ^
a n = (6 - a n=0
or equivalently, a - aw{a + N + l) = (b-a/
H
1)SJV+I,
l)5//+i. Therefore
(a + l ) - ( a + iV + l ) \
6(6+l)...(6+AT) J=<*-a-l)Sw + i
and, by 2.5.87, we see that
lim
SJV+I
= t,_°_r
3.1.14. By the foregoing problem,
U
x p o(o + l)...(q + n - l )
^
1)...(6 + n -- l 1)
6(6 + l)...(6
)
+
a
6 -- aa --l l
6-1
6-a-l'
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Solutions. 3: Series of Real Numbers
250
Replacing a by a + 1, we see that
(**)
(a+l)...(a + n)
6-1
1 +^ E6(6 + 1)...(6 + n - l )
b-a-2'
Subtracting (*) from (**), we obtain
y , (o+l)...(a + n - l )
^ n 6 ( 6 + l)...(6 + n - l )
6-1
b-a-2
6-1
b-a-1
3.1.15. Set An =
(6-1)
(6 - a - 1)(6 - a - 2)'
(a2+b)CT);;;^+1+fc)
and Sn = ± Ak. Then
k=l
X^T = akl1+b> o r equivalently, Akak+i 4 4*6 = Ak-iak.
Summing
both sides of these equalities from k — 2 to k = n, we get
(*)
4 n a n + i 4 Snb - Aib = Aia 2 -
Now note that
0 < A n a n + i = a\
a 2 • as • ... • a n + i
(a 2 4- 6)(a 3 4- 6)...(a n +i 4- 6)
1
(i + £)( 1 + £ ) - ( 1 + =£r)'
Hence, by 1.2.1,
.
0 < Anan+i <
ai
n+l
&£ -
Therefore, by assumption, lim A n a n + i = 0. It then follows from (*)
that lim Sn =
Al(fc +a2)
,
n—>oo
= fii.
3.1.16. By the trigonometric identity 4 cos 3 x = cos3x 4- 3cosx, we
get
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3.1. Summation of Series
251
4 cos 3 x = cos 3x + 3 cos #,
4 cos3 3x = cos 32x + 3 cos 3a:,
4 cos3 3 2 x = cos 3 3 x + 3 cos 3 2 x,
4 cos 3 3 n x = cos 3 n + 1 x + 3 cos 3 n x.
Multiplying both sides of these equalities by 1, - | , ^ - , . . . , ( — l ) n ^ r ,
respectively, and then summing them we obtain 45 n = 3cosx -f
( - l ) n ^ c o s 3 n + 1 x . Thus 5 = fcosx.
3.1.17.
(a) By assumption,
f(x) = af(bx) + cg(x),
af{bx) = a2f(b2x)
a2f(b2x)
= a3f(b3x)
an-lf(bn-lx)
Hence f{x) = anf(bnx)
Since lim anf(bnx)
+ acg(bx),
+ ca2g(b2x),
= anf(bnx)
+
aT^cg^^x).
+ c (g(x) + ag{bx) + ... + a71'1 g^-1
= L(x), V <*"£(&"*) =
/(x)
(x)
x)) .
-
(b) As in (a),
/ ( x ) = a/(6x) + c#(x),
a~ 1 /(6- 1 x) = / ( x ) + a " 1 ^ - 1 * ) ,
a-2f(b-2x)
=
a-1f(b-1x)+ca-2g(b'2x),
a-nf(b-nx)
=
a1-nf{b1~nx)+a-ncg{b~nx).
Thus
o/(6x) = a - n / ( 6 - " x ) - c (g{x) + a^g^x)
+ ... +
a-ng(b~nx))
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Solutions. 3: Series of Real Numbers
252
and consequently,
x \
M(x) -
af(bx)
T-g(-) =
n=0
3.1.18. We may apply the foregoing problem to the functions f(x) =
sinx, g(x) = sin3 f with a = 3, b = | and c = —4. The desired
results follow from the equalities lim 3 n sin ~ = x = L(x) and
lim 3~ n sin3 n x = 0 =
n—>oo
a
n—•oo
M(x).
3.1.19. One can apply 3.1.17 to f(x) = cotx, g(x) = t a n x , a
2, 6 = 2 and c = 1, and then use the equality
lim — cot — = —.
n->oo 2 n
2n
X
3.1.20. We apply 3.1.17 to
f(x) = arctanx,
g(x) = arctan
(1 - b)x
— j - , a — c— 1,
and use the following:
for
( 0
lim arctan(6 n x) = < TT
n-+oo
I —signx
C
for
0 < 6 < 1,
b > 1.
3.1.21. Since a n +i = a n + a n _ i , we have a n + i a n = a£ + a n _ i a n
for n > 1. Summing these equalities, we get
(*)
S,n = a n o n + i ,
n > 0.
One can prove, by induction, that
»> ^((^r-(^T''-°.
(ii)
an_1on+i-aS = (-l)n+1,
n>l.
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253
3.1. Summation of Series
Combining (*) with (ii) yields
(~l) fc
=
y > (~l) fc
'O-k-l
\
k=1
a
a
k
=
! _ V^ Qfc-iQfc+i - a !
0-k \
0-n
fc+l/
&n+l
By (i),
lim —— = lim
+T
" ^ °" " " ^ ^ ( ( ^ P - (^)" + 2 )
(iii)
1 + V5'
3.1.22. It is easy to check that
( - l ) n + 1 = an+ian+2 - a n a n + 3 ,
(*)
n > 0.
Thus
<? — V ^
(
j l
)fc
— _ V ^ f^h±l
_
V ^ Qfc+lQfc+ 2 - QfcQ/c+ 3
_
afe
+ 3> \ _ _q i a ^+2
Qn+3
Now, by (iii) in the solution of the foregoing problem,
lim Sn =
n—•oo
V5-2.
3.1.23. By (*) in the solution of the preceding problem,
arctan
arctan
G2n+1
= arctan
= arctan
^2n^2n+3
n+
^271+1^271+2 + 1
0>2n+2
0>2n
n
,
1
= arctan -^ 2 n + 3
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Solutions. 3: Series of R e a l N u m b e r s
254
Summing up these equalities gives
1
^
1
arctan — = > arctan
dl
^
h arctan
&2k
i
G2n+3
.
oo
So, E a r c t g ^ = f.
n=l
3.1.24.
(a) Note that a r c t a n ^ = a r c t a n ^ j — arctan^W, n > 1. Hence
oo
Y arctan ^
= arctan 2 4- arctan 1 4- arctan ^ = |7r, where the
71=1
last equality follows from the fact that arctan a 4- arctan ~ = |
for a > 0.
(b) For n £ N, arctan 2 -1 _L1 = a r c t a n - — arctan —W. Hence we
oo
see that Y] arctan $} _L1 = arctan 1 = W .
n=l
(c) Since for n > 1, arctan
oo
n4 _4n
2
+5
= arctan
( n -i) 2 ~ a r c t a n (n+i)2>
we get, as in (a), Y axctan n4_%%a+5 = arctan2 4- arctan2 4n=l
arctan | = ^7r 4- arctan 2.
3.1.25. To obtain the desired result one can apply the trigonometric
identity
x—y
arctan x — arctanyy = arctan
.
l + xy
It is worth noting here that the results in the preceding problem are
contained as special cases of this one.
oo
oo
n=l
n=l
3.1.26. Let Y, bn be a rearrangement of YL an- Moreover, let Sn =
ai + a>2 + ••• + a n , S' = 6i 4- b2 + ... 4- bn and S = lim S n . Clearly,
71—>-CO
Sn < S. So {Sn} converges, say to S", and Sf < S. By similar
arguments, S < Sf.
3.1.27. Since
2n
fe=i
fc2
^ ( 2 ^ ) 2 - ^ ^ - 1 ) 2
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3.1. Summation of Series
255
and lim S2 n +i = lim S2n> we get
n—• oo
oo
-
oo
E ^2 = Z-> ( 2 n 1- l )
l
i
^—v
1 ^—\ 1
2 +
n
n=l
n=l
v
22 2 ^ ^ 2 '
n=l
'
3.1.28. Following A. M. Yaglom and I. M. Yaglom, Uspehi Matem.
Nauk (N.S.) 8 (1953) no. 5(57), pp. 181-187 (in Russian), we present
elementary proofs of these well known identities.
(a) For 0 < x < | , the inequalities sinx < x < tanx hold. Thus
cot 2 x < ^2 < 1 + cot 2 x. Putting x = ^ ^ with k = 1, 2,...,m,
and summing from k = 1 to k = m we obtain
/.x \~^
9 kir
(2m-h l ) 2
cot
M >
~
7<
o-2w
^
2m + 1
7r2
fc=i fc=i fe=i
We now show that
\-^
9 ^TT
(ii)
> cot 2
w
f^
2m+ 1
\-^ 1
v^
9 kn
> To < m + > c o t «
^
fc2
^
2m + :
m(2m — 1)
- —'—'-.
3
Let 0 < t < § . By DeMoivre's law
cos nt + i sin nt = (cos t -f i sin t)n = sin n £(cot £ + z) n
= sin
n
^
Wifccotn-fct.
Taking n = 2m + 1, and equating imaginary parts, we get
sin(2m + l)t = sin 2 m + 1 tP m (cot 2 1),
(iii)
where
(iv)
Pm(s)=^
2m + l \
/2m + l\
m
^ J ^ - ^
3
x
J x ^ + .-.il.
Substituting t = ^ p i i n ( ^ ) gives P m (cot 2 ^ p j ) = 0- So,
x/c = cot 2 2^+1? & = 1,2, ...,m, are the zeros of P m and their
sum is
w
2m +1
2 fcTr _ ( 3 ) _ m(2m - 1)
X> 2m+1-r+1)- 3
TO
«=i
v
1
/
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Solutions. 3: Series of Real Numbers
256
This, (i) and (ii) imply
m(2m - 1)
<
(2m + l ) 2 ^
„2
1
Ep<m+
£v
an
Multiplying these inequalities by (2TTH-I)2
oo yield equality (a),
(b) To prove the second identity, note that
d
m(2m - 1)
next
letting m
/2m+l\
— Q V
5
/
where Xk, fc = 1,2,..., m are the zeros of the polynomial (iv). It
then follows by (v) that
£cot 4
fc7r
2m+1
m(2m - 1) ^ 2
2m(2m - l)(2m - 2)(2m - 3)
&3
/
5!
_ m(2m - l)(4m 2 + 10m - 9)
~
45
'
The inequality cot 2 x < 4? < 1 + cot 2 x (see (a)) implies that
cot4 x < J J < 1 + 2 cot 2 a; + cot 4 # for 0 < x < §. Consequently,
m(2TO-l)(4m 2 + 1 0 m - 9 )
45
<
(2m + l ) 4 y ^ 1
w5
^ fc4
„ 2m-1
m ( 2 m - l)(4m 2 4-10m - 9)
< m + 2m—
h —^—
-.
3
45
Thus (b) is proved.
Remark. It is worth noting here that the above procedure can
oo
be applied to calculate the sum of the series ^
n=l
- ^ , where fc e N.
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3.1. Summation of Series
257
(c) It follows from DeMoivre's formula, that for m = 4n, n G N,
cosm* = cos m t - (™ I cos m ~ 2 £sin2 * + ...+ sin™ t,
m\
sin mi =
_
Vi/
and consequently,
cosmm l x£ sin £ + . . . -
I
i
m
. .
I costsin771m
V™ - 1 /
-i.
£,
cot m t - ( m ) c o t m " 2 1 + ... - (m™2) cot 2 t + 1
cot mt =
(™)cotm-H-C2)cotm-H
+
....-{™_1)cott-
It follows from the last equality that
4/C7T - f 7T
Xk = c o t —
Am
are the roots of the equation
,
A; = 0, ...,m — 1,
?>"-'-(!>"
x">-|7) I "-'-(™)x"- J + - + ( m ™ 1 l « + l=0,
which implies that
(1)
> cot — ;
Since m = 4n,
E
4 b
cot —
+ 7T
Am
fc=0 fc=0 fc=2n
E
cot
v~^
4fc7T + TX
= > cot —
*-*'
Am
4/C7T + 7T
—
= m.
4m
f n
fc=0
Am
x~^
\
f—'
cot
4fc7T — 7T
—
Am
x~^
4fc7T + 7T
h > cot —
*—'
Am
.
fc=0 fc=l
This and (1) give
7T
[/«x
]
Since
cot
37T
cot
57T
h cot
4m
4m
(2m - 3)TT
H-... + cot - —
4m
77T
cot -—
4m
4m
(2m - 1)TT
cot - —
— = m.
4m
cot a - cot /3 = tan(/3 - a) [ H
),
VM
;
V
tan a t a n / 3 / '
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Solutions. 3: Series of Real Numbers
258
we get from (2) that m is equal to
7r (m
1
1
tan —
—
+
—
o
+
.
.
.
+
2m {2
tan^tan^
" tan &£* tan &£pL
J'
Hence, by the inequality ^ > ^ ^ for 0 < x < 7r/2,
. .
(3)
7r / m
1
1
\
m < tan — I - + - ^ + ... + ^ _ 3 ) 7 r (2m _ 1)7r •
\
4m 4m
4 m
4 m
/
On the other hand, we have
sin(/3 — a )
n
cot a — cot p = — ;——,
sin a sin p
and as above, using the inequality ^ < ^A^, we obtain
7T /
.
1
m = sin -—
2 m ^ s i n ^ s i n o^
.
7T /
> s m 2m
—
1
„ ^
l -2-iL?L
\ 4m 4m
+ '.". . +
h -... + s i n I
1
2
^
\
sin
(2^zl)ZL
"\
1
(2m-3)7T (2m-l)7r J '
4m
4m
/
This and (3) yield
m
tan ^
2m
m\ 7r2
2/16m2
1
1
< 2^-^
- + (2m-3)(2m-l)
- 3 + ""
r2
< 16m sin ^
7T 2
Letting m - > o o w e get
1
2-35-7
h ... = - > ( - 1 )
= -.
2 m^= 0: } 2m + 1
8
3.1.29. We have a n + i - l = a n ( a n —1). Hence a 1 _ 1 = — ^- + ^ z j Summing these equalities from n = 1 to n = N, we get
/ x
(*)
1
1
1
— + — + ... + — = 1
1
It is easy to check that the sequence {an} increases and diverges to
oo
infinity. Thus (*) implies J2 ^~ ~ 1n=l
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259
3.1. Summation of Series
3.1.30. By the definition of the sequence,
eai - 1 - a i e a 2 ,
e a2 - 1 = a2ea'\
Hence
eai - 1 = ai + a1a2ea3
= ... = ai + aia 2 + ... -f a\ • ... • a n + ai • ... • a n +ie a n + 2 .
oc
This implies that ^ 6 n = e a i — 1, because
n=l
lim (ai • ... •a n + 1 e a " + 2 ) = 0 .
n—>oo
Indeed, {a n } is bounded below by zero, is monotonically decreasing,
and converges to zero.
3.1.31. We have S n +i = Sn 4- a n + i = Sn + ^
\/2. Consider
the function /(x) = a; + ^ — \/2, x > 0. If the sequence {5 n }
were convergent to 5, we would get / ( 5 ) = 5. The only solution
of this equation is 4=. Moreover, the function x —• f(f(x)) — x is
monotonically decreasing on the interval (4=,1). So, if x E (-^, 1),
then
/(/(*))-z </(/(-I))--L=0,
and since / is monotonically decreasing on the interval (0,1), we also
have f(f{x)) > ^= for x e ( ^ , 1 ) . Finally,
— <f(f(x))<x
for
x
€
( _ l ) .
This means that the sequence {52n-i} is monotonically decreasing
and bounded below. Thus it converges, and its limit is -4=. Furthermore, lim S2n = lim / ( 5 2 n - i ) = / (~k ) = ~k- So, the sum of the
n—>-oo
n—•oc
given series is equal to -4=.
\v^/
v^
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Solutions. 3: Series of Real Numbers
260
3.1.32.
(a) Note that
9
1
-1
+
1
l
1 1
2 3
*
1
2n
/
1
1
2
1
n
\
3
1
1
1
n+ 1+—
n + ^2 + - + 2n
So, by 2.5.8 (a), we have lim S2n = In2. Clearly,
n—>oo
lim 5 2 n +i = lim ( S2n + 7;—-7 I =
n—>oo
n—•oo y
(b) We have ^
*~C
-
= 1 + ^ . Hence by (a),
oo
n=l
ln 2
ZTl "T 1 /
0
'
oo
1
n{n + l)
^ }
'
n—\
v
1
'
n
oo
^
n=l
1
'
n +1
= l n 2 - ( l n 2 - l ) = l.
(c) Denote by Sn the nth partial sum of the given series. Then
1
1
1
1
S2n = z +, n2n +, ,1 H- x +, o2n +• n2 + - + —
x +- A4n - 71 + x + 4n
As in the proof of 2.5.8, one can show that
Obviously, lim S2n+i = ln2.
lim S271 = In 2.
n—+00
n—>oo
3.1.33. We have
, 2 , 3 , 4 , 5
,
2n
, 2n + 1
5 2 n = In - - ln - + In - - ln - + ... + In
- ln —
1 2
3
4
2n - 1
2n
2-4-...-2n
3-5-...-(2n+l)
n
~ ni-3-....(2n-l)
2-4-...-2n
mJ
1 t (»)n ^
2n + l V ( 2 n - 1 ) ! !
By the Wallis formula, \ = lim ^ p x ( ff L )
• Hence lim Sin =
Inf.
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3.1. Summation of Series
261
3.1.34. We have
-^-""K^HO-STl))
= - £(-!)»-' In (l + i ) - £ ( - ! ) - ' In (l + i ) .
It follows from the foregoing problem that the sum of the series is
ln2-21nf.
3.1.35. The nth partial sum of the given series can be written as
S n = 1 + I + ... + 1 - ( I n 2 - l n l + l n 3 - l n 2
2
n
+ ... + ln(n + 1) - Inn) + 1 + - + ... + - - ln(n + 1).
2
n
Thus, by Problem 2.1.41, the sum of this series is equal to the Euler
constant 7.
X
3.1.36. [20] Write F{x) = J f(t)dt. It follows from the Taylor theo1
rem that there exist #&, yk such that k < Xk < k + ^, k + | < yk <
k + l and
-F (k + 1 ) + F(k + 1) = i/(fc + 1) - i/'(y f c ).
Summing the above equations from k = 1 to k = n— 1 yields
1/(1) + /(2) + /(3) + ... + f(n - 1) + i / ( n ) - F(n)
= | (/'(»i) - / ' ( * i ) + /'(ift) - /'(*2) + ... + /'(»„_i) - / ' ( I n . i ) ) .
The limit of the expression on the right-hand side of the last equality
exists, because the series —f'(xi) + f'{yi) — /'(#2) + f'fa) ~ ••• is
convergent (the terms have alternating signs and their absolute values
converge monotonically to 0).
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Solutions. 3: Series of Real Numbers
262
If we take /(x) = ^, then we can prove the existence of the limit
lim ( 1 + - + - + ... + - - I n n
n->oo \
2 6
n
(Compare with 2.1.41 (a).) Taking /(x) = lnx, we can show that the
sequence {Inn! - (n + | ) In n + n} converges. (By Stirling's formula
its limit is In \Z2n.)
3.1.37. Applying the foregoing problem to f(x) = ^ ,
can show the existence of the limit
(lnn)2\
i _ ^ ] = s.
2
,.
/ml
In 2
Inn
hm — + _ + ...+
n->oo \ 1
2
n
Hence
hm
n—>oo
S2n = n—>oo
hm
,
lnl
V -——
1
In 2
+ —2
x > 0, we
••• +
ln2n
2n
, /lnl
In 2
ln2n
(ln2n) 2
v
J
= n-ooV
hm - \ _ 1 + —2 + ... -f- 2n
2
In4
ln2n\
(ln2n) s
rt ,/ I-n„2
V
;
+ 2 V~2"
— +—
4 + ...+ 2n *
2
,.
/lnl
In 2
Inn
(Inn) 2
- 5 + n—UDO
hm —1 + —2 + ...+ n
^ 2y
2
,.
/In2
In2
In2 , „, \
v(ln2)
;
+ hm — + -— + . . . +
ln21nn'
v
n^oo \
,
nf
1
2
n
In 2
where 7 is the Euler constant.
3.1.38. By Stirling's formula, n! = an\/2iTn (-)n , where lim a n
1. Hence
n
_1
~ 2
n
(2n + l ) 2 n
_ 1
2 2
((2n - l)!!) e " ~ 2
(2n-f l) 2 n 2 2 n (n!) 2
((2n)!) 2 e 2 "
_ lin(2n+l)2-22-a227rn(f)2n ^ 1 ^ / ^ + 1 V "
2
a2n47rn(2f)4ne2-
2
^
2n
;
a2 *
2a2^
Therefore lim Sn = ±(1 - In 2).
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263
3.1. Summation of Series
3.1.39. Assume that the given series converges for x and y, and write
SN(X) = g
(j n --i)]fc + i
+
+ 2 + '" + ^ ~ T " riz) '
(n-l)k
N
Then SN(X) - SN(V) = ^ p 2
n- Therefore the convergence of the
U
n-l
series implies that x = y. Now, we find the unique value x for which
the series converges. We know, by Problem 2.1.41, that the sequence
an — 1 + | + ••• + ^ — ln(nfc) converges to the Euler constant 7.
Hence
SN(k-l)=aN+ln(Nk)-'£^k-E,1^L
«7V
This implies that
n=l
n=l
+ lnfc+ [In AT- V -
I.
lim 5jv(fc—1) = 7+lnfc—7= In A:. Thus x = fc—1,
N—>-oo
and the sum of the given series is equal to In fc.
3.1.40. One can easily check by induction that
a>2n = 3n + 2
for
a2n-i=3n-fl
for
n = 0,1,2,...,
n = l,2,....
Thus
, ,
n=0
°"
/
;
\3n+l
X
X
^-fv
'
W
71=1
n=0 a 2 «
( 3 n + l )/(v 3 n + 3)
n=0
3T2^-f
i
,
3n + 3 J
'
71=1
\Zn
= i + i^ ( _ 1; )-fi._^_Vif(-i)-p
V
3
2^-f
n=l
V3n
v
3n + 3 /
'
2 ^
n=l
2«-l
*
N
2 ^
7
fl
' 3n(3n + 2)
n=l
\
„=1
'
X
Un+1
x
3n + 2J
'
M
3n + 2>/
/
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Solutions. 3: Series of Real Numbers
264
6+^(
3
n=2
' 3n
+
6(JV+l)
v
7
+
2^
Un+1
'
n=l
3n +
x
On the other hand, by 3.1.32 (a), we get
• l n 2 = lim f V ( - l ) n - |
\n=l
/
=N^ool^
lim (Y(-1)»±\
v
y
3nj
\n=l
-AT-XX>
limI fVC-l)"
( — ^ -3n^-i^
' l3n+l
+
/
x
\n=0
1
:-iln2- lim fv (-l)«f_J_-_L-)V
This implies
lim
N-l
£ (_1)» ( _ ^ - _ i _ )
= f i n 2. Finally, by
lim S2N
= i6 — 3| "l n1 A2 ^+ 3i -r+ 3l1 In
2— 4\ = 4|' . Moreover, since
—
"*
(*),
N_OQ
lim
N+ l
52JV+I
= lim S2N + lim /oW^ 2 _i = lim S2JV, the sum of
the given series is 4.
3.1.41.
(a) Suppose, contrary to our claim, that the sum S of the given series
is a rational number £. Then (q-l)\p
= q\S = £
ri=l
This implies that
<
Yl
n=q+l
$ +
£
$ .
n=q+l
n\ is an integer. On the other hand,
A g!!
1
1
+
: T + ( a
^
^
^
T
l)(g
«
+
l
(
9
+
!)(*
+ 22)
) i( 9 + l ) ( g + 2)2
n=(?+1
+ ...
<7 + 2 < 3
(<Z+1)2~4'
a contradiction. Thus 5 is irrational,
(b) The same method as in (a) can be applied.
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265
3.1. Summation of Series
3.1.42. On the contrary, suppose that the sum S of the series is a
rational number 2 . Then
V ^ Ql'£n
q
E
{q-iy.P=q>s=Y; -^
n! + *-—'
q^n
,
n=\
This implies that
n\
n=q-{-l
^ff ^s a n integer. On the other hand,
Yl
n=q+l
qy-£n
^
n=<?-fl
n!
To obtain a contradiction, it is enough to show t h a t
^
^f-
is not
equal t o zero. We have
>
n=q+l
q+l
^
n! > 9 + 1
q{q+l)
>0 ,
which proves that 5 is irrational.
3.1.43. Reasoning similar to the above can be applied.
3.1.44. Suppose, contrary to our claim, that ^
^~ ~ f > p, g E N.
2=1
By assumption, there exists a positive integer k such that if i > k
then n „ n * — > 3q. Thus
fc-1
2=1
n i n 2 • ...-nfe-ig
n;
+£ n n
x 2
• ... -rifc-ig
:pnin 2 •... -rifc-i.
i=k
Moreover,
• ... • rik-iq
1 / 1
£ -niri2
< - 1+—+
1 1
i=k
11
rii
6 \
nk
1
,
+ . . . ] < 1,
riknk+i
a contradiction.
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Solutions. 3: Series of Real Numbers
266
3.1.45. If the sum were a rational number - , then for any positive
integer fci, we would have
co
J^ •—- = | k=k\
CO
J2
q n i n2
n
k=k\
nkl l
~
k\ — 1
J2 ^~- Thus the sum
k=l
would be a positive integer, and we would get
oo
1
nk
k=k
... -n f c l -i
q-ni-ri2'
Set lim - IIiL - = / > 1 and take e > 0 so small that a = / — e > 1.
K—•OO
Then there exists an index fco such that if A: > fco, then
(**)
ftfc-i
> a > 1.
Since lim - — ^ — = -foo, there is ki > k0 such that
k—>oo ni'---'Tlk-1
^2_. Thus by (**
co
1
oc
>
n
nfe ajn~^a°nki
Uk
k=kl
ft
fc
^
ni-...-nfcl_i
ft -
<
(oL-l)nkl
q-rii • n 2 • ... • nkl-i*
which contradicts (*).
3.1.46. On the contrary, suppose that J2 •£- =
fc=i
are positive integers. Then
oo
nin 2 ...n / c _i V "
f^o
1
nk
+J
nfc
2
9
, where p and g
1
> -
<*
for all k > 2. (See the solutions of the preceding problems.) Set
&k = 2 \ / ^ - By assumption, lim ak = -foo. We claim that there
k—>co
exists an ri such that a,j < ari for j = 1,2, ...,ri — 1. Indeed, if
ai < a 2 , then ri = 2. If it is not the case, then we take the least
integer r\ > 2 such that a\ < ari. In fact, there exist an infinite
sequence of integers rk with the above property, that is, dj < ark
for j = 1,2,...,^ — 1. To find r 2 we apply the above procedure to
the sequence {ak}n>ri
and so on. Denote by r the least positive
integer such that a r + J > q -f 1 for j = 0,1,2,... and a,j < ar for
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3.1. Summation of Series
267
j = 1, 2,..., r — 1. Observe also that since nr < n r + J , ar < off+j for
j — 0,1,2,.... It follows from the above that
94.92,
n^/4
2J(2r-2)
,9r-i
n2r+J
n_L-
r+7
'V+7
&_ 1 „•
~~°r+j
^^i-J-J
Hence
OO
..
CO
-.
1
nin2...Vi E
< 5 > + )"
0+1)
= -'
nr4.j
g
a contradiction.
00
3.1.47. Since the series Y]
2lL
n=l
9n
converges,
follows from the given inequality that
n
lim - ^ r = 0. It then
-*°°
9n
00
^_1 > J2 V" • Suppose that
n=ra
the set A is finite. Then there is an index m such that
771—1
OO
Q — \~^ ElL — \ ^ En -u
^m
Therefore 5 is rational. Assume now that
s = f p^ = p71=1
Then
fL
CX J
x^Pk
V^ Pk s P n + 1
k£?+iqk
"Qn+i"l
k=i *«
Multiplying the inequality by bn — q • q\ • ... • qni we get
P
O n + l r n + l = O n r n g n + l ~ 9 ' Ql ' ••• * <7n+l
< KrnQn+i - bnrn(qn+i
<7n+l
- 1) = frnrn.
This means that the sequence {bnrn} of positive integers is monotonically decreasing. Thus, beginning with some value of the index
n, all its terms are equal, which implies that the set A is finite.
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Solutions. 3: Series of Real Numbers
268
3.1.48. Clearly, we can write n! = 2a^P{n), where 0{n) is odd.
More precisely, a theorem of Legendre asserts that a(n) = n — ^(n),
where u{n) denotes the sum of ones in the binary expansion of n.
oo
Moreover, we have Y2 7T7 = ^C ^ ^ r ? where Sn = 1 if n = rifc and
fc=l
n=l
oc
<5n = 0 otherwise. Suppose, contrary to our claim, that Y2 Sn^n —
2
n=l
Take TV = 2 r >
| , p, g G N. Write g = 2 % where * is odd.
max{*,2 s + 2 }. It then follows that ^ G N . Therefore 2 s /3(iV)| =
^ i p G N. A simple calculation shows that TV! = 2N~1/3{N) (which
also follows directly from the Legendre theorem). Multiplying the
equality
N
- = >
g
by 2sp(N),
(*)
0n—
^-;
/7
^—'
n!
77.!
+
>
^-f
* — '
On-7
T7.!
n!
i
we get
2s/?(7V)^ = 2 s /3(AT)V<$„^ + 2^(Ar)
V 6
n
^.
Note that
m2n
2TB(N)Ts^-rTs
2 li(M) 2^On
n—l
n]
i-A^Wfln)'
-l
v 7
n—\
Since /?(n) divides /3(iV), we see that the first term on the right-hand
side of (*) is an integer. To get a contradiction, we will show that
oo
O<2S0(N)
£
«n£r < 1. Indeed,
n=N+l
00
iL
on
-'
°°
*—'
n\
n=7V+l
00
2S+2
^~^
ATI
- — _/\r_i ^ ;
Z. ° ^
n!
n=N+l
S 2
2 + iV + 2
iV-hl N
on
<
n!
n=N+l
w
os+2
°°
OS+2
*
AT + I
L. I
n=iV+l
//
n
x
iV + 2
\ n-JV-1
S 3
2 +
7
N+l
< -Z-,
< I-
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3.2. Series of Nonnegative Terms
269
3.2. Series of Nonnegative Terms
3.2.1.
(a) We have
&n
=
v n2 + 1 — V™3 + 1
y/n2 + 1 + $ n 3 + l
3n 4 - 2n 3 + 3n 2
(n 2 + l ) 2 + (n 2 + 1) ^/(n 3 + l ) 2 + (n 3 + 1 ) y ^ T T
Hence
an
r
lim
1
"J- =
o-
n
By the comparison test the series diverges.
(b)
lim {/a^ = lim (1
n—>oo
v
n—>oo \
n
2,
n
,, )
= -, and the root test gives
+n_r1/
e
the divergence of the series.
(c) One can verify by induction that
(2n-3)H
1
(2n-2)!!
2n-l
77^
^TTT > «
for
7
n
>
2
-
So by the comparison test, the given series is divergent.
(d) The series is convergent because lim r/a^ — K
n—>oc
(e) 1 —cos ^ — 2 sin2 ^ < ^ r , and therefore the series is convergent.
(f) lim ?/a^ = 0, which proves the convergence of the series.
n—>oo
(g) By Problem 2.5.4 (a),
vhm
n—•oc
^
_
;
—
n
1
= iIn a,
and therefore the series is divergent.
3.2.2.
(a) The series converges because \ In (l + ^) < ^ .
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Solutions. 3: Series of Real Numbers
270
(b) The convergence of the series follows from the inequality
1 . n+ 1
2
~7=ln
T < "7=7
for
T\
1
n2 _ lnn
(c) Using the inequality Inn < n, we get
the given series converges.
(d) We have
n > 1
1
1
(lnn)lnn
nlnlnn-
-
< ^ z i y - Hence
Thus the series is convergent.
(e) Applying methods of differential calculus, one can prove that for
sufficiently large x the inequality ( m i n x ) 2 < lnx holds. Therefore
1
_
1
1_
( l n n ) l n l n n "" e ( l n l n n ) 2
n
for sufficiently large n. This proves the divergence of the series.
3.2.3. Put cn — | ^ . By our assumption,
Cn+1 = 7
< T - = Cn,
U > no-
Thus the sequence {cn} is monotonically decreasing for n > no- This
implies that the sequence is bounded, i.e. there exists C > 0 such
oo
oo
n=l
n=l
c
that 0 < cn < C, n e N. Hence X] a n = ^
completes the proof of our statement.
oo
n^n < C X &n> which
n=l
3.2.4.
(a) By Problem 2.1.38, we get
/
i\n-2
a n + i _ (1 + J
/
<
/
n
n+1
x
\
2
1
_ (n+i) 2
Now the convergence of the series follows from t h e convergence
test given in the foregoing problem.
(b) Similarly, by Problem 2.1.38, we obtain
«n+i
(i + *) n
(! + £)"
(i + i) n + 1
dr
£ '
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3.2. Series of Nonnegative Terms
271
If the given series were convergent then, by Problem 3.2.3, the
oo
series J2 nn
wom<
n=i
^ a ^ s o ^ e convergent. Thus the given series
diverges.
3.2.5.
oo
oo
n=l
n=l
(a) By Problem 2.5.4 (a), the series ]£ (tfa - l ) a and £
^
ei-
ther both converge or both diverge. Therefore the given series is
convergent for a > 1 and is divergent for a < 1.
(b) It follows from the solution of Problem 2.5.4 (b) that Inn <
n(^/n1). Thus for n > 3 and a > 0,
± <(£)•<„*_,,..
This implies that the given series is divergent for 0 < a < 1. Let
us also observe that for a < 0 the necessary condition an —• 0
for a series to be convergent is not satisfied. For a > 1, by
Problem 2.5.5, the given series converges if and only if the series
oo
£ (H?)
does
-
The
convergence of the latter series follows from
n=l
Problem 3.2.3, because for sufficiently large n we have
Gn+i _ fnln{n + l) y < ^
(n-hl)lnny
\ n + l/
(c) By 2.5.5, the given series converges if and only if the series
converges. By the inequality ^ ^ < ln(l + x) < x, which is valid
if x > 0, see Problem 2.5.3, we get
( K 1 - ^ ) " 1 - 1 ) <7T«
for a>1
and
l X
ln(1 +
-'
n+l
-1
>
(2^TTF f ° r ° < a - L
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Solutions. 3: Series of Real Numbers
272
Therefore the given series is convergent for a > 1 and divergent
for 0 < a < 1. Furthermore, let us observe that for a < 0 the
necessary condition for a series to be convergent is not satisfied.
(d) It is easily verifiable that
hm
n—•oo
1-nsin-!1
2- = - .
l
-4r
nz
0
Hence the given series is convergent if and only if the series
oo
Yl T^
71=1
U
ls
convergent.
Therefore our series is convergent for
a > \ and divergent for
ot<\.
3.2.6. By Problem 2.5.5, we get
aa?. i In
n a
lim
n—>-oo
oo
a
" ?i = 1, and so our series
converges if and only if the series ]T] an does.
71=1
3.2.7 .
(a) The convergence of the series Yl ~ l n ( c o s n) foH°ws fr°m the
n=l
fact that
lim
(b) If c ^ 0, then
-In (cos j1)
w
\—
alnn +b
l i m eclnn+d
=
1
= e? ^ 0 .
n—K>o
Thus the given series diverges.
If c = 0 and ^ > 0, then the necessary condition an —• 0 for
convergence does not hold. If c = 0 and § < 0, then
a ln n + b
b_ a i n „
£
a
Therefore, in this case, our series converges if § < — 1 and diverges if 2 > — 1.
(c) We have
n^
1
(n + a)"+ b (n + 6 ) n + a " (n + a) b (l + £ ) n ( n + 6)<* (l + £ ) n '
oo
Thus the series in question converges if and only if Y2 ~^*+h- does.
71=1
n
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273
3.2. Series of Nonnegative Terms
3.2.8. The convergence of the series ^
y/a>nO>n+i is an immediate
71=1
consequence of the inequality y/anan+i < \{an + &n+i)- Moreover,
if the sequence {an} is monotonically decreasing then y/anan+i >
oo
o n +i. Consequently, the convergence of ^ an follows from the con71=1
OO
vergence of the series ^
71=1
y/anan+i.
Now, let us consider the sequence {an} defined by setting
( 1
if n is odd,
n = { 1 .. .
—7 it n is even.
a
Then
n
^
2n
fc=l
fc=l
n
1
/c=l
oo
oo
71=1
71=1
Therefore the series ]P - v /a n a n + i converges, whereas the series Yl
diverges.
a
n
3.2.9.
(a) Let us first notice that if the sequence {an} is bounded above,
say by M > 0, then
>
1 + an ~ 1 + M
oo
Therefore the series ]T) j ^ -
is divergent. On the other hand,
71=1
if the sequence {an} is unbounded above, then there exists a
subsequence {aUk} divergent to infinity. Thus
lim
/c-»oc 1 -f aUk
and therefore the necessary condition for convergence fails to
hold.
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Solutions. 3: Series of Real Numbers
274
oo
(b) The series ^
n=l
T^~
can
converge or diverge. To show this let
us consider the following example:
(1
an = < 1
if n = ra , ra = l , 2 , . .
• ,
^
,
.
.
.
,
otherwise.
oo
The series J2 an diverges. We have
^1-ffcafc
^ 1 - f k2
+ k2 ~
£^k
Hence, in this case, the series ]T
n=l
Q
1+ ^a
fr[l
+ k2'
converges. If we take
oo
oo
an = - , we see that both series V a n and V -r—^— can din=l
n=l
verge,
(c) The convergence of the series in question follows from the inequality
1 + n2an
n2an
n2
(d) If the sequence {an} is bounded above, say by M, then
(Xr),
Q"n
1 + a2n - 1 + M2 '
Therefore, in this case, the series in question is divergent. But, if
oo
for example an = n 2 , then the series Yl \+na2
n=l
n
conver
ges-
3.2.10. For any positive integers n and p,
n+p
Since lim
n
<tP~
n
= 1, the sequence of partial sums of the series
oo
J2 f^ is not a Cauchy sequence. Thus this series diverges.
n=l
n
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275
3.2. Series of Nonnegative Terms
On the other hand,
&n ^
bn
&U
bn ~ bn—\
Onbn-i
bnbn-l
1
-*-
bn—l
bn
and therefore
n+p
n+p
i \
/
i
1
Sn
Sn+V
1
V °^< T f J _ _ J_) = J-_ J _ < J_.
, i ~ ^ i ^fc
k=n-\~l *-
u ^ \Sk-l
k=n-\-l x
$k)
'
oo
^
Sn
Hence the series ^Z §2" is convergent by the Cauchy criterion.
n
n=l
3.2.11. We have
q qP
q qP
Let p be a positive integer
t
such that - < /?. Then, for sufficiently
large n, the inequality
a„
holds. Therefore it is enough to establish the convergence of the series
with terms — ^ — . To this end we will show that the inequality
C
D
OP
ndn-l
\
OP
\°n-l
OP
n
°
is satisfied. The last inequality is equivalent to
This follows immediately from the easily verifiable inequality 1—xp <
p(l — x), which is valid if 0 < x < 1. Indeed, it is enough to set
x = (-jF^- J P . Therefore (see the solution of the foregoing problem)
the convergence of the series in question is established for (3 > 0.
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Solutions. 3: Series of Real Numbers
276
3.2.12. Let us first assume that a > 1. Then for n > 2,
an
CQ
iJn
a
— O QOt — 1 '
OnOn_i
Thus the convergence of the given series follows from the last problem.
Now, let a < 1. Then for sufficiently large n, we have §^ > f^. This
and Problem 3.2.10 imply the divergence of our series for a < 1.
3.2.13.
(a) By assumption, the sequence {rn} is monotonically decreasing
and tends to zero. Moreover,
Hence for any positive integers n and p,
&n+l
Q"n+p
,
7*n
Tn
^n+p—l ~~ rn+p
Tn+1 ,
7*n+p—1
^n-j-p—l
^*n
^"n ~~ 7*n+p
7*n
-j
^n+p
^*n
Given n, we have lim (1 - —^j = 1. Therefore the divergence
r
p—XX) \
n J
of our series follows from the Cauchy criterion.
(b) We have
^n
f*n— 1
y/Tn-l
7*n
( y ^ n - l ~ V ^ X ^ n - l + y/K)
y/rn-i
y/rn~l
< 2( v /r n _i - Vr^).
Applying this inequality one can show that the sequence of partial
sums of the series ]T ^ n _ is a Cauchy sequence. Thus the
71=1
^
series converges.
3.2.14. We first assume that a > 1. Then, for sufficiently large n,
The divergence of the series in question is now derived from part (a)
of the foregoing problem.
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277
3.2. Series of Nonnegative Terms
Let a < 1. Then there exists a positive integer p such that a <
1 — ^. Hence
p
tin
r
@"n
n-i
(rn_i)
^Vi__p
Tn—i
p
r
^-i
Applying the inequality 1 — xp <p(l — x), which is valid if 0 < x < 1,
with a: = f — ^ J P , we get
^-<p(rlx-ri
r
\
n-l
The convergence of the given series follows from the Cauchy criterion.
3.2.15. For 0 < a < 1,
a n + i l n rn
5—^
= 0.
hm
Therefore the convergence of the series ^ a n + i In r n follows from
Tl=l
the foregoing problem.
3.2.16. We know (see, e.g., Problem 2.1.38) that
i \
n
/
i \
n
+i
Assume that g > 1 and let £ > 0 be so small that g - e > 1. Then
there exists no such that n l n ^ 2 L > g — e for n > no- Thus by (*),
n ln-^-
> ( ( / - £ ) > nln fl + - V
and consequently,
Q>n+1
(n+l)9" E
1
Tl9"
Therefore the series ^2 an is convergent by the test proved in Probn=l
lem 3.2.3. Similar arguments apply to the case g = +oo. Analogous
reasoning can be used to prove the divergence of the series if g < 1.
The following examples show that the test is inconclusive if g = 1.
oo
Taking an = £, we see that g = 1 and J ] £ diverges.
71 = 1
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Solutions. 3: Series of Real Numbers
278
On the other hand, letting an =
^
, we get the convergent
oo
series J ] a n (see Problem 3.2.29). To show that in this case g = 1,
n=l
let us first observe that
nln
^
=,lnfl
lV
+
+
2nInln(r+1).
As the first term of the sum tends to 1, it is enough to show that
lim 2nln lj^~L = 0- To this end, note that
n—>oo
m n
U m f l n ( n ± l ) y = 1 . m / + ! n f £ ^ = e0 = 1
Inn
/
n->oo \
In7i
3.2.17.
(a) We have
lim nIn —— = lim n(y/n + 1 — \/^) In 2 = -foo.
n—*oo
fln_|_i
n—+00
The convergence of the series follows from the preceding problem.
(b) Similarly,
(
i \
n
1+ -J
nJ
•In2 = l n 2 < l .
Hence the series diverges.
(c) Likewise,
lim n In —— = In 3 > 1,
n-+oo
an+i
which proves the convergence of the series.
(d)
lim n In —— = In a.
n->oo
a
n +
i
Therefore the series is convergent for a > e and divergent for
a < e. For a = e the series in question is the harmonic divergent
series.
(e) We have (see the solution of 3.2.16)
r
i an
,.
, ln(n+l) .
lim n In
= hm n in —^
• In a = 0.
7i-+oo
an+i
n->oo
Inn
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279
3.2. Series of Nonnegative Terms
Therefore, by the convergence test given in the foregoing problem,
the series is divergent for any a > 0.
3.2.18. Since
lim n l n
n-^-oc
an
Q>n+1
n
i_
1
+1 = l n - ,
= lim nma
n—*oo
a
the series is convergent for 0 < a < \ and divergent for a > \
(compare with 3.2.16). If a = ^ then (see, e.g., Problem 2.1.41)
I
l i m e_2
n—>oo
.ix
-i-l
n_
—
n
=
e
-
where 7 is the Euler constant. The comparison test and the diver00
gence of the harmonic series ^
71=1
^ imply the divergence of the series
n
in question for a = \ .
3.2.19. Applying the inequality yqL_ < ln(l + x) < x, which is valid
if x > —1, we get
\an+1
1-L(_£JI
J
I
an
'—
<
n
In
1
H
1)
\
ttn+1
.
1/ < n
\«n+l
By this inequality, the Raabe test and the test given in Problem 3.2.16
are equivalent for a finite r. It also follows from the above inequality
that if lim n l n - ^ - = +00, then lim n (-?*
1) = +00. We
will now show that the other implication is also true. Indeed, if
lim n ( - ^
1) = + 0 0 , then for any A > 0 there is no such that
-**
CLn + l
1 > - for n > n0. Hence
n
n l n _ ^ L _ = In (l + - ^ - - l V > In f 1 + - V
a
n
ttn+1
V
n+l
/
\
J
• A
™~*o°
Since A can be arbitrarily large, we see that lim n l n - 2 * - = +00.
Similar arguments apply to the case r = — 00.
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Solutions. 3: Series of Real Numbers
280
3.2.20. Since the sequence {an} is monotonically increasing,
V^T
)
n an
n
By the Raabe test, the series diverges.
3.2.21. By the definition of the sequence,
n-l
-
an = a\e
We will first show that
S < +oo, then
£
k=1
Ofc
for
n = l,2,....
oo
oo
Yl an *s divergent. Indeed, if X) a n
n=l
lim a n = a\e~s
> 0, and so
n—+ 00
=
n=l
lim a" > 0; this
n—*oo
would contradict the necessary condition for the convergence of the
oo
series. Therefore the series ]P a% diverges and, by the foregoing,
lim an = 0.
n=l
n—>oo
Now assume that (3 > a. We will prove that in this case the series
in question is convergent. To this end, we will show that
(*)
a~a > a(n - 1) for n > 1.
This inequality is obvious for n = 1. Assume that it holds for an
arbitrarily fixed n. Then, by the definition of the sequence, we obtain
a - £ = a~aea<
> ^ " ( 1 + « < ) = a~a + a > an.
Thus (*) holds for any positive integer n. This inequality is equivalent
(for n ^ 1) to
a£ < ( a ( n - 1 ) ) - S .
Hence, by the comparison test, the given series converges for (3 > a.
Let /? < a. It has already been shown that lim an = 0. Therefore,
n—>oo
for sufficiently large n, 0 < a n < 1. Thus a" < a(*. This inequality
oo
and the convergence of the series ^ a% imply the convergence of
n=l
n=l
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3.2. Series of Nonnegative Terms
281
3.2.22. Note that
lim n ( —
n-+oo
yan+i
1 J = lim
-a = a.
n->oo n + 1
J
By the Raabe test, the series in question is convergent for a > 1 and
divergent for 0 < a < 1.
For a = 1 the given series reduces to the divergent harmonic series
2-^ n+l*
3.2.23. By
( an
-0
b
= hm
—- =
a
n—>oo
'V n+1
n-*oo (n-\- l)a
and the Raabe test, the series is convergent for b > a and divergent
for b < a. In the case where b = a, the convergence of the series
depends on the sequence {bn}. Indeed, if {bn} is a constant sequence,
lim n
a
oo
then the given series is the harmonic series Yl ^+1*
n=l
We will now show that if bn = a -f
"
convergent. In fact,
CLn
=
2a
T f ,,s,
In(n+1) 7
then the series iIS
ni
(2+I^)(3+I^)•....(n + l +
R
^IT),
Thus
a n (n — l)ln(n — 1) — an+\n\nn
= an
I'(
IM i
-i\
(n - 1) ln(n - 1)
{n + l)nlnn
2
\
•
Calculation shows that
,.
//
IM /
T\
hm I (n - 1) ln(n - 1)
(n+l)nlnn \
——^— s — = 1.
Therefore, for sufficiently large n,
an(n - l)ln(n - 1) - a n + i n l n n > (1 - e)an > 0.
Hence the positive sequence an(n - l)ln(n - 1) is monotonically
decreasing and so is convergent. This, in turn, implies the convergence
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Solutions. 3: Series of Real Numbers
282
of the series with the terms a n (n—1) l n ( n - l ) - a n + i n l n n . By the last
oo
inequality and by the comparison test, the series ^ an converges.
n=l
3.2.24. By assumption,
an((n-
l ) l n n - 1) - an+1nlnn
= {^n -
l)an.
Now if 7 n > T > 1, then
(*)
a n ((n — 1) Inn — 1) — a n + i n l n n > (r — l ) a n .
Combining (*) with the inequality (n — 1) ln(n — 1) > (n — 1) Inn — 1,
we get
(**)
Q>n(n — l)ln(n — 1) — a n + i n l n n > (r — l ) a n > 0.
This means that the sequence {an(n — 1) ln(n — 1)} is monotonically
decreasing and so convergent. Therefore the series with the terms
an(n — 1) ln(n — 1) — a n + i n l n n is convergent. By (**), the series
J2 an is also convergent.
If 7n < r < 1, then an((n— 1) Inn — 1) — a n + i n l n n < (T — l ) a n .
Hence
an(n — 1) ln(n — 1) — a n + i n l n n <
T + In I 1
j
J an.
Since
lim [ r + l n f l - - )
U r - K 0 ,
the sequence { a n + i n l n n } is monotonically increasing (except for
finitely many n). Therefore there is M > 0 such that a n + i n l n n >
M. Thus a n + i > ^ j ^ , which proves the divergence of the series
3.2.25. We have
{
lim n
n—»oo
'V
an
a
n+1
-0
= lim
^ 4 - = a.
n->oo 1 _ « _ 2*.
n
nx
Therefore the Raabe test implies the convergence of the series for
a > 1 and its divergence for a < 1. In case a = 1, the divergence
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283
3.2. Series of Nonnegative Terms
of the series follows from the test given in the foregoing problem,
because
an
where 7 n =
$ n Inn
^B?
n
nx
< T < 1 for some T.
nlnn'
n
3.2.26. We will apply the criterion of Gauss from the preceding problem. We have
a n +i _ n2 + (a + f3)n + a/3 _ _ 1 + 7 - a - / ? _ i?n
«n
n 2 + (1 + 7)71 + 7
n
n2
Therefore the series in question converges if a + f3 < 7 and diverges
if a + f3 > 7.
3.2.27. As in the foregoing proof, we will use the Gauss criterion.
We have
a
n+l
-.
n
2
an
n
n2'
Hence the series converges if p > 2 and diverges if p < 2.
.
00
3.2.28. Let S n , 5 n denote the nth partial sums of the series Y
n=l
a
n
OO
and ]T 2 n a2^, respectively. Then for n < 2k,
n=l
Sn < ai+(a2+«3)+---+(«2 fc +---+ a 2fc+i-i) < ai-h2a2+...+2/ca2fc = 5 fc .
For n > 2k,
^ > « l + f l 2 + («3 + ^4) + ••• + ( ^ - i + l + ••• + a2fe)
> a 2 + 2a 4 + ... + 2k'1a2k
= -5fc.
Thus the sequences {5 n } and {Sn} are either both bounded or both
unbounded.
3.2.29.
(a) We will apply Cauchy's condensation test (3.2.28). Since the
condensed series is
00
V
cyn
-
°°
=v
1
-
^ 2n(ln2n)a
^ (nln2)a'
v
n
=
l
'
n
l v
'
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Solutions. 3: Series of Real Numbers
284
the given series converges for a > 1 and diverges for 0 < a < 1.
oo
If a < 0, the divergence of ]T n(\nn)a
f°M° ws immediately from
71=2
the comparison test,
(b) By the equality
y^ 2
00
- =^ vn In 2 ln(n-In 2)
lnln2
on
n
ln2n
°°
1
n
71=2
V
71=2
and by (a), the given series is convergent.
'
3.2.30. Reasoning similar to that in the proof of the Cauchy condensation test (3.2.28) can be applied. For n < gk)
Sn < Sgk < (ai + ...a 5 l _i)4-(a 9 l +...-ha 5 2 _!) + ...-h(a^+...-ha P f c + 1 _i)
< (ai + ...a^_i) + (92 ~ gi)a>gi + ••• + (0fc+i - 9k)a9kFor n > gk,
cSn > cSgk > c(a9l+1 + . . . + a92) + ... + c{a9k__1+1 + . . . + a9k)
> c{g2-gi)ag2+...+c{gk-gk-1)agk
>
(g3-g2)ag2+...+{gk+i-gk)a9k.
These inequalities prove our assertion.
3.2.31.
(a) It is enough to apply the Schlomilch theorem (3.2.30) with gn =
3n.
(b) Applying the Schlomilch theorem with gn = n 2 , we get the
oo
oo
71=1
71=1
equiconvergence of the series Yl an and ^ (2™ + l) a n 2 - Since
lim ( 2 n + 1 ) Q " 2 = 2,
n—>oo
na n 2
oo
oo
71=1
71=1
the series ^ (2n + l)a n 2 and ^ na n 2 are also equiconvergent.
(c) Compare with (b).
oo
(d) By (b), the series £ ^
n=l
oo
and J2 #r
n=l
are
equiconvergent. The
latter is convergent, e.g. by the root test. To determine the
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3.2. Series of Nonnegative Terms
285
divergence or convergence of the series Yl 2 ^ '
oo
S
71=1
S 3^
n=l
~T^IT5
a
n=l
and
the Cauchy condensation theorem or the test given in
(a) can be applied. We will now study the behavior of the series
with terms aln1in n . If a > 1, then the convergence of this series is
00
n
equivalent to the convergence of ^
- ^ . It is easy to check that
fl
71=1
the last series diverges, e.g. by the root test. This establishes
00
the divergence of the series ]T Qln1ln n for a > 1. Observe that
if 0 < a < 1, then the necessary condition for convergence is not
satisfied.
3.2.32. By Problem 2.4.13 (a), there exists a n o O such that
( a n ) ^ < e~ 1 _ £ ,
n> k.
Hence j ^ l n a n < — 1 — e, and so a n < ^rW- The comparison test
00
yields the convergence of ^
71=1
an.
3.2.33. Analysis similar to that in the solution of the preceding problem gives
an < —
r-r—
n(mn)i+£
for
and for an
n >k
e > 0.
00
Therefore, by Problem 3.2.29(a), the series J2 an is convergent.
71=1
3.2.34.
S2n 0 +fc_;L~~ S ^ o - l
+
(^2n0 + l +
~ ( a 2 n o + O2 n o+l ~h ••• + Cl2no +
••• " f a 2 n 0 + 2 _ 1 )
+
••• +
( a 2 n 0 + fc-l +
1
-l)
. . . "f
< 2 no a 2 n 0 -f 2 n o + 1 a 2 n 0 + i + ... + 2 n ° +/c - 1 a 2 n 0+te _ 1
Hence, for sufficiently large fc,
x
2 Tl o+ fc _i
( -9) Y2
n=2 n o
an g
/2no_i
- [ ^2
\ n=7i 0
an
n
o+ f c _i
n=n 0 +fc
fl
\
n
o_i
- XI « ~"E f l n "
2
/
2
n—n0
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Solutions. 3: Series of Real Numbers
286
Thus the sequence of partial sums of ^2 an *s bounded, and so the
n=l
series converges.
oo
3.2.35. Assume that the series JZ a™ converges. Then
n=l
2n
lim
7
n—>oo ^—'
fc=n-fl
2n+l
ak = lim
7
n—>oo ^—^
/c=n+l
a& = 0.
Hence, by the monotonicity of {a n },
2n
^2
2n
ak
- X^
k=n+l
and
2n+l
a
1
a2n = na2n =
k—n-\-l
k>
2n+l
2^
a2n
+!
=
2^2na2n^
'fl2n+1'
9 4- 1 ^
It follows from the above that lim nan = 0.
n—>-oo
Let a n =
n l n A + 1 x. Then the series with the terms an diverges
and
lim nan = lim —,
n-»oo
3.2.36. Let
( 1
I —
— = 0.
n ^ o o l n ( n -f 1)
for n = A:2, k — 1,2,...,
—77 otherwise.
v nz
oo
The series Y] an is convergent but the limit lim nan does not exist.
n
n=i
-"°°
3.2.37. The condition we are looking for is the convergence of the
oo
oo
n=l
n=l
series ]T} y/a^. Indeed, if JZ V&n converges, then we take bn =
Now assume that there is a sequence {bn} such that both series
J2 bn and £
n=l
n=l
f^ converge. Then
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287
3.2. Series of Nonnegative Terms
and so the series Y2 y/a^ converges.
3.2.38. Assume that there is a sequence {an} such that both series
oo
oo ,
£ an and J2 ^h~
converge. Let
n=l
A=
Then
Y
nseA
G N : — < -^— 1
ns
njans J
\ns
t
^- < -foe and so
s
and
A ' = N \ A.
^~ = ~^oc (°^ c o u r s e A can be an
Y
nseA'
empty set).
Now observe that aUs > ^- for ns € A ; . Therefore the series
oo
Y2 a«n diverges, contrary to our assumption.
n=l
3.2.39. We have
oo
1
^
oo
1
n
an
n=l
-
oo
^-j nan
n=l
^
nan '
n=l
oo
We will show that the convergence of the series Y2 ^Qn
T+ 11 implies the
na
n
n=l
oo
divergence of Yl ^~ • By the Cauchy criterion, there is k £ N such
71=1
U
n
that for any positive integer n,
£
*±l < I . Thus ^
2=fc+l
£
^f
<
2 = fc+l
| . Therefore, for n > fc,
/c+n
- 1 7 ,
E
i
fli±i 1 k + n
1
ncti
4
n
2
By the relation between arithmetic and geometric means,
Gfc+n+l
Gfc+i
^ 1
< -,
2
,
and so
^ Clfc-j.1
a/c +n +i <
-zr-2n
Thus
(fc + n + l)afc +n +i
>
( H n + l)a f c +i'
Therefore the series Y2 ~^~ diverges.
71=1
na-p
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Solutions. 3: Series of Real Numbers
288
oo
3.2.40. Of course the series ^ cn can diverge (e.g. if an < bn for
n=l
n e N). Surprisingly, it can converge. Indeed, take the series with
terms
ii.i_i_JLi_i_J_i _
2
' 2 2 ' 2 2 ' 2 2 ' 2 2 ' 2 2 ' 72'
S 2^1'"
and
X
_l_jLJ_J_i_l.JlJL
_L
' 22 ' 32 > 4 2 > 52 ' 5 2 > ^2 ' g 2 > 3 2 ' - ' ^2 ' - '
V
v
8 2 -f 1 times
'
Each of these series contains infinitely many blocks of terms whose
sums are greater than 1. Therefore each of them diverges. In this
00
case cn = 4_-, and so J2 cn converges.
n=l
3.2.41. We will use the Cauchy condensation theorem (3.2.28). The
0 0
h
divergence of _T] n *s equivalent to the divergence of the series with
the terms
71=1
62n=min{a2n, _-L_j .
In turn, the series J2 ^
diverges if and only if the condensed series
71=1
with terms
2n622n = m i n J 2 n a 2 2 n ,
J^
diverges. We will now show that the latter series diverges. Indeed, if
oo
oo
71=1
71=1
a series ^ dn is divergent, then ^ min{d n , c}, where c > 0, is also
divergent. If min{d n ,c} = c for infinitely many n, then the series
oo
J2 min{d n ,c} diverges. If min{d n ,c} = c for finitely many n, then
71=1
OO
the divergence of the series follows from the divergence of £3 d n .
71=1
3.2.42. We have
1 _
Q n
^n+l
__
Q
n + 1 ~ Q>n <
Gn+1
~~
Q"n+l "~ Q>n
Oi
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3.2. Series of Nonnegative Terms
289
This and the convergence of the telescoping series ]T (a n +i — a n )
imply the convergence of the series in question.
71=1
3.2.43. We have
-.
Q"n
^n+1
&n+l
&n
&n+l
Setting bn — an+i — an and Sn = b\ + .... + 6 n , we obtain g ^ Q
a n + i-a n ^ rp^ ug ^ divergence of our series follows from 3.2.10.
=
°
dn + l
3.2.44. If the sequence {an} is unbounded, then the convergence of
the series in question follows from Problem 3.2.11. To see this one can
apply arguments similar to that given in the solution of the preceding
problem. Now assume that the sequence {an} is bounded. Then
&n+l "~ an ^
1 /
\
— <
^ ( « n + i ~ an).
an+ia%
a2a(?
Hence the convergence of our series follows from the convergence of
oo
the telescoping series J2 ( a n+i — &n).
n=l
3.2.45. It is enough to take c n = j ~ y where Sn is the nth partial
oo
sum of Yl an,
an
d apply 3.2.10.
71=1
3.2.46. One can set cn = -?==> where rn = a n +i 4- a n + 2 + ••-, and
use 3.2.13 (b).
3.2.47. The sequence {rn} is monotonically decreasing. Hence by
3.2.35, lim nrn = 0. Therefore
n—>oo
lim nan=
n—+oo
lim n ( r n _ i - rn) = lim ( ( n - l ) r n _ i - n r n + r n _ i ) = 0.
n—•oo
n—>oo
3.2.48.
(a) Since
lim an = +oo, a n > 2 for n large enough. The con-
n—>oo
vergence of the series in question follows from the inequality
^r < ?pr, which holds for sufficiently large n.
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Solutions. 3: Series of Real Numbers
290
(b) As in (a), n can be chosen so large that -^~ < 3 ^ . Thus by
3.2.17 (c), our series converges.
(c) The series can be either divergent or convergent. Its behavior
depends on the sequence {an}. If an = Inn, n > 2, then the
00
series ]T
n=l
1
Qln lnn
°n
diverges (see 3.2.2 (e)). On the other hand, if
an = n, then for n > e e ,
-T4_ =
^Inlnn
T T l T - <
gin l n n - l n n
J - ,
where
a > 1.
^a '
In this case the series in question is convergent.
3.2.49. The series diverges because the necessary condition an —• 0
for convergence is not satisfied (see 2.5.25).
3.2.50. We assume first that p = 0. Then by 2.5.22, lim y/nan =
n—>oo
y/S and so our series diverges. Now suppose that p > 0. Then
lim an = 0. Hence lim ^ ^ = lim ^ ^ • -^ = 0. The series
n—>oo
a
n—KX>
™
n—>oo
a
n
™
converges by the ratio test.
3.2.51. Observe that an £ (n7r,rar4- | ) . Hence 4r < ^ 3 , and so
00
the series Yl ~t? c o n v e r g e s 3.2.52. Set bn = y/a^; then bn € (n7r,n7r + f ) . Hence the series
00
00
]T} ~ = J2 W converges (see the solution of the foregoing problem).
3.2.53. The series diverges because lim nan = 2 (see 2.5.29).
n—>oo
3.2.54. For simplicity we introduce the following notation:
Ln = ai + a 3 -f ... 4- a 2 n - i
and M n = <22 + a 4 + ••• + «2n-
By the monotonicity of {an},
(*)
L n > M n and Ln - ax < Mn.
n
Hence 2Mn = Mn + Mn> Mn + Ln - ax = J2 ak- Thus
(**)
lim Mn — +00.
n—>oo
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3.2. Series of Nonnegative Terms
291
Combining (*) and (**), we arrive at
Ln _
_ Ln - Mn
j2i_
^
3.2.55. By the definition of fcn, we have 0 < Skn - n < ^-. It is
known that lim (Skn — ln fcn) = 7 , where 7 is the Euler constant (see
n—>oo
2.1.41). Therefore
lim (n — ln kn) = lim (n + 1 - ln fcn+i) = 7.
n—• oo
n—KX)
Hence
lim ( " i - i n ^ ± 1 ^ = 0 ,
n-»oo y
kn J
and so
lim — — = e.
3.2.56.
(a) [A. J. Kempner, Amer. Math. Monthly 23(1914), 48-50] A jfc-digit
number from A can be written in the form
10fc~1a1 + 10 /c_2 a 2 + ...4-afc,
where
0 < a* < 9, i = l,2,...,fc.
For a given k, there are 9k /c-digit numbers in A, and each of
them exceeds 10 fe_1 . Therefore
^
°° gk
nGA
k—1
(b) As in (a) we have
y,
1
n €A
^
9
fc
fc=l
Therefore if a > log10 9, then the series XI ^
converges. More-
nGA
over, since
y^
1 ^
1
n
o c
>
>
y>
2 ^
9fc
( 1 Q fc _
y,
l)a
>
Z-,
9fc
10^'
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Solutions. 3: Series of Real Numbers
292
the series J2 —• diverges if a < log 10 9.
nGA
Remark. Let Ak denote the subset of positive integers that do not
contain the digit k in their decimal expansion. In much the same way
one can show that the series ^ —• converges if a > log10 9.
n €A k
3.2.57. Assume that —oo < g < 1 and take e > 0 so small that
g + e < 1. Then, for sufficiently large n, l n ^ - < (# + £)lnn and
Q>n > T^TF- Therefore the series diverges. If g = —oo, then (for n large
enough), In -—- < —1 -Inn. Thus an > n and the series diverges. The
oo
same proof works for g > 1. Let us consider two series: Yl ^ and
oo
J2 nin^n' ^ ^ ^ r s t
n=2
one
diverges and the second converges, although
for both g is equal to 1.
3.2.58. The equivalence of these tests has been shown in the solution
of Problem 3.2.19. By 2.5.34, if the Raabe criterion is decisive, then
so is the criterion from the foregoing problem. To show that the
converse is not true, we consider the series with terms an defined by
setting a 2 n-i = ^z, a 2 n = 4^7.
3.2.59. Let bbnn =
= yV22 ++ y\f\2 + ... + \/2 . Then bn = 2 cos ^ S T (comn—roots
pare with 2.5.41). By the definition of {a n }, we have an = 2 — 6 n - i ,
and so an = 2 sin ^ r < <pr- Thus the series in question converges.
3.2.60. Assume K is a positive number such that
(ai - a n ) + (02 - a n ) + ... + (a n _i - an) < K
for
n € N.
Hence for every n £ N we have 01 + ... + a n — na n < iC. Let m £ N
be arbitrarily chosen. By the monotonicity of the sequence {an} and
by its convergence to zero, there exists n 0 £ N such that
a>n < 2 a m
(*)
for
n > n0.
We have
ai -I-... + a m - rna n + a m + i + ... + a n - (n - m)a n < AT.
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293
3.2. Series of Nonnegative Terms
Again, by the monotonicity of {a n },
am+i 4-... -f an > (n — m)an
and
a\ + ... 4- am > mam-
Therefore m(arn — an) — mam — man < a\ +02 + ... + o m — man < K.
This and (*) imply \mam < ^ ( a m — Q>n) < K- Finally,
Sm = ai 4- &2 4- ... 4- a m = S'm - mam 4- ^ a m < K + raam < 3K.
3.2.61. Prom the relations
« n = Gn+1 + &n+2 + &n+3 + ... ,
a n + i = a n + 2 4" G n +3 + ...
we gather that a n + i = \an. Now, by induction, an = ^ , n G N.
3.2.62. [20] Let rn,fc = a n + a n + i + ... + a n+fe , n = 1,2,.., fc =
0,1,2,..., and let lim r n ^ — vn,n — 1,2,... Assume that s G (0,5)
k—>oo
and that a n i is the first term of the sequence {a n } for which a n i < s.
Either there exists a k\ such that r^^ < s < r nij fc 1+ i, or rni < s.
In the second case we have s < a n i _ i < r n i < s, and so r n i = s.
In the first case we determine the first term an2 with ri2 > rii + fci,
^ni,fci 4- fln2 < s - Either there exists a &2 with
r
ni,ki
~r 1*712,k2 ^
s
— Tm,k\
4" ^n2,/c2 + l >
or rniifc1 4- r n 2 = 5. This procedure can be repeated, and if the first
case occurs at every step, then s == r^^ 4- rn2j/e2 + ....
3.2.63. [20] Suppose, contrary to our claim, that there is k G N such
that ak = 2p+
00
^2
n—k+l
an, where p > 0. Then ak — p = p+
00
^
an —
n=k+l
00
£ ] €nan, where £ n equals either zero or one. By the monotonicity
n=l
00
00
of { a n } , en = 0 for n < k. Hence ak - p = ]T £nfln <
n=l
afc — 2p, a contradiction.
3.2.64. By the Stolz theorem (see 2.3.11), we obtain
1
^
lim a 1 5 1 - + a 2 y + - + a „ 5 - ^ lim
n—oc
ln^n
^
n-^oo - I n (1 - a
n
n=fc+l
S
n
)
^
=
L
The last equality follows, e.g., from 2.5.5.
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Solutions. 3: Series of Real Numbers
294
3.2.65. Set an = 1, n G N.
oo
3.2.66. Since Q i+ a 2+-+"n > «i, the series V
fll+fl2+ +flw
-
n=l
is diver-
gent for any positive sequence {an}. The divergence is independent
oo
of the behavior of the series J2 an71=1
3.2.67. By assumption,
2
a 2 <ai,
n
2
a
]T
^ ^271T ^
n-l
a f c
-
Hence, by induction
2
E«>^(>n)( 1+ ?)-( 1+ ^)"'Moreover,
1+2 ) ( 1 + 2 * - (
3.2.68. Put c n = ^ £
1 +
2^ H ^
<e
'
= n ( ^ 1 ) " , n G N. Then
c\ -... • c n = (n + l ) n
(*)
^
and
cn < ne.
Using the geometric-arithmetic mean inequality, we arrive at
1
,
. aici -f ... + a n c n
$/ai •... • a n = — — ^/oici •... • a n c n <
n+ 1
n ( n + 1)
Therefore
iV
iV
Ei
^
n=l
n=l
aici + ... + a n c n
i
ra(n
/ 1-2
l
2 -l 3
-,—7; + TT~^ + •••
1 1
+ a2C2 2
-r—r
-—- + ... +
- 3 + 3-4
= aici
v
+ l)
7
N(N l + l)
+
1
\
1
iV ,v
.
.
,
.
.
,
,
.
+
...
+
CIMCNN{N + 1)J
N(N + 1)
< axci 4- a2C2- + C13C3- + ... + a^CN— < 2oi + ea2 + ••• + ea^.
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295
3.2. Series of Nonnegative Terms
The last inequality follows from (*) . Letting N —> oo, we obtain the
desired inequality.
3.2.69. Writing
( n + l ) n - . . . - ( n + A ; - l ) n ( n + A;)n
n71-1 • ... • (n + k - 2) n ~ 1 (n + k - l)™"1
' n -f k^
n ( n + l ) .... • (n + fc- 1),
n
we get ci •... • cn = (n + l ) n •... • (n + /c)n. Hence, as in the solution
of the preceding problem, we obtain
N
E
n=l
N
v-^ aici + ... + ancn
yai - ... a n < 7
^ n ( n + l ) - . . . - ( n + k)
= aici
a 2C 2
+
<
+
-
-—
,
N + ... +
1 - 2 • ... - (lH-fc )
JV(i V + 1) •.. . • (iV + Jb)
1
1
—
TT
rr + .. . +
2 • 3 •.. • (2 + k)
'" N(N + 1) •... • (N + fc)
+ a C
^ ^7V(iV + l ) . . . . . ( i V + fc)
K^ aiCl + 2.3.... 1 (l + ^) a 2 C 2
-
+
iV(7V + l ) . . . 1 . ( A r +
fc-l)a7VCiV)'
The last inequality follows from Problem 3.1.4 (a). Since
l(Z + l ) . . . . - ( Z +
c
l + k^1
fc-l); i
letting N —> oo, we obtain the desired inequality.
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Solutions. 3: Series of Real Numbers
296
3.2.70. Let Tn = ai-ha 2 -h...-f a n and let Sn denote the nth partial
sum of the series in question. Then
c
SN =
1 , ^n2(Tw-rw-i) ^ 1 , ^n2(Tn-rn-!)
^7+ 2 ^
i
_ 1
T2
^ N 2
n
- ^7 + 2^ — T ^ T ~ , —
N
2
^
~~ ^7
5
— J7T
V^
n=2
n
22
11
W
^ -- il
/
iVN
,. 11\2
N2
2
_ 1
V^ ( n + 1)
V^
—
2 ^ T 7 ~~ 2 ^ T~ ^7 2-/ T
21^ T
i V n- 1= 2 TV
TV
nN=-2l
n = l rt
n=z
rt
^ ^n
^
1
5
^ 2n ^
1
+
7
7
2 ^ T " 2-^ T " — ^7 2 ^ T " 2l^ T7"*
^
n=l
n=2
n=l
Moreover, by the Cauchy inequality (see 1.2.12),
/
N
\
\n=l
'*/
2
o
N
N
nn
n=l
„n== il " n
with M = Y] —. Hence
n=l
TV
E fT ^ ^ .
n=l
Consequently, 5/y < f- 4- 2\/S/v v M + Af, and so
5TV<
f\/M+W2M+-M
.
3.2.71. The arithmetic-harmonic mean inequality (see, e.g., 1.2.3)
yields
2k
1
T
2fc-l
>
-
2k
]T)
{nan - (n -
l)an-i)
n=2fc-1+l
2fc~1
2^02*: — 2
fc-1
1
a2fc-i ~~ 2a2fc
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3.2. Series of Nonnegative Terms
297
Hence
, n nan - [n - l)a n _i
^
4a2*
Therefore
The divergence of the series follows from the Cauchy condensation
theorem (see 3.2.28).
3.2.72. We will show that the series £] i r is divergent. If it were
71=1
not, then there would be an n such that
P n
OO
Y]
m=n+l
- ^ < i . Let a =
Pi *P2 * ••• 'Pn- Then the number 1 + ka with fcEN can be written as
a product of primes. This unique factorization does not contain any
of the numbers pi, ...,p n - Hence
oo
oo
1
/
oo
i \
°°
E i i < E E j- <E
/ i \ *
)-i.
a contradiction.
3.2.73. It is enough to apply the results from the foregoing problem
and from Problem 3.2.71.
3.2.74. We get
oo
Epkr
n*Sgo ^ ~ T
- ^ (i + g £ + ££. + ...)
=
n^S,
1 ( 1 + 21 + 21 + ,..)
1
=
2'
k=2
because the sums in parentheses tend to 1 as n tends to infinity.
Indeed,
o nn +
+ il
n+i
n
o n9- f+li
+ _
' 471+I
+
' *"
^
=
9n+1 V
-i
Z_-•
*
k
= 3 jUn+1
*
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Solutions. 3: Series of Real Numbers
298
Moreover,
CO
-.
CO
1
Ek+
l
n l
=
+
¥^
^
1
^
V"—\
1
^
1
CO
^~^V
1
(2/c)
n+1 +
1
(2k + l ) n + 1
1
1^
^
1
=
^
1
Therefore
9n+l
V^
1
V
V 33/ /
<
V1 2 - J
k=3K
fc=3
and so
CO
2n+l
^+ 2: n
V^
-1
• 0-
/c=3
CO
3.2.75. Assume first that the series ^ a n converges.
n=l
Then the
convergence of the series in question follows from the inequality J ^ <
CO
n
a
^ . If the series J2 n diverges, then there exists a strictly increasing
1
n=l
sequence {nm} of positive integers such that S n m - i <m<
Then
^
^
m(m + 1 )
m
Tnm = Si + ... + Snm > Sni + - + Snm > - ^
^.
SUrn.
Hence
CO
E
^ m + 1-1
CO
a
n
_
7
rpa
n=n2
\T^
n
J
m=2
CO
i
m=2
~nm
CO
Y^
_^fe_
<
7 j
rpa
— 7
k=nrn
&
C
Q
\T~^ ^ n m + i - l ~ *->nm-l
J
ra=2
n
rpa
™>
CO
<
m=2
Therefore the series in question is convergent if a > \. This series can
be divergent if a < ~. Indeed, it is enough to take an = 1, n G N.
3.2.76. By Problem 3.2.35, lim -&- = 0. Take 0 < K < 1. Then
n—>oo
Gn
there is n 0 such that n < Kan for n > no. Hence
lnfe an ^ , t. / 1 \ lnfc n
> In
# 7 On
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3.2. Series of Nonnegative Terms
^
Thus the convergence of the series ^2
n—l
0 0
Q£
299
i k
Q
Qn
implies the convergence
k
^ lJL-". To prove the other implication put
an
n=i
an < nk+2}
l ! = {n<EN:
and
I2=N\Ii.
Then for n G I i we have l n a n < (k + 2)Inn, and so the convergence of J2 %-^ implies the convergence of Y] " Q n . Moreover,
nGli
nGli
for sufficiently large n in I2,
lnfc an
a^__
CLri.
U"n
1
fc + 2
Hence ]T ^ - ^ < 00, because r i ? > 1.
^
nEl2
On
'
fc+1
3.2.77. We have
¥>(n)-l
y(l)-l
£ /(*) = E /(*) + (/^C1)) + /M 1 ) + 1) + - + /(V(2) " 1))
fc=l fc=l
+ ... -f (/(y>(n - 1)) + f(ip(n - 1) + 1) + ... + f(<p(n) - 1))
<p(l)-l
< E
fc=l
n-1
/(*) + £/(?(*))(¥>(*+ !)-¥>(*))•
k=l
Inequality (1) is proved. The proof of (2) is analogous.
3.2.78. Assume first that
f(<p{n))(ip{n+l)-<p{n))
<q< 1.
Then, by (1) in the preceding problem,
<p(i)-i
k=l
Therefore, in view of ip(n) > n, (1 - q)Sn-i
00
convergence of 2Z fin)
71=1
1S
tp(l)-l
<
£
fe=i
/(*0-
Tne
proved. To prove the second part of the
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Solutions. 3: Series of Real Numbers
300
statement we can use inequality (2) from the foregoing problem and
proceed analogously.
3.2.79. One can apply the result from the preceding problem with
<p(n) = 2n.
3.2.80. We apply the result from Problem 3.2.78 with (p(n) = 2 n .
3.2.81. Apply the result from Problem 3.2.77 with
(p(n) = 3n,
(p(n) = n2
ip(n) = n 3 ,
and
respectively.
3.2.82 .
(1) We have anbn — a n + i 6 n + i > ca n +i. Therefore {anbn} is a monotonically decreasing sequence with positive terms, and so it is
oo
convergent. Thus the telescoping series £ {anbn — an+i&n_|_i)
n=l
oo
converges. The convergence of £ an follows from the Comparing
son test.
(2) We have
a
n+l .
l
bn + i
a n
bn
oo
Therefore the divergence of ]T} an follows from the test given in
71=1
Problem 3.2.3.
3.2.83. To derive the d'Alembert test (the ratio test) we take bn =
1 for n = 1,2,.... Setting bn = n for n — 1,2,..., we get the
Raabe test. Putting bn = n l n n for n = 2,3,..., we arrive at the
Bertrand test.
3.2.84. [J. Tong, Amer. Math. Monthly, 101(1994), 450-452]
oo
(1) Let S = Yl a n a n d let
n = 1
,
0n =
S-
n
£afc
k=\
CLn
=
rn
CLn
•
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3.2. Series of Nonnegative Terms
301
Of course, bn > 0 for n e N. Moreover,
0n+i =
On
=
= 1.
(2) In this case set
n
E ak
h
—
Q
— —
(ln
k=1
CLn
oo
Then the series J2 jT diverges (see, e.g., Problem 3.2.10). Moren=l
over,
On
a
0n+i =
n+l
—
a
&TI+1
= — 1.
&n+l
n+l
3.2.85.
(a) It is enough to apply the ratio test to each of the series:
oo
oo
a
oo
2^ftl+/cn,
n=0
2_^ kn,
n=l
•••>
/ ^Q(fc-l)+fcnn—0
(b) It is enough to apply the Raabe test (see 3.2.19) to each of the
series given in the solution of (a).
3.2.86. By assumption, there exists a positive constant K such that
Vn<K-
Inn
, n>2.
Let us define the sets of positive integers Ni and N2 as follows:
Ni = in:
an < \
j
and N2 = N \ N i .
For sufficiently large n G Ni,
(1)
1
K
x
an-^<an-^=an^
!jk-
ln-L
(eK\~^t~
=( _ j
e2K
< -g-.
Furthermore, for sufficiently large n €N 2 ,
(2)
^
< a ^=
an
—)
\anJ
<n^=e2K.
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Solutions. 3: Series of Real Numbers
302
Combining (1) and (2) with the convergence of the series ^2 am
we
arrive at
^jT a^"^71 < +oo
and
nGNi
^
alT^
< +oo.
n €N 2
3.3. The Integral Test
3.3.1. For fe- 1 < x < fc, Jfc > 2, we have / ( x ) > /(fe). On the other
hand, for k < x < k + 1, we have f(x) < f(k). Hence
pk+l
/
pk
f(x)dx
< f(k) < /
k = 2,3,....
f(x)dx,
Summing both sides of the above inequalities from k — 2 to k = n,
we get
y"
/(x)dx < /(2) + /(3) + ... + f(n) < J"
f(x)dx,
which proves the integral test.
3.3.2. Note that 4 is a positive and monotonically decreasing function. Therefore, by the integral test, convergence of the given series
is equivalent to boundedness of the sequences
< f f'(x) dx > and
{] %$**}. Since
[" f'(x)dx
and
= f(n)-f(l)
f Q^- dx = ln/(n) - ln/(l),
either both sequences are bounded or both sequences are unbounded.
N+l
3.3.3. We have SN-IN-(SN+i-IN+i)
0. Moreover, / ( n ) <
n
/ f(x)dx
= f
f(x)dx-f(N+l)>
N
< f(n-
1) for n = 2,3,...,7V.
n-l
Summing these inequalities from n = 2 to n = AT, we get SV / ( l ) <IN<SNf(N). Hence 0 < /(iV) < S N - / * < / ( l ) , which
completes the proof.
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303
3.3. The Integral Test
3.3.4. Convergence of the given sequences follows from the foregoing
problem. It remains to show that the limits of these sequences belong
to (0,1).
(a) Since f(x) = \ is a strictly decreasing function on the interval
(0, oc), SN - IN < S2 - h < f{l) = 1 for N > 2 and
f(2) + f(3) + ... + f(N-l)
+ f{N)
/(s)dx,
>f(2) + f{3) + ... + f(N-l)>J
or equivalently, SN — f(l)
> IN — h- Finally,
0 < 1 - I2 < lim (SN - IN) < S2 - h < 1N—>-oo
(See also 2.1.41 and 3.1.36).
(b) The proof is analogous to that in (a).
3.3.5.
oo
(a) Convergence of the series ]T n(\nn)a
n
ls e ( u v a e n
n=2
l i l t t ° bounded-
ness of the sequence J x(\nx)a ^x- For a ^ 1,
2
1
(lnn)~a+1
J
ax —
a
J x(\nx) ~~
-a +1
2
(ln2)~ a + 1
-a-hi
Thus the series converges if a > 1 and diverges if 0 < a < 1.
Clearly, if a < 0, then the series diverges. Finally, if a = 1, then
n
n
2
2
1 l^t^c^x = ln(lnn) — ln(ln2). Hence the sequence J ^j^dx
is
unbounded and therefore the series diverges,
(b) In this case, we have
1
-dx = ln(ln(lnn)) - ln(ln(ln3)).
/ 3 x In x In (In x)
i:
Thus by the integral test the series is divergent.
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Solutions. 3: Series of Real Numbers
304
3.3.6.
(a) We have
N
N q
_ V"^ ^ n + 1
Gn+l
ST.
—
TV
q
rSn+1
rlnr
= In In SV-f i — In In S\ —• oo.
N—>-oo
(b) As in (a), we have
N
N
Sn In2 Sn
^2
^
Q
N
Q
i
rSn
Sn In2 Sn " ^ 7 ^ x In2 a;
1
1
1
In SN + In Si N^OO In S i '
2
3.3.7. If
H/{x)f{<p{x))
JJ—
< q < 1 tor
x >
XQ,
then
|
f(t) dt = J V'(t)f(v(t)) dt<qf
V?(x0)
^o
f(t) dt.
x0
Hence
p<p(x)
(1 - g) /
(
/
/(<)* < q
fV(x0)
/
JXQ
nx
/
JX
/(*)*
\ •
ry{x)
/(«)* - /
\
pv(x)
/(*)* - /
rv(xo)
f(t)dt \<q
OO
J
Thus by the integral test, the series Yl f(n)
f(t)dt.
JXQ
^s convergent.
71=1
Now if
<p'(x)f{<p(x))
—
>1
tor
x>X(h
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3.3. The Integral Test
then
J
305
f(t) dt> J f(i) dt. As a consequence,
<p(x0)
x0
V>{x)
<p(x0)
I f(t)dt > J f(t)dt.
X
XQ
Moreover, since for any n there exists kn £ N such that n < ip(ri) <
n + kn, we have
rn+kn
r<p(n)
/
f(t)dt> /
Jn
Jn
/(*)*> /
rv{xo)
f(t)dt.
J XQ
Therefore {/n} is not a Cauchy sequence, and consequently, it is not
bounded. So by the integral test the series diverges.
3.3.8.
(a) If lim (~9(x)jTxT
~ 9'(x))
x—+00 ^
> 0, then there exist xo and 6 > 0
'
such that
~9(X)-FT-V
- 9*(x)
> <$
Therefore —{g(x)f{x))' > 8f(x),
ficiently large n, we get
f
f(x)dx<-\
Jxo
G
r'
for
x >
x0.
x > XQ. Consequently, for suf-
-{f{x)g{x))'dx
J XQ
= -$(9{xo)f(xo)-g{n)f{n))
<
-^g(x0)f{x0).
Thus by the integral test, the series converges.
(b) As in (a), we get —(g(x)f(x)Y
< 0 for x > XQ. Thus the
function gf is monotonically increasing on [zo,oo), and consequently, g(x)f(x) > g(xo)f(xo) if x > x$. This means that
f(x) >
f(x ) g Xo)
l ( x)
for x > x0. Therefore the sequence J
f(x)dx
l
n
is unbounded because, by assumption, the sequence J - ^ dx is
unbounded.
l
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Solutions. 3: Series of Real Numbers
306
3.3.9. It is enough to apply the result in the preceding problem to
g(x) = x.
3.3.10. In 3.3.8, we substitute g(x) = xlnrr.
3.3.11.
(a) Set
oo
ff(t)dt
x
^ =
Then -g[x)^-sf{x)
-jw
= \>0.
(b) Put
/
n
f(t)dt
n
1
1/2
1/2
Then J ~zh\dx = In J f(t)dt — In J f(t)dt,
1
X
which means that
n
the sequence / ~^h\dx is unbounded. Moreover,
-9(x)j^l-g'(x) = -l<0.
3.3.12. We will apply the test proved in 3.3.9.
(lnx)-(lnx>>\ x> 1, we get
Taking f(x)
=
- x § ^ = (lnx)7-1(7lnlnx + l).
If 7 > 1, then lim (In a:) 7 - 1 (7 In In x + 1) = +00, and consequently,
X—KX>
the series is convergent. On the other hand, if 0 < 7 < 1, then we
have lim (In # ) 7 - 1 (7 In In x + 1) = 0, which means that the series is
£—•00
divergent.
3.3.13. Set
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307
3.3. The Integral Test
One can show that Km f ~xf}^y
) = 1. So we cannot apply the test
given in 3.3.9. Thus we will apply the test proved in 3.3.10. For
sufficiently large x,
fix)
JKJ
f(x)
because Xm^ ( ^
1\ ,
lnx
*
x
l
n
x
=
x)
m-i n x
lnx
7(minx)
^
r^2 + l > 2
- j ^ y ) = +oc.
3.3.14. We have (A n+ i — An) ^
TW y <
j
Hence
jrr^dt.
An
00
/
A
\
Z"00
1
1
Thus we have proved that the series X] (1 — y*^-) jry—y is conn=l ^
n
/
n
vergent. Let {Sn} and {Sfn} denote the sequence of partial sums of
the series given in the problem and the sequence of partial sums the
above series, respectively. Then
N
y(-l( T 7 ^ -
< >
.
L'
, 1<
1
which implies convergence of the given series.
3.3.15. By the monotonicity of / ,
(*)
/ ( A n + i ) ( A n + i - An) < f ^
f{t)dt < /(A n )(A n + i - A n ).
(a) By the left inequality and by our assumption, we get
M^/(A
n=l
n + 1
)</
f(t)dt<
00.
^Ai
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Solutions. 3: Series of Real Numbers
308
(b) The right inequality in (*) implies the divergence of the series
oo
£ /(An)71=1
OO
3.3.16. Assume first that the series J2 j ^ y converges. Then, by
n=l
oo
the integral test, the improper integral J jr^dt
also converges. In-
tegration by parts and then integration by substitution give
1
,
,.
V*J
t
1
f°° tf'lt)
,
-|
tj .
1_
ffOO n-lf
- ^TT +
+ I
—^-dt.
* - - / ( *(*)
)
/(I)
m) ' Jf Jfii)~2
lim -^r
We will now show that
2t
The convergence of the improper integral implies lim J yr^- dx = 0.
t-*oo? / W
=
Since ^jrb
2t
2*
jm) IJ dx < j JJ-T dx, the equality (**) holds. Thus
the improper integral /
t?
^ converges.
l
Moreover, we have
£(^TTj^SA
—*=h
—dt<oo>
which means that the series ]jn fn+i?a converges. Obviously, the
n=l
^
/
_ 1
(n)
i
y
1=1
series 72^
also converges.
n$
To prove the implication in the other direction, assume that the
00
series ^
f_ 1{
J
na
^
converges. In much the same way we can show that
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3.4. Absolute Convergence. Theorem of Leibniz
the integral
f
dt converges, and consequently, / jWjdt
* t2
309
a so
^
converges. Thus by the integral test the series ]T JT^T is convergent.
71=1
3.3.17. First observe that the function (p can be defined, in the same
way, on the whole interval [e, oo). Then ip{x) = 1 for x G [e,e e ),
<p(x) = 2 for x G [e e ,e eC ) . For simplicity set e 1 = e and ek = ee
for k > 1. Thus we have
for
<f{x) = k
Let
x<G [e fc ,e fc+1 ).
1
x(lni x)(ln 2 x) •... • (ln^x) x)'
fix
then
1
/(*) = x(lni x)(ln x) •... • (In* x)
2
for
x€
[ek,ek+1)
Now, by the integral test, our series diverges because, for n > ek,
e fc
+
r fc
e2
e3
i
-dx = fc — 1.
Jgfe-i x(lnx)(ln 2 x) • ... • (ln/b-ix)
3,4. Absolute Convergence. Theorem of Leibniz
3.4.1.
(a) We have
an
a.
n+1
Thus the series converges absolutely if \a\ < 1 and diverges if
\a\ > 1. If |a| = 1, then the series diverges because
lim
n—•oo
lim
n—•oo
an
n+ 1
= lim
n
i
^°°( +^r
1
e
*
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Solutions. 3: Series of Real Numbers
310
(lnx)a
for x > 0. Then f'{x) = ( l n x ) a " 1 2 ( a " l n : c ) < 0
for x > max{l, e }. Thus by the Leibniz test the series converges
for every a E l . We will now decide whether the series converges
(b) Set f(x)
a
oo
absolutely, that is, whether the series Y2
n=2
n
converges. By
the theorem of Cauchy (see 3.2.28), the convergence of this series
oo
is equivalent to the convergence of ^ n a (ln2) a . Thus our series
n=2
converges absolutely if a < — 1.
(c) If a > 0, then by the Leibniz test the series converges. If a < 0,
then
^-irsm^^-ir 1
OO
OO
I
sin • a
,
n
Applying the Leibniz test again, we see that the series converges
for all a G i The series does not converge absolutely if a ^ O ,
because
sin L-L
lim — j - s - = a.
n—>oo —
n=l
n=l
-4a-8
(d) The series converges if and only if — 1 < -^+6a-16
< X> t h a t 1 S >
if a G [—4,|)U[3,oc). Clearly, the series converges absolutely if
ae ( - 4 , | ) U ( 3 , o o ) .
(e) Since
= 0
lim
if \a\ > 1,
n—>oo
the series converges absolutely if \a\ > 1. If \a\ < 1, then the necessary condition for convergence is not satisfied because lim
n—->oo
\a\"
= +OC.
(f) Observe that
fin r))lnn
lim ^ a}- = lim n]nlnn-a
n—>oo
n-^oo
fi
= +oc.
Thus the necessary condition for convergence is not satisfied.
3.4.2. If \a\ < 1, then for sufficiently large n,
,n-l
a'1
\n — 1
< \a
na ~ + Inn
n l
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3.4. Absolute Convergence. Theorem of Leibniz
311
Thus the series converges absolutely. If \a\ > 1, then
a71'1
nan~l -h Inn
I (
1
\
n \1 + nn
Therefore for sufficiently large n the terms of the series are positive,
and, by the comparison test, its divergence follows from the divergence
oo
of E£n=l
3.4.3. Assume first that an > 0 for all n G N. Using differentiation,
one can show that sinx > x — %- for x > 0. Hence 1 - s i n a n < \ai.
oo
Since a^ < an for sufficiently large n, the series Y a^ is convern=l
gent, which in turn implies the convergence of the given series. If
we drop the assumption an > 0, then the series can diverge or converge. Indeed, take an = (—l) n ^r with a > 0. Then the series
J2 f 1 - ^m^n 1 diverges if 0 < a < \ and converges if a > \ .
n=i
n
^
'
3.4.4. No, as the following example shows:
(~l)n
n
1
n In n
(-l)n
n
,
3.4.5. We have an — pn — qn and \an\ = pn -f g n . Note also that
oo
pn and qn are nonnegative. Thus both the series Yl Pn
oo
oo
a
an
n=l
oo
d Yl Qn
n=l
a
diverge, because Yl n converges and Y \ n\ diverges.
71=1
71=1
3.4.6. Set Sn = ai -f- ... + a n - By the foregoing problem, we have
l i m ^ =
Qn
n ¥OC
lim (l +
%L)=l.
-*°° \
QnJ
n
3.4.7. The series does not converge absolutely. We will show that it
converges (conditionally). To this end we group the terms with the
same sign and we obtain
(-D[?]
n=l
3 , ^ ,
n=l
^
n
( l
N
,
1
i-
1
'
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Solutions. 3: Series of Real Numbers
312
Thus the convergence follows from the Leibniz theorem.
3.4.8. Clearly, the series converges absolutely if a > 1 and diverges
if a < 0. We will show that if 0 < a < 1, then the series converges
conditionally. Observe that the first three terms of the series are
negative, the next five terms are positive, etc. Now grouping the
oo
terms of the same sign we get the alternating series ]T] (—l) n A n ,
where An =
(n+l)2-l
Yl
k=n2
n=\
^ . Moreover, for a ^ 1,
~dt = 4" + :4-((" + i)2"2" ~ ^2_2a)-
An < 4 - + /
;
n2a
Jn2
ta
n2a
l-au
'
Hence (see 2.2.3), lim An = 0 if h < a < 1. For a = 1 we have
n—•oo
^ Y < i4 n < ^ ^ , and consequently, lim A n = 0 for \ < a < 1.
We will now show that for such a the sequence {An} is monotonically
decreasing. Indeed,
(n+l)2-l
An - A n + i =
]T
(n+2)2-l
— -
/c=n 2
(n+l)2-l
= V
^
k=n2
(n+l)2-l
1fca
y (1
v
fe_n2
^-—'
fca
V k<*
2^
(n+l)2 + l
V
I
(fc' + 2n + l ) a
^
k'=n2
j ^
fc=(n+l)2
V
1
{k + 2n+l)aJ
ffc-4- 9r7
y
((n + 2 ) 2 - 2 ) a
((n + 2 ) 2 - l ) a
2n
A=:0 v
,n 2 + fc)« ((n + l ) 2 +
fc)°y
((n + 2 ) 2 - 2 ) a
> v(2n + 1)y
((n + 2 ) 2 - l ) a
V("2 + 2 " ) a
1
1
((n + 2) 2 - 2) a ~ ((n + 2 ) 2 - l ) a '
((n + 1 ) 2 + 2n) a
where the last inequality follows from the monotonicity of the function
9(x) =
1
(n +x)a
2
1
( ( n + l ) 2 + z)a
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3.4.
Absolute Convergence. Theorem of Leibniz
313
on the interval [0,2n]. Hence, for sufficiently large n,
An-An+i
2
n2a
>n-
2o
1
> ( 2 n + l ) (n 2 + 2n) a
+
((n+l)2
1
+ 2n)aJ
(n + l ) 2 a
•+a((' r-o-^r)-Nn
(2a-l) >0,
because (1 + : r ) - a > 1 - ax and (1 + x)~a < 1 - ax + ^ ^ x 2 for
a,x > 0. (These two inequalities can be proved by differentiation.)
oo
Thus by the theorem of Leibniz, the series Yl (—l) n A n converges if
2
n=l
< a < 1.
If 0 < a < | , then since A n > (2ra + 1) (n2+2n)° > ^ e
necessar
y
oo
condition for the convergence of ^ (—l) n A n is not satisfied.
n=l
3.4.9. As in the solutions of 3.4.7 and 3.4.8, we group the terms of
the same sign and rewrite the series in the following form:
§("1)n"1l[e«-i] + l + - + H J We also observe that
[e"- 1 ] + 1
[en]
^ [e"]
[e»] "
Moreover, since
b n f 1 .J^l
,
). 1 .I,
the necessary condition for the convergence of the series is not satisfied
and therefore the series diverges.
3.4.10.
(a) Observe that the series can be written in the form
oo
£(-l)Mn,
n=0
2n+1-l
where An = £
i
fc=2"
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Solutions. 3: Series of Real Numbers
314
Since An > 2% n + l l _ 1 —> ^, the series diverges,
(b) As in (a), the series can be written in the form
2n+1-l
oo
£(-l)M„,
where An = £
n=l
^-^.
£=2™
Moreover,
0 < A , < 2" 2 n In
* 2n
This implies lim An = 0. We will now show that {An}
n—>oo
mono-
tonically decreases. Indeed,
2n + 2_][
An+1 =
^
2
A: I n A: =
2n-y
=
-l
(2r*+1 -h «) ln(2"+ 1 H- Z)
^
!
x
n X
^
n+ 1
n
l
V(2 + +2J) ]n(2 + +21)
+
n
(2 +* +2Z +1) ln(2 n + x +2/ +1)
2n-l
"^ ^
2n-l
(2^+i + 2Z) ln(2^+ 1 -f- 2/)
<
^
(2 n + /) ln(2 n + I) =
An
'
3.4.11. We have
[~ir( ,f ^sin^^ ( -irfi-, \;l)n A
K
)
n
(-l) + v^
v^
V
= ( - l ) n s m — + - ^--sin^=
y/n
n —I
yn
By the Leibniz test, both the series
> (-l)nsin—=
^
v^
n=2
v
n
(-l) + VV
^-sin—.
n —1
vn
and
^
v^
> -—^-sin—-7r
v^
^-i n - 1
v^i
n=2
converge. But the series
- J L — sin
«n- 1
n=2
y/n
diverges, so the given series also diverges.
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3.4. Absolute Convergence. Theorem of Leibniz
315
3.4.12.
(a) The series converges absolutely (see 3.2.1 (f)).
(b) Convergence of this series follows from the test of Leibniz. The
series converges conditionally (see 3.2.1 (g)).
(c) Clearly, the sequence { y/n}, n > 3, is monotonically decreasing
and therefore the series converges. However, it does not converge
absolutely (see 3.2.5 (b)).
(d) Convergence follows from the monotonicity of {(l + - ) } and
from the fact that the limit of this sequence is e (see 2.1.38). To
prove that the series does not converge absolutely, we use the
inequality
ln(l 4- x) < x - -x2 + - x 3 , x > 0,
with x = ^, and we get (l + ^ ) n < e 1 - ^ " ^ ^ . jjence
e— ( 1 H— )
> e (l - e " ^
+
^ ] > e (l - e~^)
for
n > 1.
It follows from 2.5.4 (a) that, for sufficiently large n,
4n (l - e~^j
oo
Therefore the series ^2 ( e ~ ( ^ + n )
n=l
> -.
) ^i v e r g e s -
(e) The convergence of this series follows from the monotonicity of
the sequence {(l + ^)
} and from the fact that e is its limit
(see 2.1.38). In view of 3.2.5 (c), the series does not converge
absolutely.
3.4.13.
(a) The function
(lnx) a
/ ( * ) = xb
x G (es+oc),
monotonically decreases to zero as x —• oo. Therefore by the
Leibniz test, the series converges. We claim that if b > 1, then
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Solutions. 3: Series of Real Numbers
316
the series converges absolutely. By the theorem of Cauchy (see
3.2.28), it is enough to show the convergence of
n=l
Now by the root test this series converges if b > 1 and diverges
if 0 < b < 1. Clearly, if 6 = 1 , this series diverges.
(b) Note that
(lnnVnn
e(lnn)(lnlnn)
b
^lnlnn
b
n
n
nb
Hence the necessary condition for convergence is not satisfied.
3.4.14. By the monotonicity of {a n }, we have
Tin — ( « 2 n + l ~ a 2 n + 2 ) + ( a 2 n + 3 ~ ^271+4) + ••• > 0,
?"2n+l = ( - G 2 n + 2 +
a
2 n + 3 ) + ( ~ « 2 n + 4 + «2n+5) + ... < 0
and
T*2n = « 2 n + l + ( - « 2 n + 2 + a 2 n + 3 ) + ••• < 0>2n+lj
- ^2n+l = «2n+2 + ( ~ ^ 2 n + 3 + ^271+4) + ••• < a 2 n + 2 -
3.4.15. Note that
n
E
n
= an+i - ai —>
{ak + ak+\) - 2}ak
-ax.
3.4.16. Observe that
n
n
*Y^{aak + bak+i + cak+2) - (a + 6 + c) ^
fc=l
afc
k=l
= 6(a n + i - a\) + c(a n + i -f a n + 2 — ai — a 2 ) —• - 6 a i - c(ai + 02).
3.4.17. By assumption, there exist positive constants c and C such
that for sufficiently large n, c < \an\ < C. Hence
]_1
1I
< "o | a n + l ~ « n | ,
l^n+1 ~ a n | < C^
^n+1
"n
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3.4. Absolute Convergence. Theorem of Leibniz
317
Thus our claim follows from the comparison test.
3.4.18. Let Sn and Sn denote the nth partial sum of Y2 an and
oo
71=1
n
^2 (<*>n — Gn+i)> respectively. Then
71=1
fc=l fc=l fc=l fc=l
= - ( n + l ) a n + i + S n +i,
which proves our claim.
3.4.19. Convergence follows from the Leibniz test.
3.4.20. If \a\ < 1, then the series converges absolutely. Indeed, since
| sinx| < |x|,
I
i •
•
a
•
a
\
/
l
in
n! sin asm - • . . . • sin — < a .
I
2
nl _ ' '
We now turn to the case \a\ > 1. We claim that in this case the
series diverges because the necessary condition for convergence is not
satisfied.
In fact,
for
a there exists no such that
-^ < the
1.
and using
Then, setting
C =
(n 0a- fixed
1)! sin
• s i nu
a sin •
n 0 —1
'
n0 —
> 1 - ^-, x > 0, we get
inequality
n! sin a sin •
Cnn
sin
n sin
• sin
n0
n
1
1
/
1 \
> Cn 0 • ... • nsin — • ... • sin — = C TT ( 1 — 2-7777 I
n0
n
^
V
6/c /
/c=n 0
^n('-p)fc=no
c
(n0-l)(n+l)
n0
non
3.4.21. By 2.5.4 (a),
_
lim
n—»-oo
Vb+yt
^5-1
= lim
Vb-1+
ltfc-1
n—>oo
In o
2
(In 6 + In c) = In —=.
V
Vte
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Solutions. 3: Series of Real Numbers
318
Hence if a > Vbc, then, beginning with some value of the index n,
the terms of our series are positive, and by the comparison test it
diverges. If a < \/bc, then the terms of the series are negative and it
also diverges. For a = \/&c, we have
y/b+ v 7 ^
E
i
^
•-Y
2
n=l
,
(V6- v^y
n=l
Since
lim
n—»oo
V b - l - V5+l'
(ln6-lnc)2
2n
the convergence of our series follows from the convergence of Yl
3.4.22.
(a) By 1.1.14, there exist a sequence of integers {pn} and a sequence
of positive integers {qn} for which
Pn
1
Qn
Hence |cosp n | = |cos(7rgn - pn)\ > c o s ^ = 1
Thus
(|C0Spn|)P->
1
2<7n /
2
2 sin _J_ >
2qn
<?n 2gn
This means that the subsequence {{cospn) } of {cos n n} does
not converge to zero. Therefore the necessary condition for convergence is not satisfied.
Pn
(b) By Problem 1.1.22, we know that the sequences {pn} and {qn}
mentioned in (a) can be chosen in such a way that all the terms
of {qn} are odd. Then by the inequality
Pn
1
we get | sinp n | = | cos (§<?n - pn) \ > cos ^ > 1 - ^ r . Thus, as
in (a), the sequence (sinpn)Pn does not converge to zero, and
consequently, the series diverges.
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3.4. Absolute Convergence. Theorem of Leibniz
319
3.4.23.
(a) By assumption (see also 2.4.13 (b)), there are no and a such that
Q-n+1
—1
> a > 0 for
n > TIQ.
Hence ^ ± < ^ ^ < 1, which means that, beginning with the
value no of the index n, the sequence {an} monotonically decreases. We will show that lim an = 0. It follows from the
n—->oo
above that
Qn+l
An
an
a n _i
ftnp
+1
71 • ( n -
a no
( a + n) - ... - (a + n 0 )
lim , n ' ^—Y""710 x = 0. Indeed,
Now it is enough to prove that
hm -
n- ( n - 1) . . . . . n 0
n - + o o [OL + n) - ... • (a + n o )
1) • ... • n Q
1
r
= hm
n
r- = 0,
7
n-*oo / |
i a\ ,
- i l l
Q
I
because (see 1.2.1)
1 + — . . . . . ( l + - J > 1 + — + ... +
> oo.
\
no/
\
nJ
no
n n-^<x>
oo
So by the Leibniz test, the series ^ (—l) n a n converges.
n=l
(b) By the assumption n ( ^
1] < 0, the sequence {an} monooo
tonically increases and consequently the series Yl ( — l) n<2 n din=l
verges, because the necessary condition for convergence is not
satisfied.
3.4.24. By assumption, lim n ( —^
1) = a. For a ^ 0, the test
n—»oo
\an+i
/
proved in the foregoing problem can be applied. For a = 0, the
necessary condition for convergence is not satisfied. Indeed, we have
Q
_I_ _ _I_ °± °2L
n-i
an
ai a2 a3
an
Ql
V' I ^ A
-111+e
2i+^-"V -"(^1)
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320
Solutions. 3: Series of Real Numbers
Moreover, there exists (3 such that \j3n\ < (3. Hence
°"a(i+T&)(i+5X')...(i+^)^'
CX O
where A=
£ ^T+In=l
3.4.25. By 2.5.34, the existence of the limit lim n In -9^- is equivalent to the existence of lim n ( - ^
1) , and both limits are equal.
Set an = ^ . Then lim n l n - ^ - = p - \. Hence, by 3.4.23, the
n
n—>oo
°n+i
^
series converges if p > ^ and diverges if p < \. In the case P — \
the necessary condition for convergence is not satisfied because, by
Stirling's formula, lim an = y/2n.
n—>oc
3.4.26. Let Sn = a\ + a2 + ... + an. We use the so-called summation
by parts (Abel's transformation) to obtain
n-l
axpi + a2p2 + ... -f anpn = ] T Sk{pk - Pk+\) +
fc=l
and we get
aipi + a2p2 + ... + anpn
Pn
\-^
= on - y bk
£j
pk+i - pk
.
Pn
Now it is enough to apply the Toeplitz theorem (see 2.3.1).
oo
3.4.27. Apply the result in the last problem to the series 7]T}
1 = 1 anbn
and take pn = -^.
3.4.28. The result is contained as a special case in the preceding
problem.
3.4.29. If the series were not absolutely convergent, then the subseries of all positive terms and the subseries of all negative terms
would diverge (see 3.4.5).
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3.4. Absolute Convergence. Theorem of Leibniz
321
3.4.30. [20] No, as the following example shows.
oo
Take a conditionally convergent series ]T) bn and set
71=1
a i = &i, a2 = a$ = —, a± = a$ = ... = a 9 = —,...,
a
l!+2! + ... + (n-l)!+l = a l!+2! + ... + (n-l)!+2
K
— ••• — ai!+2! + ... + (n-l)!+n! = " T J ••• •
n!
00
a
Then the series ]T n converges conditionally. But for each k > 1
n=l
and / > 2 the subseries
afc 4- a/c+i + ak+21 4- ...
converges. Indeed, for n > / there are y- terms of the form
Grouping these terms, we get the convergent series
^.
1 °°
4- y / ^ 6n-
constant
/
n=no
3.4.31. Consider the series
1
/
2v 2
+
_1
2^2
1_
¥2 +'"
_J_
n ^ + -
+
+
__1
^
1_
^
+
'
3.4.32. Yes. Consider the series
+
1
21n2
+
__1
21n2
1_
In 2 + ""
+
1
1
4- ...4nlnn
nlnn
1
4Inn
n times
Then
1+ E j ^ n
n=2
+ A )
if A: is even,
71=1
n=2
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Solutions. 3: Series of Real Numbers
322
By the Cauchy theorem (see 3.2.28) the series ]T] h ^ diverges f° r
n=2
every k G N. On the other hand, the series ]T
i=2
if k > 2.
nk-^lnkn
converges
3.4.33. [20] Suppose, contrary to our claim, that lim £i+£2+---+£n =
n—>oc
2a > 0. Then, by 2.4.13 (b), there is no such that for n > no,
(*)
£i + s2 + ... +sn > an.
Set £ n = e\ 4- £2 + ••• + sn- Using summation by parts, we get
n-1
eidi + e2a2 + ... + fnfln = J ^ ^(ajk - ajfc+i) + ^ n a n .
Therefore, by (*),
eiai +^2«2 + ••• + snan
no
n—\
>^Ek(ak-ak+i)
+a
fc=l
]P
k(ak - ak+\) + anan
/c=no + l
n
= constant + a
a
\ ,
fc>
/c=no+2
a contradiction.
3.4.34. [20] Set En = ex + £ 2 + ... + £ n , n e N. The sequence {£„}
has the property that between two terms with different signs there is
a vanishing term. We consider two cases:
(1) finitely many terms of {En} vanish,
(2) infinitely many terms of {En} vanish.
(1) is contained as a special case in 3.2.35. In case (2), by the Cauchy
criterion, for each e > 0 there is no such that if n > m > no, then
/ ^ {{Ek - Em) — (Ejfc-i — Em)) ak
e >
(*)
k=m-\-l
k=m+\
/ ^ {Ek - Em){uk — Q>k+\) + {En — 2J m )a n +i
k=m+l
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3.4. Absolute Convergence. Theorem of Leibniz
323
Assume that Em = 0 and that the terms i£m+i> £ , m + 2 , ...,En have
the same sign. Then (*) and the monotonicity of the sequence {an}
imply that
\Enan\ < £,
n > m + 1.
3.4.35. The proof is analogous to that of 3.4.33. We set En —
Pibi + ... +PnK and assume that lim P l 6 l + ' n + P n 6 n = 2a > 0. Then
n—>oo
for n > no we have p\bi -f ... + pnbn > an, and consequently,
&i + ... + bn = — (pih) + ... + — {pnK)
Pi
Pi
1 \
^r,
(I
= > E^
tl
VPfc
1
A
+ -c/n — > constant + a y
Pk+J
1
—,
k
kJ^+2P
Pn
a contradiction.
3.4.36. We first show that if p = q, then the series converges. We
have
2
+...
+
p/
2p
\p+l
(_D'+I (
I
+ ... + -
Therefore 5/ p is a partial sum of an alternating series. By the Leibniz
test the limit lim Sip exists. Clearly, each partial sum of the form
Sip+ki fc = 1,2, ...,p — 1, tends to the same limit as / —• oo.
Assume now that our series converges. Then, by 3.4.34,
l i m nP~n(i
n-^oo np + nq
=
P~g =
p+q
0
which implies that p = q.
3.4.37. We note first that if conditions (i)-(iii) are satisfied, then for
any convergent sequence {a n }, the transformed sequence {bn} is well
defined. Now the proof runs in much the same way as in the solutions
of Problems 2.3.1 and 2.3.26.
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324
3.5. The Dirichlet and Abel Tests
3.5.1.
(a) Since
§1J J1
^
= ^ ( 1 — cos(2n)), it is enough to consider the series
OO
OO
H
y^(-l)n-
n' = ^l
n.
and
..
Y " ( - l ) n - cos(2n).
n' = lJ
n.
By the Leibniz test, the first series converges. Convergence of the
second one follows from the Dirichlet test (see, e.g., [12], p. 105).
Indeed, by the formula (which can be proved by induction)
71
(1)
V cos ka =
sin —
cos ( n+1 ) Q
1
—n—-—
for
a 4 2/TT, I £ Z,
we obtain
n
£(-l)fccos(2A:)
^ c o s f ( 7 r - 2)k\
fc = l
k=l
(^~2)n
(n+l)(7r-2)
<
cosl
cosl
Therefore the sequence of partial sums of ^ (—l) n cos(2n) is
n=l
bounded. Moreover, {^} tends monotonically to zero. Thus the
OO
series Y2 {-l)n^
n=l
cos(2n) converges.
(b) The sequence
n
of arithmetic means of {^} converges to zero (see 2.3.2). Moreover, it is easy to check that the sequence {an} monotonically
decreases. By the formula (which can be proved by induction)
(2)
2 J s^n ka '•
fc=i
s i n
^sin(g+j>
sin§
for
a^
2/TT,
I e Z,
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325
3.5. The Dirichlet and Abel Tests
we get
sin § sin £±i
yjsinfc
<
Sill;
k=i
sin 1 •
Therefore by the Dirichlet test the series converges,
(c) Observe that
,2
72T
COS
7T
n+ l
= COS
T17T ~
-(-1)
n+l
= (-l)nCOS
n+1
7T
n+ 1
cos •
n+l
Thus the given series can be rewritten in the form
oo
i 1c
o s
7TZT
ln 2 n
n=2
The convergence of the above series follows from the Abel test
(see, e.g., [12], p. 106), because the series J2 (~ 1) n + l
1
In2 n
converges (by the Leibniz test) and {cos j ^ } is a monotonic and
bounded sequence.
n=2
(d) We have
T17T-
sin J
n a + sin *f
The series
Li
sin-
1+
£
sin
a > 0,
converges (by the Dirichlet test). Now we will study the series
(with positive terms)
oo
E
sin 2 a
n
n—1 -1 '
na
There exist positive constants c a and Ca such that
oir, 2 JUL
~2a
n
<
1+
< Ca- 2 a '
rt
n ^ 4k, fe € N.
Therefore the series converges if a > ^ and diverges if 0 < a < \.
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Solutions. 3: Series of Real Numbers
326
3.5.2. We have
N
.
//
,. n1 \
NTV
•
^
ii
Y^SmnCOSn
^ - i In Inn
(n+^J
In Inn
n=2
AT
TV
..
ii
Y^COSnSmn
^-^ In Inn
n=2
n=2
By formula (2) given in the solution of 3.5.1(b) and by the Dirichoo
let test we see that the series ^T ^ ^
n=2
converges. Since the se-
quence {cos^} is monotonic and bounded, the series Yl
n=2
lnlnn
i
OO
converges, by the Abel test. Finally, the convergence of J2
n=2
In I n n
follows from formula (1) given in the solution of 3.5.1(a) and from the
Dirichlet test.
3.5.3.
(a) We have
2_]sm (&2fl0 s m (ka)
^ [ c o s (k(k - \)a) - cos {k(k + l)a)]
U=i
k=l
= | l - c o s ( n ( n + l ) a ) | < 2.
Thus the convergence of the series follows from the Dirichlet test,
(b) As in (a), the Dirichlet test can be applied.
3.5.4. In view of the formula
cos n sin (na)
n
l s i n ( n ( a + l))
2
n
1 sin (n(a — 1))
2
n
'
the convergence of our series follows directly from the Dirichlet test
(use formula (2) in the solution of 3.5.1(b)).
3.5.5. If a = kn, k e Z, then all terms of the series are equal to zero.
If a ^ kit then, by the inequality |sinx| > sin 2 x = | ( 1 — cos2x),
we get
E
n
71 = 1
| sin (na) |
1^
n
~ 2^
71=1
1
1 ^ cos (2na)
n ~ 2 ^-j
n
'
71=1
Hence in this case the series does not converge absolutely.
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327
3.5. The Dirichlet and Abel Tests
3.5.6. Assume first that 0 < a < 7r, and set m = v^ . Then, for
sufficiently large n,
E
fc=i
sin (ak)
k
sin (ak)
k=l
+
E
sin (ak)
k=m+l
Since |sin£| < \t\ for t ^ 0,
E
(«)
fc = l
sin (ak)
Jfc
ka
— = ma < 0 r .
fc=i
Moreover, from (2) in the solution of 3.5.1(b) and from the inequality
sin* > H, 0 < t < f, we get
sin (afc)
(**)
fc=m+l
<
(m + 1 ) | sin | | < £ v ^
Combining (*) with (**), we see that the desired inequality holds
for a G (0, TT). Clearly, since the sine function is odd, it also holds
for a G (—7r,0). Moreover, since sinA:7r = 0 and the sine function is
periodic, the inequality holds for every a G R.
3.5.7. The convergence of the series follows from the Abel test, beoo
cause the series ^ (~~l)n~7^ ^s convergent and {arctann} is a mono71=1
tonically increasing and bounded sequence.
3.5.8. By the Abel test the series converges. Indeed, ^2 (—l) n ^
71=1
converges, and the sequence { -\/lnx} is bounded, strictly decreasing
if x > e, and strictly increasing if 1 < x < e.
3.5.9.
oo
(a) Observe first that, by the Abel test, the series J^ ^
oo
n=\
converges.
Moreover, since the series J2 an is convergent, the sequence
71=1
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Solutions. 3: Series of Real Numbers
328
{fn}, Tn = ^ a&, tends to zero. Hence for p> n,
k—n
p
k=n
r_n+ y
Tk
bk
k=n
r
v
V^
T--T ^
t^h
rk ~ rk+i
k=n
1
(l
k=n
r
k+l
bk
T
P+l
<£ (— + 1 - 1 + 1 " ) = ^
where en = sup |rjt|. Consequently,
k>n
<2en^-=o(±-
07c
bk
/c=n
0n
\0n
(b) See 3.4.26.
3.5.10. Note that
Y]{k + l)cn+k = V " ——
~—'
fc=0
-(n + *: - l)cn+fc_i
f—' n -j- K — 1
fe=i
Thus the Abel test implies the convergence of ^ (fc + l)cn+fc for each
fc=0
n G N. Setting r n = nc n + (n + l)c n +i + ..., we get
t n = ]£^(fc + l)c n + f c = ^ ( f c - n + l)cfc
fc=0
oo
k=n
oo 1
oo
Zkck = rn ~ (n - 1) ^
= ^2kck-(n-l)^2
fc=n fc=n
1
-(r f c - r fc +i)
k=n
= ^ n + (n-l) ±
k=n+l
(~\)rk.
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329
3.5. The Dirichlet and Abel Tests
Hence
1
°° / 1
1\
| t n | < - | r n | + sup |r f c |(n-l) £
[-—- - k=n+l
1,
.
,
< - F n + SUP \rk\
n
fc>n+i
.ft"!
n
•
This together with lim rn — 0 yields lim £n = 0.
n—>-oo
n—>oo
3.5.11. By summation by parts,
5„ = Y.n °ibi = n—1
E ^
2=1
- 6 t l ) + ^«&n.
1=1
where An denotes the nth partial sum of the series ]T an. Given
n=l
e > 0, there is no such that \bi\ < e for i > no- So if ra > n > no
and \An\ < L, then
ra —1
&n\
l^Aitf-bt+J-AX
I i=n
+ Amlt
m —1
i—n
(
m-1
£
\bt - fti+illfc?"1 + 6?" V i + ... + b&\ + \bkn\ + \bkn
i=n
<L(kek-lT^2\bi-bl+1\
+ 2ek).
Hence the convergence of Yl anbn follows from the Cauchy criterion.
3.5.12. By summation by parts,
n-l
(*)
S„ = ^aibi
= Y Mbi ~ bi+i) + Anbn,
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Solutions. 3: Series of Real Numbers
330
where An denotes the nth partial sum of ^ a n . Since the series
n=l
oo
^2 (^n — bn+i)
converges absolutely and the sequence
is
{An}
n=l
oo
bounded, the series J2 An(bn
rc=l
- 6 n +i) converges absolutely.
oo
convergence of J2 fin — &n+i) implies that
n=l
n
The
lim bn exists, because
^°°
(6i - 62) + (62 - 63) + ••• + fin-i - K) = h - bn. Consequently,
n
00
lim Anbn
also exists and, by (*), Y] anbn converges.
n=l
^°°
3.5.13. For 0 < x < 1 the sequence {xn} is monotonically decreasing and bounded, and therefore the Abel test can be applied. For
— 1 < x < 0 both the sequences {x2n} and {x2n~1} are monotonic
and bounded. Consequently,
00
an<
^ a,2nx2n
i
n=l
00
Yl &2n-i£ 2 n _ 1 are
n=l
convergent. The convergence of our series follows from the equality
00
00
00
] T a n x n - ^ja2nX2n
n=l
+ ^a2n-_ix2n_1.
n=l
n=l
3.5.14. Observe that if x > xo, then
00
n=l
00
1
n=l
Now it is enough to apply the Abel test.
3.5.15. We have
OO
n=l
1
OO
n\an
^v—v ^n
an
x(x + l)...(x + ri) ^ nx
v
'
v
n=l
I T
nln^
x(x + l)...(x + n)'
v
7
v
'
n
Note that for sufficiently large n all the numbers , +%l (x+n\ have
the same sign. We will show that they form a monotonic sequence.
To this end, observe that the ratio of the (n + l)th term to the nth
term is
( n + l ) ( n ± l ) * ^ c (*+l)ln(l+£)
x+ n+1
~"
1 + £±I
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3.5. The Dirichlet and Abel Tests
331
Now set Rn = e ( x + 1 ) l n ( 1 + - ) - 1 - s±I. By the result in 2.5.7 we see
that
^ = ( , + l)(ln(l + I ) - I ) + I ( ,
+
i(,
+
l)3ln3(l
+
+
l)M^(l + i
I)+...
i ( x + i . i, . 1 ) 2 . „n
n»V
2 + 2<* + 1 > + ° V n
i,
. „ . , / . .1
where O (an) denotes an expression whose value divided by an remains bounded as n —> oo. This implies that for sufficiently large n,
i? n is positive if x(x + 1) > 0 and negative if x(x + 1) < 0. Consequently, for all sufficiently large n the ratio of two consecutive terms
°^ 1 x(x+i)nX(x+n) f *s e ^ n e r greater than 1 or less than 1. We will
now show that this sequence is convergent for x ^ O , —1, —2,.... To
this end, write
x
\n
n\n
x(x + l)...(x + n)
x x
+ n7'_
1+ f
Assume first that x > 1. For such a; we have (1 + i)
Consequently,
k
fc=l
k=l \
\
/
> 1 + f.
/
where all terms of the sum are positive. Moreover, note that
lim
* l n ( l + l ) - l n ( l + f)
fc-^OO
Jy
=
zQr-1)
2
As a result, the existence of the limit
lim in TT i i l i L
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Solutions. 3: Series of Real Numbers
332
follows from the convergence of the series ]T] -^. Thus the sequence
under consideration converges for x > 1.
Assume now that x G (0,1). Then for such x we have (l + £) <
1 -f | . Therefore one can apply the above reasoning to the sequence
with terms
Finally, we consider the case a: < 0, x ^ —1, —2, —3,.... Let k0 be
a positive integer such that 1 + | > 0 for k > /c0. To show that the
sequence
-'(i+i)*
_+k
k = k0
converges, note that
1 \
1+ - J > 1 + |
for
k > k0
and proceed as in the case x > 1.
3.5.16. It follows immediately from the Abel test that for \x\ < 1
oo
oo
the convergence of J2 anXn implies the convergence of ]T
n=l
n=\
anx2n
(see 3.5.13). Since {1_1x2n} is monotonic and bounded, the equality
V^
n=l
XH
\V
n—l
n
1
,
2n
1
x
^
/
oo
n
and the Abel test imply the convergence of ^2 an i-xn •
n=l
oo
3.5.17. [20] No. Let ^ 6n be a conditionally convergent series. Put
2
n=l
°°
F(x) = 2X and define the new series ]T an by setting
n=l
b\
a i = 0 2 = -7T,
2
brn
_
«fc = — —
—
— for F ( r a - l ) < k < b\m).
v
F(m) - F(m - 1)
'
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333
3.6. Cauchy Product of Infinite Series
This series is also conditionally convergent. Now we show that every
subseries of the form
(*)
Q>k + au + afc/2 + ...
converges. Note first that for any positive integer m there exists a
unique tm, tm — log/ — ^ , such that
kltm
<F{m)<kltm+1.
It follows from the definition of tm that, beginning with some value
of the index m, the subseries (*) has tm - tm-\ terms of the form
F(m)-F(m-i)- Grouping these terms, we transform (*) into the series
00
+
- +
— 6m.
1)
constant + > ——— —
^—' F(m) - F(mrn=ni
x
v
J
'
This series converges by Abel's test, because the sequence with terms
t"m
Crn
£m—l
~ F(m) - F(m - 1)
is monotonically decreasing. Indeed,
(2m- l)logj2- 1
(2m+l)log/2 + l
Cm+1 <
2 m 2 - 2( m - 1 ) 2
2( m + 1 ) 2 - 2 m 2 '
Hence for sufficiently large m we have c m + i < c m , because
Cm >
,.
lirri
( 2 m + 1)log, 2 + 1
m=£o(2ra-l)log,2-l
•
2m2-2^m-1)2
2
2(^+!) -2^
2
„
= f)
3.6. Cauchy Product of Infinite Series
3.6.1. Assume that the series ^ an converges absolutely. Let
n=0
00
Bn and Cn denote the nth partial sums of Y
n=Q
respectively. Then
a
00
n > Yl bn
n=0
an
An,
00
d Yl cn >
n=0
Cn = a0bo + (ao&i + GiM + ••• + («o^n + ai&n-i + ... -f a n 6 0 )
= a>oBn + &iBn-i
-f ... + ttnBo-
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Solutions. 3: Series of Real Numbers
334
Write
B = Bn + r n ,
where
lim rn = 0.
n—->oo
Consequently,
Cn = BAn - (a0rn + a i r n _ i + ... +
anr0).
We will now show that
(*)
lim (a0rn + a i r n _ i 4-... + a n r 0 ) = 0.
n—>oo
To this end, let e > 0 be chosen arbitrarily and let ra, M be such
that
\rn\ < m for n > 0,
oo
M = ^|P \an\.
There exist A; 6 N and I G N such that if n > fc, then \rn\ < -^
and if n > I + 1 , then |a/ + 1 | +... + |a n | < ^ • Therefore for n > I + k
we get
\a>orn + a i r n _ i + ... -f a n r 0 |
< (|ao|kn| + ... + |a/||r n _j|) + (|az+i||r n _j_i| + ... + |a n ||r 0 |)
< (|a0| + K| + ... + H ) ^
< MWT7
+
^a/+1' + "' + W ) m
+ 7T-™ = e,
2M
2m
which proves (*).
Note that it follows from the above analysis that if both the
series converge absolutely, then their Cauchy product also converges
absolutely.
3.6.2.
(a) It follows from the Mertens theorem that if |x| < 1, then the
oo
Cauchy product of the series ^ xn with itself converges. Moreover,
cn = xn + xxn~l
+ ... + xn = (ra + l ) x n .
Hence
oo
n=l
(b)
/
^
\ 2
xl-x
T^-T^-
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335
3.6. Cauchy Product of Infinite Series
(c) The series is the Cauchy product of ^
n(n+i)
n=l
£ • The
with £
n=l
sum of the first series is 1 (see 3.1.4 (b)), and the sum of the
second is e — 1 (see 2.5.6). Therefore, by the Mertens theorem,
the sum of our series is e — 1.
3.6.3.
(a) We have
=
C n
V—
1
' ^ k\'2n-k(n-k)\~
v
k=0
=
'
1 AMfc 1 = 1 /
n\^\k)
2"-k~
n\\
k=0
x
7
+
x
l\ n
2J '
/
By 2.5.7 the sum of the Cauchy product is e*.
(b) The Cauchy product is the series
y> 1 A(~3)fc
Z ^ 3n+l 2 ^
k '
n=l
k=l
By 3.1.32 (a), its sum is - \ In2.
(c) We have
2n+l
2n l
J2 (-!)*(* + l)(2n + 1 - fc 4-1)
C2n+i = x +
k=0
= x 2 n + 1 f J2(~l)k(k
2n+l
+
I]
+ l)(2n + 1 - k + 1)
\
(-l)*(fc + l)(2n + l - f c + l)J
= x2n+1 fet-1)^ + l)(2n + 1 -fc+ 1)
x
n
fc= 0
,
\
] T (-1)* (fc' + l)(2n + 1 - k' + 1) I = 0.
Moreover, since fc'=n
C2n+i = 0, we get
/
fc'=0
2n
c2„ = x2n ^ ( - l ) 2 " - f c ( f c + l)(2n - k + 1)
fe=0
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336
Solutions. 3: Series of Real Numbers
/2n-l
=x2n (J2 (-vk(k+x)(2n
^ fe=0
- 1 - *+1)
+ £ ( - l ) f e ( f c + l) + (2n + l ) )
k=Q
2n
'
= x (0 + (-n) + (2n + 1)) = (n + l)z 2 n .
Finally, by 3.6.2 (a),
oo
(1-X2)2'
n=0
3.6.4. Observe that the series 53 Anxn
is the Cauchy product of
71 = 0
OO
OO
53 £ n with 53 a n £ n . Therefore it is convergent for |x| < 1 and its
n=0
n=0
sum is ^
OO
53 «n^n-
n=0
3.6.5. To prove the equality given in the hint it is enough to equate
the coefficients of xn in the formula (1 + x)n{l + x)n = (1 + x)2n.
Consequently,
Cn = (
_ i
f
^ l
(
n !
f f ^ 2 , (.x)-^
)2~W
1
/2n\
W W
3.6.6. In view of the relation
C
_/ll-3-...-(2n-l)
1 1 l-3-...-(2n-3)
+
" ~\a
2-4-... -2n
2 a + 22 • 4 •... • (2n - 2)
1
l-3---(2n-l)\
a + 2n 2-4-... • 2n / '
it is enough to prove the equality
ll-3-...-(2n-l)
1 1 l-3-...-(2n-3)
+
a 2-4-...-2n
2 a + 2 2 • 4 •... • (2n - 2)
1 l-3-...-(2n-l)
(a + l)(o + 3)...(a + ( 2 n - l ) )
+ ...+ a + 2n 2-4-... • 2n
a(a + 2)(a + 4)...(a + 2n)
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337
3.6. Cauchy Product of Infinite Series
To this end, we decompose the right-hand side expression into partial
fractions:
(q + l)(q + 3 ) . . . ( a + ( 2 n - l ) )
a(a + 2)(a + 4)...(a + 2n)
=
a0
a
o?i
a+ 2
an
a + 2n
Multiplying both sides of this equality by a(a -f 2)(a -f 4)...(a -f 2n)
and substituting a = 0, a = —2,..., a = —2k,..., a = - 2 n , we get
(2n-l)!!
(2n)!! '
-l(2n-3)!!
ai =
- 2 ( 2 n - 2 ) H '•"'
(-2fc -f l)(-2fc + 3)...(-1)1 • 3...(2(n - k) - 1)
Oik =
-2k{{-2k -f 2)...(-2) • 2 • 4...(2(n - fc))
_ (2fc-l)!!(2(n-fc)-l)!!
(2fc)!!(2(n-*))!!
_ (2n-l)H
an
~
(2n)!! '
a o =
which gives the desired equality.
oo
oo
n=0
n=0
3.6.7. Let Ani Bn, Cn denote the nth partial sums of Yl an> J2 ^n
oo
and Yl cm respectively. It is easy to check that
71=0
Cn = aoBn + aiBn-i
+ ... + anBo.
Therefore
Co + C\ -f ... -f C n = Ao-Bn -f AijB n _i + ... -f AnjBoDividing both sides of the last equality by n-f 1, using 2.3.2 and 2.3.8,
we obtain C = AB.
OO
OO
n—\
n=l
3.6.8. Let Jl cn be the Cauchy product of £ ( - l ) n _ 1 ^ with itself. Then
^ = (~l)n * h
1-n
+ XT
rr + ... + 77
r — r r -f ... +
' 2 ( n - l ) ' **'
fc(n-fc-hl)
'"
n-ly
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Solutions. 3: Series of Real Numbers
338
Since
k(n-k
+ l)
n+1
-H
:
I
\k
n-k + 1
for
k = 1,2, ...,n,
we can write
/
1X „-1
2 /
n+1\
1
1
2
1
n
3
We know that £ ( - l ) n _ 1 £ = In 2 (see 3.1.32 (a)) and that the series
71=1
OO
E (-l)
n - 1
n=l
^ (l + £ + - + £)
is
convergent (see 3.4.19). Thus, by
the result in the preceding problem,
N
71=1
/
3.6.9. If Yl cn i s the Cauchy product of the series ^ ( ~ l ) n 1 ~T^
n=l
n=l
with itself, then
Cn
= (-lr- 1 (—L= +... + _ /
+ ...+ *
Since each term in the parenthesis is greater than ^, we see that
oo
\cn\ > 1 for n > 1. It then follows that Yl cn *s a divergent series.
n=l
3.6.10. We have
cn = a0bn + ai6 n _i + ... + anb0 > a0bn,
oo
and consequently, if the series ]T bn diverges, then the Cauchy prodoo
ra=0
c
uct Y, n also diverges.
n=0
3.6.11. No. Consider the following two divergent series:
71-1
( 9 n -I
71=1
X
'
71=1
X
7
X
i+l
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339
3.6. Cauchy Product of Infinite Series
Then
n-l
cn = a0bn + b0an + ^
a>kbn-k,
where a 0 = bQ = 1, a n = - ( | ) n , 6 n = ( | ) n
(2 n + ^r+r) • Hence
:
/3
«,-iTr'(*+2 ^
2/
\
/
\2
n+1
-(l)-g(^+i^).(5y
n — 1 n—1
3.6.12. Let .An,iE?n,Cn denote the nth partial sums of the series
Y2 ani
n=0
id ^^2 c nni, respectively. Then
Yl bn and
n=0
n=0
Cn = ^O^n + a\Bn-i
+ ... + CLUBQ.
As a result,
n
22 ak(bn + 6n-l + ... + &n-/c + l)
= Q>\{Bn — Bn-i) + CL2{Bn — Bn-2) + •• + an(Bn — BQ)
— Bn{An
— ao) — a\Bn-\
— a2#n-2 — ••• — «n^o = BnAn
OO
OO
n=0
n=0
— Cn.
3.6.13. Let J ] c n be the Cauchy product of the series ] ^ (—l) n a n
oo
with £ (-l) n 6 n . Then
n=0
cn = {-l)n{a0bn
+ ai6 n _i -f ... + anbQ).
oo
Assume first that the series Y
71= 0
c
n converges. Then lim cn = 0. By
" - ^
the monotonicity of the sequences {an} and {6 n }, we get
|cn| > bn(a0 + ... -h an)
and
|c n | > a n (6 0 + ... + bn).
lim an(bo + &i -f ... + 6 n ) = 0
and
lim 6 n (a 0 + ai + ... + a n ) = 0.
So
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340
Assume now that these two equalities hold. Then, by the preceding
problem, it is enough to show that
n
l i m y ; ( - l ) f c a f c ( ( - l ) " 6 n + ( - l ) " - 1 6 „ _ 1 + . . . + ( - i r - f c + 1 6 „ - f e + i ) = 0.
fc=l
Note that
| ( - l ) n 6 n + ( - l r ^ & n - i + ... + ( - l ) n - f c + 1 6 n _ f c + i | < 6 n - f c + 1 ,
and consequently,
n
£(-l)*a f c ((-l) n 6 n + ( - l f - 1 ^ ! + ... + (-l) n - fc+1 6 n _, +1 )
n
n—fc + 1 •
Now we show that lim V" akbn-k+i = 0. Indeed,
2n
0 < ^ J afc62n-fc+i < («i + ... + a n )6 n + (&i + ...6 n )a n ,
fc=i
which implies that
2n
lim Y] a^n-k+i
n
-*°°k=i
2n-l
= 0. In much the same way
we show that lim Y] dkb2n~k = 0, which completes the proof.
n
-*°° fc=i
3.6.14. Observe first that it is enough to consider the case where
both a and j3 do not exceed 1. We will now show that
lim — ( l + — + ... + _
] =0
if and only if a + j3 > 1. By the Stolz theorem (see 2.3.11),
lim — [ l + _ + ... + _ ) =
Hm
= lim
nP(na - ( n 1
n->oona+0(l_
l)a)
Q
( 1 - 1i )\ Q \)'
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3.6. Cauchy Product of Infinite Series
341
By L'Hospital's rule,
1
ta+P
lim
—.
;
TTGTZ = lini
7- r—
t-o+ 1 - (1 - *)"
x _> +00 x*+p (1 - (1 - 1) )
—
hm
t-o+
—7—.
— 7 -
a^l-*)""1
Hence
lim
n-00 n a + ^ (1 - (1 - ± ) a )
f 0
if a + /3> 1,
l
-
i f a + /J = l,
[ +00
if a + j9 < 1.
Now the desired result follows from the preceding problem
3.6.15. Assume that the series ^ anbn converges. By the result in
n=0
3.6.13, it is enough to show that
lim an(b0 + b\ 4- ... + 6n) = 0 and
n—>oo
lim bn(a0 4 ai 4 ... 4 a n ) = 0.
n—•oo
For an arbitrarily fixed e > 0 there is fco G N such that afc0+i&fc0+i +
afco+2&/co+2 + ••• < f • Thus for n > fc0,
M&i 4-... 4- 6n) < an(&i + .- + bko) 4- -•
On the other hand, since lim an = 0, there is ni > fco such that
n—• oo
On<
o 2(6
7 m0 —
Z T4-o
Tfco
T)
+ ...
if
n > n
i»
which in turn implies that an(bo 4 ... 4- bn) < e for n > n\. Hence
we have proved that lim an(bo 4-... 4 6n) = 0.
n—>-oo
Assume now that the Cauchy product is convergent. It then
follows from 3.6.13 that lim a n (6 0 4 ••• 4 bn) = 0. As a result, for
sufficiently large n,
n—+oo
(n 4- l ) a n 6 n < an{b0 4 ... 4 6n) < 1,
and consequently,
(anbn)1+a < ( - 4 - "
\n + l
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Solutions. 3: Series of Real Numbers
342
3.7. Rearrangement of Series. Double Series
oo
3.7.1. Let Sn = a\ + a,2 + ... -f an be the nth partial sum of ^
n=l
Then
&i -h 62 + ... + bn = Smn
for
a
n-
n > 1.
Since each subsequence of a convergent sequence converges to the
same limit, lim 5 m = lim Sn.
n—>oo
n—+oo
3.7.2. Denote by {T n } the sequence of partial sums of the rearranged series. Then
r
- - (x -1) - 1 + ( I -1) - 1 + +
\2n-l
4n-2J
An
1 1 1 1
1
1
= 2~ - T4 + T;
«
+
.
.
+
•
6 8
4n-2
An
_ 1 / 1
1_ 1
1
_ _1_\
+
+
+
~2V
2 3 4 '" 2n-l
2nJ "
Therefore, by 3.1.32 (a), we get lim Tsn = \ In 2. Of course, lim T^n
= lim T3 n+ i = lim T3n+2- It then follows that
n—•oo
n—•oo
,
1
1 1
- 2 - 4
+
1 1 1 1
3 - 6 - 8 + 5 - "
=
l, „
2In2-
3.7.3. Let {Tn} be the sequence of partial sums of the rearranged
series. Set f(n) = 1 + 1 + 1 + 1 + ... + ^
+ I . Then
Ta+0-l
1
1 1 1
1
+ - + ... + — - - - - - - . . . - —
= / ( 2 a - 1) - | / ( a - 1) - ±/(/?) = / ( 2 a ) - \f{a)
- ±f(J3).
Now we will prove by induction that
Tn{a+0) = f(2na) - ±f(na)
- |/(n/J).
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343
3.7. Rearrangement of Series. Double Series
As we have already shown, the equality holds for n = 1. If it holds
for an n € N, then
r ( n + i ) ( a + « = / ( 2 n a ) - \f{na)
v
2J
1
1
+ ...+ 2 ( n + l ) a - l 2n/? +
,
1
2
- \f{n(3)
+ 7-^—
+
2 v '
2na + l
2na + 3
1
2n/? + 4 '" 2(n + l)/?
= / ( 2 n a ) - ^ / ( n a ) - i/(n/3) 4- /(2(n + l ) a - 1)
- i / ( ( n + l ) a - 1) - / ( 2 n a ) + ± / ( n a ) - ± / ( ( n + 1)/?) +
\f(n(3)
= /(2(n + l ) a ) - i / ( ( n + l ) a ) - ± / ( n + 1)(3.
Hence, by 2.1.41,
lim T n(a+/ 3) = lim I f(2na) - ln(2na) n—>oo \
n—KX>
-f(na)
Z
+ l l n ( n a ) - i / ( n / 3 ) + iln(n/3)
+ lim [ ln(2na) - -(ln(na) + ln(n/?))
n—»oo y
2
2na
1 a
= lim In ,
= In 2 -h - In —.
n
->°°
y/n2a(3
2 /*
Obviously, for k = 1,2,3,..., ( a + /?) - 1, we have lim 7n(a4-/3)+ifc —
lim T n ( a + / 3). Consequently, t h e sum of the series is In 2 -f \ In %.
3.7.4. Note that this result is contained as a special case (a = 1 and
/? = 4) in the preceding problem.
3.7.5. It is enough to apply the result in 3.7.3 with a = 4 and
(3=1.
3.7.6. Consider t h e series
<
1J
1
1 1 1 1
- 2 + 3+ 5-4
+
1 1 1 1
7+ 9 + l l - 6
^
/
-
i\n-l
obtained by rearranging the terms of 2J -—n—
71=1
+
m su
ch a way that
n, n = 1,2,3,... , positive terms are followed by one negative term.
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Solutions. 3: Series of Real Numbers
344
Collecting the terms of series (1) in the following way:
,
l
1\
- 2
(I
1 1\
+
+ 3 5-4
+
(\
1
+
7 9+
1
1\
TI-61+-<
we get
^
2-j I n 2 _
n=l
N
n + 1
+
n
2_
n + 3
+
- + n2
+ n
_
2n)'
1
'
Let Sn and T n denote the nth partial sums of the series (1) and (2),
respectively. Then
in
= Q(n+l)n , ^ >
>
fc=l
—
;
— —-
N
>
'
-
>
-
>
+00.
fc=l
3.7.7. Grouping the terms of our series, we rewrite it in the form
y>/
n=1
1
N>/4n-3
1
>/4n- 1
1
\/2n/ '
Moreover,
1
\/4n-3
y^n- 1
\/2n
v/(4n - l)2n + >/(4n - 3)2n - y/{4n - 3)(4n - 1)
x / 4 n - 3 \ / 4 n - lV2n
/
2 v 2n - V4n - 1
2\/2n - \/4n
2 - >/2
\/4n - l>/2n
\/4n - l>/2n
\/4n - 1'
Thus lim Ssn = +00, where {5 n } denotes the sequence of partial
n—+00
sums of the rearranged series. Consequently, the series diverges.
00
3.7.8. Assume that the series Yl an converges absolutely, let Sn
71=1
denote its nth partial sum, and set S = lim Sn. Denote by
n—KX)
{Tn}
the sequence of partial sums of a rearranged series. It follows from
00
the absolute convergence of ^
such that
(1)
n=l
a
n that, given e > 0, there is n e N
K + i | + K + 2 | + -.. <e.
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345
3.7. Rearrangement of Series. Double Series
Let m be so large that all the terms ai,a2,...,a n appear in T m .
Then, by (1),
\S - Tm\ <\S-
Sn\ + \Sn - Tm\ < 2e.
3.7.9. [4] Assume first that / > 0 and let n = d 4- u, d > u\ then
oo
rearrange the series ^ (—l) n _ 1 /(n) so that the nth partial sum of
71=1
a new series is
Tn = Td+U = (/(l) - /(2) + /(3) - ... - /(2u))
+ (f(2u + 1) + /(2tx + 3) + ... + /(2d - 1)).
This sum contains w negative terms, and all remaining terms, d in
number, are positive. The sum in the second grouping contains d —
u terms, and consequently, this sum is between [d — u)f(2u) and
(d — u)f(2d). Since the sum in the first parenthesis converges to S
as u —• oo, the change in the sum is equal to the limit of the second
parenthesis. Set v(u) — d — u. Then
(1) i/(u)/(2rf)</(2u + l) + /(2u + 3) + ... + / ( 2 d - l ) < i / ( t i ) / ( 2 t i ) ,
and the monotonicity of the sequence {nf(n)}
u
<o\
1 ;
« + !/(«)
implies
<r /(2ti + 2f(tQ)
/(2u)
'
Choose u(u) such that
(3)
lim v(u)f(2u)
= I.
u—+oo
(One can take, e.g., v(u) = I j4^y .) Then lim ^ ^ = 0, because
I = lim --^-2uf(2u)
Thus (2) implies that
squeeze principle give
lim
and
f{2u
f+£)
U))
lim 2uf(2u) = +oo.
=
L A s a result
> (!) and the
lim (/(2u + 1) + / ( 2 u + 3) + ... + /(2d - 1)) = I.
u—>-oo
So, we have proved that lim T2u+v(u) — S + 1.
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346
Solutions. 3: Series of Real Numbers
Now note that if 2u + v(u) < k < 2{u + 1) + v{u + 1), then
0 < Tk - T2u+l/(u) + f(2u + 2) < T2u+2+v{u+i) ~ T2u+v{u) + f{2u + 2).
Since f(2u + 2) —> 0 as ^ —> oo, we see that lim Tk = S + 1.
/c—>oo
In the case where / < 0, we can interchange d and u and proceed
analogously.
3.7.10. Given e > 0, beginning with some value no of the index n,
we have
(i)
£z£</(n)<£±£
n
n
Consider the rearranged series whose nth partial sum is (see the solution of 3.7.9)
Tn = Td+U = (/(l) - /(2) + /(3) - ... -
f(2u))
+ (/(2u + 1) + f(2u + 3) + ... + /(2d - 1)).
Moreover, assume that the number d of positive terms is such that
lim f = k. Then, in the case where d > u,
1
1
1
+
^
—
«
+
.
.
.
+
•
2w 4- 1 2u -f 3
2d - 1
-|l
+
^2u
i + ... + ^ - l n ( 2 u - l )
2u-f2
2d-2
J
2d - 1
By 2.1.41 each of the first two parentheses tends to the Euler constant
7. As in 2.5.8 (a), we may show that the third parenthesis tends to
! Infc.Hence
Consequently, (1) implies that
lim (/(2« + 1) + f(2u + 3) + ... + f(2d - 1)) = \gIn fc.
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347
3.7. Rearrangement of Series. Double Series
Thus the change in the sum S of the series is ^g\nk.
reasoning can be applied to the case d < u.
Analogous
3.7.11. It is enough to apply the rearrangement described in the
solution of Problem 3.7.9 with v{u) = l[(2u)p}.
3.7.12. Take the rearrangement described in the solution of Problem
3.7.10 with lim 4 = a.
u—>oo
u
oo
3.7.13. No. Indeed, let J2 ank be a rearrangement of a divergent
oo
k=i
a
series Yl n- The monotonicity of the sequence {an} implies that
71=1
aWl -h an2 + ... + aUrn < a\ + a^ + ... H- CLm.
So, it is not possible to accelerate the divergence of this series.
3.7.14. [20] Choose a subsequence {arn} of {an} such that aTn <
m i n ( 2 - n , Q n - Qn-i)> n = 1,2,..., where Q0 = 0. Then
ari -h ar2 + ... + aTn < Qn
and
ari -f ar2 + ... + a rri < 1.
Thus since lim Q n — +°o, the sequence {Qn — (^n + a r +...-ha r n )}
n—^oo
oo
also diverges to infinity. Now, we add the terms of YJ an which do
n=l
not appear in the sequence {arri} to the sum ari -f a r2 + ... + a r n in
such a way that
o i + a 2 + ... + a r i _i-f-a r i 4-a r i + 1 + ...-f-aj + arfe +a r f c + 1 -h... + a r n < Q n ,
and a^ is the last term allowed. That is, if we add a term which does
not appear in the sequence {arn} and whose index is greater than i,
then the above inequality does not hold.
3.7.15. [W. Sierpinski, Bull. Intern. Acad. Sci. Cracovie, 1911,
oo
149-158] Let J2 Pn
n=l
an
d
oo
Y2 Qn be the complementary subseries of
n=l
oo
a conditionally convergent series Y, an consisting of all successive
71=1
nonnegative and negative terms, respectively. Let a be an arbitrarily
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Solutions. 3: Series of Real Numbers
348
oo
chosen real number. Since the series ^ pn diverges to +00, there
71=1
exists a least index k\ for which
Pi +P2 + - +Pkx > <?Next we choose the least index n\ for which
Pi +P2 + .»+Pfci +91 +02 + ». +9ru < CT>
Then we find the least index &2 for which
Pi + P 2 + -..+Pfci + 9 l + 9 2 + . . - + 9 m + Pfci+l + •••+Pfc 2 > cr
and the least 722 such that
P l + P 2 + - . . + P / e 1 + 9 l + 9 2 + ...-h9n 1 +Pfci + l + . - . + P / c 2 + 9 n 1 + l + -.. + 9n2 < <*.
Continuing this process, we define two sequences fci,&2,... and ni,
ri2,... and the corresponding rearrangement of our series. Let Sn be
the nth partial sum of this rearrangement. Then
Sn < a
for n < k\ but Sn > a
for k\ < n < k\ + n\.
Furthermore,
Sn < a
for
A:m -h n m < n < Arm+i -f n m ,
Sn> a
for
A:m+i + n m < n < fcm+i + n m + 1 ,
where m = 1,2,.... By the definition of the sequences {km}
{n m } we also get
and
\Skm+i-i+nm ~cr\< Pfcm+1,
ISkm+l+Tlm -<*\ <Pfem+ i»
l5'fcm+1+nm+/ - cr| < Pfcm+1
\Skm+i+i+nm+i -v\<
for
/ = 1,2,.., n m + i - n m - 1,
\Qnm+11 for
/ = 1,2,..., fcm+2 - fcm+i - 1.
Since lim pn = lim qn = 0, we conclude that
n—>oo
n—•GO
lim Sn = a.
n—>oc
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3.7. Rearrangement of Series. Double Series
3.7.16. Denote by {Sm}
oo
a
of Yl n
n=l
an
d
349
and {Tm} the sequences of partial sums
oo
S a>nk, respectively. Since {n^ — k} is a bounded
k=l
sequence, there is / € N such that k — I < n^ < k + / for all k G N.
If m > I and n^ < m — /, then k — I < rik < m — l. Hence A; < m,
and consequently,
(1)
{1,2, . . . , r a - / } C { n i , n 2 , . . . , n m } .
Indeed, if s is a positive integer not greater than m — /, then there
exists a unique A; G N such that s = rik- It then follows from the
above that k < m, or in other words, s € {™i,n2, . . . , n m } . By (1),
we see that each a n , with n = 1,2,..., m — /, appears in T m . On the
other hand, if k < m, then n/- < A; + / < m -f J, and consequently, all
the terms a n i , a n 2 , ...,a n m appear in 5 m + / . Hence
| 5 m - Tm\ < |a m -M-i| + ... + \am+i\ for
Therefore
m>L
lim 5 m = lim T m .
m—>-oo
m—>oo
If the sequence {rik — A:} is unbounded, then the examples given
in Problems 3.7.2 - 3.7.6 show that the rearranged series may diverge
or may change the sum of the series. Now we give an example of a
rearrangement that does not change the sum of the series. To this
end, we take a sequence {rik} obtained by the permutation of positive
integers that interchanges n\n+l) with n ^ ' and leaves the other
integers unchanged. Since n(n2+3) _n{n+i) — n^ the sequence {rik-k}
is unbounded. Moreover,
0,
G n (n+3)/2
_
a
n(n+l)/2,
if
m=^±3),
•r
"
n(n+l) ^
2
-
m
<
. n(n+3)
2 "
3.7.17. [R. P. Agnew, Proc. Amer. Math. Soc. 6(1955), 563-564]
m
We will apply the Toeplitz theorem (see 3.4.37). Set Sm = ^2 ak
fc=i
m
an
d
Tm — JZ ank. Assume that m is so large that 1 G {ni,ri2, . . . , n m } ,
and arrange all the members of the set {ni,n 2 , ...,n m } to form an
increasing sequence
1 , 2 , 3 , ..., A ) , m > a i , m + l ? ^ l , m + 2, ...,/?l, m >
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Solutions. 3: Series of Real Numbers
350
where
0 < /?0,m < a i , m < /?l,m < « 2 , m < ••• < /?j m ,m-
Hence the partial sum T m of the rearranged series can be written in
the following way:
Tm = S(3Qrn + (SpltTn - Sairn)
+ ... + (S(3jrn,m -
Sajrnrn).
oo
Consequently, Tm = £ Crn,kSk, where
1,
if fc = /?/, m , J = 0, l , . . . , j m ,
0
otherwise.
Since lim /?o m = +00,
00
m—>-oo
'
lim c m & = 0 for every k G N. Moreover,
m—+00
'
00
J2 cm,fc = 1 for ra = 1,2,..., and £ |c m , fc | = 2 S m - 1, where B m
fc=l fc=l
denotes the number of disjoint blocks of successive integers in the
set {ni,ri2, . . . , n m } . Finally, by the Toeplitz theorem, lim Tm —
m—>oo
lim Sm if and only if there is N such that i ? m < AT for all ra G N.
3.7.18. Assume that the series ^ c n is absolutely convergent and
71=1
its sum is S. Then for any e > 0 there is &o such that
00
|ci + c2 + ... + cfco - S| < -
and
]P
|cj| < - .
Let ra, n be so large that for each / G {1,2,..., &o} there exist i and
fc, i G {1,2, ...,ra}, A: G {1,2, ...,n}, such that Q = a ^ . Then
| 5 m > n - 5 | < | c i + c 2 + ... + cfco - 5 | + ^
|cz|<£.
Z=fc0+1
Hence the convergence of the double series to S is proved. Likewise,
the absolute convergence of this double series can be established.
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3.7. Rearrangement of Series. Double Series
351
3.7.19. Set
CO
OO
i,fc=l
ran
n=l
n
fc=l
i=l
k=l
Arbitrarily fix e > 0 and / G N. Take ra, n so large that all the terms
of 77 are in S^n and \S* - S*mJ < e. Then T,* < S ^ „ < S* + e,
OO
which means that the series E cn is absolutely convergent. Denote
71=1
OO
by T n and T the nth sum and the sum of E c n , respectively. To
71=1
prove the identity
OO
/
OO
==
a
^
i,k
/
c
j
m
n=l
i,k=l
fix e > 0 and take Z so large that
|77-T*|<!
Tn
n
If SWi,?! = E E a*,fc
an
i=lfc=l
d if
m
and
>n
|7}-T|<|.
are so
l a r g e that all terms of T\
are in 5 m ? n , then
|5 ro ,n-r|<|r-r«| + |r*-7ri<e.
3.7.20. This is a corollary of the two preceding problems.
OO
/
OO
\
3.7.21. Assume that, e.g., the iterated series E I E \a>%,k\ )
OO
i = i \fc=i
OO
'
/
c
°n-
verges, and set E \ai,k\ = 0"i and ^ j j = a. Therefore, each of
fc=i
'
i=i
OO
the series E a ^ , « = 1,2,..., converges, and
OO
E a*,fc
fc=l
| 5 * | < (Ji.
This and the convergence of E U{ imply the absolute convergence of
OO
OO
2=1
OO
E Si. Consequently, E ^i = E
i=l
i=l
/
OO
E
i = l \fc=l
a
i,k
\
/
.
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Solutions. 3: Series of Real Numbers
352
3.7.22. Let £
and S^n
aijk = S, ]T) \ai}k\ = S*, and set S m , n = £
i,fc=l
m
n
2,/c=l
= 5] ^
£ «;,*;
2 = 1 /C = l
|fli,fc|- We first show that the iterated series
i=ifc=i
oo
/
oo
2=1 \fc=l
/
converges to 5*. Indeed, given £ > 0, there is n 0 such that S* —e <
S^n < S* for m,n> UQ. Let m be fixed for the moment. Then the
sequence { S ^ n } is monotonically increasing and bounded. Thus it
is convergent, lim S ^ = S^, and consequently, 5* —e < S ^ < S*
n—•oo
'
for m > no- This means that
lim ( lim 5* n ) = S*. We know from
771—KX> 71—»00
'
the preceding problem that absolute convergence of the iterated series
oo
implies its convergence. Thus £ a ^
fe=i
converges for each z, say, to
Si. We will now show that for every e > 0 there is mi such that
|(Si + S 2 + ... + S m ) - S | <e
for
m > mi.
By the absolute convergence of the double series,
|Sm,n-S|<-
and
|S^>n-S*|<-
for
m,n>mi.
Therefore, for m > mi,
|(Si + S 2 + ... + S m ) - S | =
i=l
771
fe=l
(X)
£ £ °M
S \hm,n ~~ &\ H~
<\Sm<n-S\ +
\S*-S^n\<e.
= 1 fc=n+l
The proof of the convergence of the other iterated series is analogous.
3.7.23. Note that the series ]H (a n> i + a n _i, 2 + «n-2,3 + • •• + ai,n)
n=l
is an ordering of the double series. If one of the series
X) ^'^
i,fc=l
X ^
n=l
1
' + K - l ^ l + |On-2,3| + -. + |fli,n|)
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353
3.7. Rearrangement of Series. Double Series
converges, then our claim follows directly from 3.7.18, 3.7.19 and
3.7.22. Hence it is enough to show that the absolute convergence of
one of the iterated series implies the absolute convergence of any oroo
/ oo
dering of the double series. To this end, assume that J ] I Y2 \ai,k\
i=i
\k=i
converges, say, to 5*. Let {cn} be a sequence obtained by an enumeration of the infinite matrix (ai,/c)i,fc=i,2,...« Then for / G N there
exist m, n so large that
771
n
\ci\ + \c2\ + ... + \Cl\<^2Yl
\o>i,k\ < S*.
i=l
oo
Thus the series ^
c
k=l
n converges absolutely, which in turn implies
n=l
the absolute convergence of the double series (see 3.7.18).
m
m
3.7.24. Since £ (£) = 2 " \ we get
k=0
E OT = W-
Hence
n,k=0
k-\-n=m
£
(m + 1)!'
n\k\(n + k + l)
nkz=Q
fc-f 71=77 1
Consequently, by 3.7.23,
ex?
.,
y^
^
oo
ra=0
k+n—m
_ 1 v^ 2m+1 _ 1
m
v
t
*-** 4 ^ n!fc!(n +fc + 1)
n\k\(n-\- k + 1)
2
1
=v v
-
_ y ,
m
'
m=0
v
2
'
where the last equality follows from 2.5.7.
3.7.25. We have (see 3.7.23)
nfc(n +fc-f 2)
^
^ n ^ n +2 U
n,fc=l
7i=l
i
^
^
71=1
/
n(n + 2) v
V
'
N
N
/e=l
i
i
2
3
n +fc+ 2y
'
i
"'
n+ 2
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Solutions. 3: Series of Real Numbers
354
1
--t( -2^ \ n
-\{
n,
1 +
i + 2i + ...+ n +' 2
+ 2/ V
n+
i
i /,
i
2+ 3+ 2 l
1 +
i
i
2
+
i\
i
3+ 4
+
i /i
4+ 5
3
i
3.7.26. It follows from 3.7.23 that
y,
_ ~ _fc!_ ~ /
n\k\
„ ^ 0 ( " + fc + 2 ) ! " ^ 0 f c + l ^ 0 l ( n +
=
^
(n + 1)! \
n!
_fc!_ _ 0 ! _
=
fc+l)!
^
(n + A: + 2 ) ! j
_1__
Hence, by 3.1.28, the desired equality is proved.
3.7.27. Observe that the sum of each row series of the matrix is
finite. Indeed, the sum of the first row is x, of the second is x(l — x),
of the third is x(l — x) 2 , etc. Moreover,
x + x{\ - x) + x(l - x)2 + .. = 1.
On the other hand, the sums of the column series are alternately equal
to 1 and — 1. Therefore the other iterated series diverges. By 3.7.23,
we conclude that the iterated series cannot converge absolutely.
3.7.28.
(a) The absolute convergence of J2 x%
i=0
an
oo
d J2 V i m ply the absofc=0
oo / oo x% k \
lute convergence of the iterated series Yl I Yl
i=0 \k=0
OO
/
OO
\
OO
•
E (^E l * V l J = E M (T^M)
V ) •> because
)
\
= (i-NKi-M)" Consequently,
the given double series is absolutely convergent.
(b) Considering the iterated series, we see that our series converges
if and only a > 1 and 0 > 1.
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3.7. Rearrangement of Series. Double Series
355
(c) Collecting the terms for which i 4- k = n, we get
oc
oo
1
i
(i + k)P ~~ ^}n
^
y
i,k=l
}
~ nP'
n=2
Thus the double series converges if p > 2 and diverges if p < 2.
3.7.29.
(a) It is enough to calculate the sum of the iterated series. We have
EE
OO
/
OO
^
\
O
O
/
.
.,
,
\
-j
(b) As in (a), we compute the sum of the iterated series:
OO
.j
The last equality follows from 3.1.32(a).
(c) As in (b), we have
i
~ / ~
m
\
=
~
i
=
i
n
3.7.30. Since S m , n = J2 ^2 ai,k — bm.n, w e s e e that
i=lk=l
^1,1 = 51,1 = frl,l>
&l,n = Si,n — 5 i 5 n _ i = 6i, n — &l, n -l>
«m,l = £m,l
—
*SVn-l,l = ^ m , l
_
&m-l,l>
n > 1,
m
> 1-
Similarly, for n , m > 1, we get
Q>m,n
=
^m,n
*^m— l,n
v*^m,n—1
^m — l,n— 1/
= ^m,n — ^m-l,n ~ (&m,n-l ~ &m-l,n-l)>
n, 772 > 1.
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Solutions. 3: Series of Real Numbers
356
3.7.31. We have 5 m , n = ( - l ) m + n ( ^ + i ) . So, for e > 0, there
is no such that if m,n> n 0 , then
| ^ r n , n | < £•
Therefore the double series converges to zero. However, both the
iterated series diverge. Indeed,
X>,fc = sitn - s^n = (-i) i + "| + ( - D i + " ~ ,
oo
which implies that every series Yl ai,k, i £ N, diverges.
k=l
3.7.32. We have
oo
/ oo
\
oo
1=1 \ f c = l
/
2=1
_\x\
E EM" -Er^tpBy the ratio test, the series on the right-hand side of this equality
converges. This means that the iterated series converges absolutely.
Thus by 3.7.23,
OO
OO
i,fc=l
k=l
fc
Collecting the pairs (z, k) with the same value of the product zfc, we
get
oo
oo
i,fc=l
n—1
because the number of divisors of n is equal to the number of the
pairs (z, k) for which ik = n. Moreover, for n = 2,3,...,
^n—l,7i—1
^n,n
= x + x2n + ... + x ( n " 1 ) n + /
n
1
+ / ( n ~ 1 ) + ... + x n ' 2 + xn
-xn
Obviously, S\^ = x = 2y5^ + x. Hence, on account of
Sn,n
=
{Sn,n
~ S V i - l . n - l ) "+" ( S n - l , n - l ~ S n - 2 , 7 1 - 2 ) + ••• + S l , l >
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3.7. Rearrangement of Series. Double Series
357
we see that
OO
OO
L.
E 1 _ ajfc
r,2
n
OO
v ^ n2
^ xr^ * ~ x
ar
fc=l
Z-j i _ xn
z-j
n=l
n=l
3.7.33. As in the solution of the foregoing problem, we show that the
iterated series converges absolutely. Thus the first equality follows
directly from 3.7.23. To prove the other equality we consider the
ordering of the double series described in the solution of 3.7.32.
3.7.34.
(a) By 3.7.23,
oo
oo
S
Z^ p
=
oo
1
Z ^
+
n=2
p=z2
oo
1
Z _ ^ + -"
=
n=2
1
2 ^ fc(jk_i)
v
k-2
= L
'
(b) As in (a),
oo
oo
p=2
k=2
1
1
v
J
3.7.35. Let B denote the set of all integers which are not powers.
Then
A = {kn : n <E N, n > 2, k e B }.
Since ^
oo
= £ i , n > 2, applying 3.7.23 and 3.7.34, we get
3=1
oo
1
^
n- 1
nEA
oo
1
z-f z-* n j
oo
1
Z_^ Z ^ Z ^ fcn?
nGAj =l
oo oo
fe€Bn=2j=l
oo oo
^
Z_^ Z—• Z—• ^n>?
feGB j = l n = 2
^
Z—/ Z—• £ n
k=2n=2
3.7.36. [G. T. Williams, Amer. Math. Monthly, 60 (1953), 19-25]
The left-hand side of the equality is equal to
v^Y> ( 1
W
1
j V ^ o J L 2 - ^ lv^2 j 2 n - 2 +
1
fc4
1
j2n-4 + "' +
1
fc2n-2
1\
J2 j "
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Solutions. 3: Series of Real Numbers
358
Summing the expression in the parentheses, we get
'
N
(**)
lim y ^
N
j2-2n
^2-2n
1
-!)^r
/
Note t h a t
N
N
,-2-271
Z-/ Z-/
j=lfc=l
k^j
N N
h2 _
K
1
X
l
-J A2TL—2 JU2
* -J '
N
A2-2U
N
2L^ 2-J h2 _ A2
J
j=lk=lK
k^j
N N
1
A2
J
1
_\^V"
N
jL2-2n
/i2
. V^V"
s J /
k^j
N
k2-2n
u2 _ AI
' 2-*I2^I
fc=iK
k&
j=i
x
3
1
J L>2n—2 A2
U2
k^j
N
N
Z - / ^ ' 2 n - 2 Z - / L2 _ A2
J
j=l J
k=l
Hence
iim
N-^oo
(* * *)
N
AT
,;2-2n _ £ 2 - 2 n
y
^
+(n-l)^
fc2_j2
\fc= l
j=l
AT
lim
j
JV
j
N
n—
2—1 -i2n-2 2—1 J.2 _ A2 + V
N—>oc
,•=1
J
3
k=l
x
L
• J 2 - f ~^
J=l
Now, observe that
N
__^
N
1
l
^
~~' — k - j
fc=i
j-l
=
J
-
~f-rfc
+
,^
k=j+i
iV-j
1
_i
k- j
J
W
1
^
l
\_
^ fc + j
fc=i
J
iV+j
f-1 k ~ ,^
2j
J
N+j
k + 2j=~
2-<k
fc=l
i
+
l
"
+
2s
^
- +—
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3.7. Rearrangement of Series. Double Series
1
2j
+ l + N-j
\N-j
1
359
+ 2 + ...+ N + j
Thus by (***),
N f N
.7 = 1 \
„-2-2ra
i,2-2n
fc2 - j 2
k=l
2
0
+ (n-l)7
^[fn~l \N-j
hi?"
Moreover, since 0 <
\
jV _
1
. +1 + ... + j ^ .
<Y,p^T
[N_j
*
i
+1
+ l+'"+
< N^Jj+1,
N+j)'
we see that
+ ••• + JTT])
i
" i
i
j=l'
<
Z> j2n-2
JV + :
T
N_j
+ 1^
+
H^
* + i£P
N-j
2^jN_j
+
l
+ l)
where 7 is the Euler constant (see 2.1.41). Finally, by (*),
N
jJ^L
AT->oo
N
V - V2 - / J _ _2 Jn _2
l^l^\k
j=l k=l
x
J
j ~
1
1
+ k4 J j 2 n - 4
+
1 1
'" + k2n~2 Jj2
1 N 1
1
= lim ( n + -2 ) <—<
V - ^j - = ( n + - ) C ( 2 n ) .
j = l
J
3.7.37. Substituting n = 2 in the identity given in the foregoing
problem, we get
C(2)C(2 ) = (2 + i)C(4) .
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Solutions. 3: Series of Real Numbers
360
Since C(2) = 4- (see 3.1.28 (a)), we obtain (see also 3.1.28 (b))
n=l
Likewise, taking n = 3, we find that
°° 1
7T6
Similarly,
oo
cm-Ei'
n
9450
8
n=l
3.8. Infinite Products
3.8.1.
(a) We have
± x
V
kz J
x±
k2
2n
n-+oo 2
(b)
A(fc-l)(fc 2 +fc+ l)
1=1 (fc + l)(fc2-fc + l)
n
(fc - l)((fc + l ) 2 - (fc + 1) + 1) = 2(n 2 + n + 1) _ ^ 2
k=2
(k + l)(k2-k
+ l)
"
3n(n + l)
n ^ 3 "
(c) For x = 0 the value of the product is 1. If x ^ 2 m (f + /CTT) ,
then cos ^ ^ 0 and sin ~ ^ 0. Hence
c s x =_
n
sm x
1 sin 2 ^ r _ s m x
n
~ 2 sin ^ n~Z£> ~ 7 ~ '
:
n ° 2*^ ~ ri9
J-A 2 sin^-
fe=l fc:
TT
(d) On account of the formulas
sinh(2x) = 2sinhxcoshx
and
lim
x-+0
sinhx
X
= 1,
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3.8. Infinite Products
361
as in (c), we get
n
oo
„=1
( sinh x
I
x
cosh—- = <
2
"
if x ^ 0,
x
\ l
ifx = 0.
(e) We have
.,
n
2
i + x ) = [[
i_x2k
2fe+1
=
^^
^
Y^-
fe=0 fc=0
(f)
nfi+ir^r )-n
, J- V
x
P r (fc + l ) 2 _ 2 ( n + l)
_ 2_
11
fc(fc + 2) ra + 2 n-oo
fc(fc+2W
«=i
''
fc=i
(g) Since
n
a
= afc=1
fc
the continuity of the exponential function and 3.1.32 (a) imply
that J ]
a
=a"ln2-
"
(h)
n
n
fc=1l
I
E
e fc=1
efc
+ i
=
A
k
e
=
™+!
Z,
i
k=1
,
fc~lnn
•
n
™ + l'
T h u s by 2.1.41,
oo
1
n i ^
e
n=l ~
n
where 7 is the Euler constant,
(i) We have
P
n
_fr
(3A:)2
y
(3fc - 1)(3* + 1)
-rr
_33n(n!)3
£ J (3* - l)3fc(3ifc + 1)
Using the Stirling formula
n\ = a n v27rn ( — ) ,
\ 6 /
(3A:)3
where
(3n + 1)!'
lim an = 1,
7i—>oo
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362
Solutions. 3: Series of Real Numbers
we get
33n(27r)3/2n3n+3/2e-3n
J^o
}^L (27T) 1 /2(3 n + !)3n+l + l / 2 e - 3 n - l
Pn
\ 3n /
0
= 2ire lim
n-^^\Sn + lJ
N
3/2
\3n+lJ
0
3\/3'
3.8.2.
(a)
In
(-l)k\
Zn
3 2 5 4
/
1
n—»oo
_ 3 2 5 4
2 n - l 2n - 2 _
~ 2 ' 3 ' 4 ' 5 ' •" ' 2n - 2 ' 2n - 1 ~ '
P
2n_1
(b) We have
1\
T W ,
fc=l
^ 3 4
n+1
'
so that f| (l + ^) diverges.
n=l
oo
(c) The product fj (l — ^) diverges, because
n=l
fc=2
A; /
1 2 3
2 3 4
n - 1 _ 1_
n
n n->oo
3.8.3. Note first that for nonnegative a n ,
(1)
ai + a 2 + ... + a n < (1 + a i ) ( l + a 2 )...(l + a n ) .
Moreover, the inequality e x > 1 -f x, x > 0 , gives
(2)
(1 + ai)(l + a 2 )...(l + an) < e -i+-2+...+a n>
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363
3.8. Infinite Products
The inequalities (1) and (2) together with the continuity of the expooo
nential function show that the convergence of the product Yl ( l + a n )
71=1
OC
a
is equivalent to the convergence of the series J2 n •
n=l
oo
3.8.4. Assume that the series ^ an converges. Then for sufficiently
71=1
OO
a
large TV, ]T) n < \- It follows from 1.2.1 that
n=N
n
n
k=N
<
Pn
k=N
n
n
k=l
k=N
Since Pn = Yl (1 — a^) = PN-I
^
J2ak> o'
J\(l-ak)>l-
Yl (1 ~afc)> w e
see
that the sequence
> is monotonically decreasing and bounded below.
p
quently, it converges, say, to P. Then P e [ i , l ] . Thus
Conse-
lim Pn =
n—>oo
PN-IP
^
0.
To prove the other implication, assume that
oo
^ an diverges.
n=l
If the sequence {an} does not converge to zero, then the sequence
{1 — an} does not converge to 1 and the necessary condition for
oo
convergence of Yl (1 ~~ an) is not satisfied. So we may assume that
n=l
lim an = 0, and consequently, 0 < an < 1 beginning with some
n—>oo
value iV of the index n. In view of the formula (see 2.5.7)
X2
X3 \
(X4
X5
we get 1—x < e~x for 0 < x < 1, because all the terms in parentheses
are nonnegative. Hence
71
™
(1 - afc) < e *="
?
n
> AT,
fc=iV
and consequently,
diverges.
n
lim
n
Yl (^ ~ ak) = 0. Therefore,
->°°k=N
oo
f] (1 — an)
n=l
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Solutions. 3: Series of Real Numbers
364
3.8.5. Note that
2n
J\(l + ak) = JJ(1 + a2fc-i)(l + a2k)
Thus, by 3.8.4, the product converges.
3.8.6.
(a) Since cos £ = 1 - (1 - cos £) and 1 ^ 1 - cos £ > 0, n € N,
we can apply the result in 3.8.4. Thus the convergence of the
oo
product follows from the convergence of the series Yl (1 ~~ cos n)
n=l
(see 3.2.1 (e)).
(b) As in (a), the convergence of the product follows from the convergence of XI (* ~
n s m
n=l
n) ( s e e 3-2-5 (°0)-
(c) We have
tan
/TT
1\
1+ tani
^
S
' 7r 1 \
1 + tan 2 t a nn
- -f - =
?• = 1 + — t a n: r
\4
ny
l-tau£
1
Since ^ - ^ V > 0 for n > 2 and
1—tan £
—
2 tan i
n.
1 - t a n 2^
..
hm
j — - = 2,
n—+oo
by 3.8.3 the product diverges.
(d) In view
0f
lim
n—+ 00
l-m° i (l + ^)
=
l
the
convergence of the product
n
follows from 3.8.4.
(e) The divergence of the product follows from the divergence of the
oo
series £ ( $ n - 1) (see 3.2.5(a)).
1=1
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365
3.8. Infinite Products
(f) Since lim
n-+oc
n
lfn = 1, it follows from 2.5.5 that limn^oo -*f
T
=
Vn-l
1. Thus the convergence of the product follows from the converoo
gence of the series Yl ^ •
n=2
oo
3.8.7. By assumption, the series J2 an converges and, without loss
n=l
of generality, we can assume that \an\ < 1. Since
,.
a - l n ( l + an)
1
hm n
'- = -
m
(1)
n—>oo
alL
1
and the series ^2 a n converges, the convergence of Y2 an
n=l
1S
equiv-
n=l
oo
alent to the convergence of ]T) ln(l + a n ), which in turn is equivalent
oo
n=l
to the convergence of JJ (1 + an).
Note that if Y2 an diverges, then by (1),
71=1
1
an — ln(l + an) > -an
for sufficiently large n.
Thus the series ^ ln(l + a n ) diverges to — oo, which means that
oo
n=l
Y[ (1 + CLn) diverges to zero.
n=l
3.8.8. The result follows immediately from 3.8.7.
3.8.9. Apply 3.8.7 or 3.8.8.
3.8.10. We use the equality
|ln(lH-an)-an + |a£|
n m
n-^oo
i
^
\an\6
1
=
77
3
and proceed as in the solution of 3.8.7.
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Solutions. 3: Series of Real Numbers
366
3.8.11. No. By the test in the foregoing problem, we see that the
product given in the hint converges if ^ < a. On the other hand, the
series
_1_
2"
J_\_l_
2?<*) 3"
(\_
\2a
1 \
3**/
(}_
\3<*
1
4^
and
\ 2°/ + ( ^
+
2^J +\ ¥J
+
3^J + \ T j +"*
+
(^
both diverge if a < \.
3.8.12. Observe that if lim an = 0, then
n—»oo
lim
1 ln(l + On) - an + \a\ - \a\ + ... + ^ a * I =
1
3.8.13. By the Taylor formula,
ln(l + an) = an-
a% = a n - 6 n a £ ,
+ g
where | < O n < 2, if |a n | < ^. Thus if ni,ri2 are sufficiently large
and ri\ < ri2, then
a
5^ ln(l + an) = J2an-&J2
n=ni
n-
where
n=ni
n—n\
G €
( 9' 2 ) '
^
'
CO
Hence the convergence of ]T a n follows from the Cauchy criterion.
n=l
00
3.8.14. If the products r i ( l + a n)
00
then Yl ( I
n=l
n—l
-a
an
00
d n (^ ~~a^) both converge,
n=l
00
n ) also converges. Consequently, the series ^2 an
con_
n=l
verges (see 3.8.4). Now the desired result follows from the preceding
problem.
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367
3.8. Infinite Products
3.8.15. Yes. Indeed, since {an} decreases monotonically to 1, we
can write an = 1 -f-a n , where {an} decreases monotonically to zero.
The convergence of the our product is equivalent to the convergence
of
oo
^-l)"-
1
k i ( l + a„).
71=1
Clearly, by the Leibniz test this series converges.
3.8.16.
(a) Since
lim (an -f bn) = 1-1-1 = 2, the necessary condition for
n—>-oo
convergence is not satisfied.
oo
(b) The convergence of \\ c?n follows from the convergence of the
oo
series ^
n=l
(c),
n=l
lna^.
(d) The convergence of the products follows from the convergence
of the series
oo
^
oo
oo
ln(a n 6 n ) = Y] In an + Y] In bn
n=l
and
oo
n—1
m
=
oo
X^ 7T Yl
n
n=l
n=l
oo
lnCLn
~ Yllnbn'
n=l
n=l
oo
3.8.17. Suppose that V} x2n converges. Then lim xn = 0, and the
n
n=i
->°°
convergence of both the products follows from 3.8.4 and from the
equalities
lim
n-+oo
l-COS£n
-
1
= -
2
x^
.
and
,.
lim
n-^oo
1
J-1
TJL~~
£^
1
~ ~6
Assume now that one of the products converges. Then lim xn = 0,
OO
71—>0O
x
and the convergence of Yl r\ follows also from the above equalities.
71=1
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Solutions. 3: Series of Real Numbers
368
3.8.18. Observe that
s
*n»r:-nr:-i
S
sk
k
fc=2
k
fc=2
3.8.19. See Problem 3.1.9.
3.8.20. See Problem 3.1.9.
3.8.21. Apply the foregoing problem with
oo
Yl an converges, that is,
3.8.22. Assume first that the product
n=l
n
lim Pn — P 7^ 0, where Pn = Y\ «fc- This implies that there is
a > 0 such that \Pn\ > a for n e N. The convergent sequence {Pn}
is a Cauchy sequence. Thus for every e > 0 there is an integer no
such that |Pn+/c - Pn-i\ < ea if n > n 0 and k e N. Therefore
Pn +fc
Pn-l
-1 <
£a
Pn-l\
for n > no.
<e
Assume now that for every e > 0 there is an integer no such that
(*)
\anan+i • ... 'Q>n+k - 1| < e
for n > no and k G N. Taking e = ^, we get
(**)
1
Pn_l 3 f
—^
< 2
- tor n > no2 <~ —
Pno
Next using (*), with e replaced by 3[ p •, we find an integer n\ such
that
2e
Pn+k
for n > n i , k e N.
1 <
3|Pn
Pn-l
Hence if n > max{no, n i } , then
\Pn+k — P n - l | <
2E
3
Pn-l
P
1
< e.
n0
This means that {P n } is a Cauchy sequence. Moreover, it follows
from (**) that its limit is different from zero.
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369
3.8. Infinite Products
3.8.23. We have
na-* 2 *)
2n
2 n
fc=i
2
™
fc=l
i
-22*
l-x-
fc
fc=1
2n
n(i-^)
na-* 2 *)
n a-*2fc)
2n
fc=i
n(i-^2/e) fc=i
iic 1 -^' 1 )
A;=i
k=n+l
n(i-x 2fc - ] )
Now, the desired result follows from the Cauchy criterion (3.8.22).
3.8.24. This is a consequence of 3.8.3.
3.8.25. Note that for a i , a 2 , ...,a n € R,
|(1 + ai)(l + a 2 )...(l + an) - 1| < (1 + h | ) ( l + |a 2 |)...(l + |a n |) - 1
and apply the Cauchy criterion (3.8.22).
3.8.26. Set P n = ( l + a i ) ( l + a 2 ) . . . ( l + a n ) , n € N. Then P n - P n - i =
P n _ i a n and
P n = Pi + (P 2 - Pi) + ... + (Pn - Pn-!)
= Pi + Pia 2 + P2a3 + ... + P n _ i a n .
Thus
P n = (1 + a x ) -f a 2 (l -f ai) 4- a 3 (l 4- ai)(l 4 a 2 )
+ ... + a n ( l + ai)(l 4- a 2 )...(l -f a n _i),
or equivalently,
P n = (1 4- ai) -f (a 2 4- a x a 2 ) + (a 3 + a x a3 4- a2a3 4- aia 2 a 3 )
4- ... + (an 4- a i a n + ... 4- an-\an
4- aia 2 a n
H-... H- a n _ 2 a n _ i a n + ... 4- aia 2 ...a n -.ia n ).
oo
Note that absolute convergence of \\ (1 4- an) implies the absolute
n=l
oo
convergence of the series 1 + ai + V] a n ( l + ai)(l 4- a 2 )...(l 4- a n _ i ) .
n=2
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Solutions. 3: Series of Real Numbers
370
This series is an ordering of a double series whose terms form the
infinite matrix
/
ai
a\a2
a3
&2&3
a2
€L\o>s
^1^2a3
a4
a\a±
<X\Q.2&\ 0,1(13(14
... \
02(2304
\
/
By 3.7.18 the double series converges absolutely, and by 3.7.22 the
iterated series given in the problem converges. Consequently, the
desired equality holds.
00
00
n=l
n=l
3.8.27. By the absolute convergence of ]T) an, the series ]P anx
converges absolutely for every x € R. Now it is enough to apply the
result in the preceding problem.
00
3.8.28. Obviously, for \q\ < 1 and x G R, the product J] (l + 0naO
n=l
converges absolutely. Taking an = qn in the foregoing problem, we
00
get f(x) = [7 (1 + QUx) = ! + ^x
+ A2x2 + .... Now observe that
71=1
(l+qx)f(qx)
= f(x). So equating coefficients of like powers, we obtain
Qn
q
A1 = and An = An„l1- q
1 - qn
Finally, by induction, we can show that
for
n = 2,3,....
n(n+l)
n
Q
A - (1 - q)(l - qt)2 . ... . (1 -
py
S±n. —
OO
[ j {l + q2n~lx)
3.8.29. Set f(x)=
n=l
and note that {l + qx)f(q2x)
=
/(#), and apply reasoning similar to that in 3.8.28.
3.8.30. W e have
n=l
\
00
fc=i
/
.
x
7
0
fc=l
0
/
00
fc=i
\
fc=l
A
fe=i
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3.8. Infinite Products
371
The absolute convergence of J2 AkXk and Yl ~$ i m P ^ e s the absolute convergence of their Cauchy product (see the solution of 3.6.1).
Observe that this Cauchy product is an ordering of the double series
corresponding to the matrix
/
A1A1
A2AX (x + I )
h43^i(x2 + ^ )
A2A2
A3A2 (x + i )
A 4 A 2 (x 2 + ^ )
V
A3A3
A4A3 (x + ±)
V 3 ( x 2 + ^)
...\
...
...
/
Therefore, by 3.7.18 and 3.7.22, we get
OO
OO
A
E ^ ^ E ^F = ( ^ 1 + ^ 2 + A3A3 + •••)
fc=l fc=l
+ ...) (x2 + -5 J +....
+(A2A1 + A3A2 + ...) (x+-\+{A3A1+A4A2
3.8.31. [4] By 3.8.30,
n=l
^
Setting
'
00
^
n—\
•
'
2n—1 \
F(X)=n(i+q^x)(i+q—)
and using the equality qxF(q2x) = F(x), we get
P i = B0q,
Bn = Bn-iq
n
~ ,
and inductively,
Bn = B0qn\
n=l,2,....
Thus
To determine P 0 we may use the results in 3.8.29 and 3.8.30. Put
oo
n
Pn = I K 1 " ?2*) and P = ft (! ~ 4 2n )- Then
n=l
k=l
2
n
an2
a(^+i)
2
+i
J30</ - Sn = ^n + >Mn+l + »• = V
P n + ^Pi P n + 1
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Solutions. 3: Series of Real Numbers
372
Hence
q2n
q4n
PnB0 - 1 < - ^ + ~p2 + ••• •
Now, letting n —• oo, we get #o = j>.
3.8.32. Apply 3.8.31 with
(a) x = - 1 .
(b) x = l.
(c) x = q.
3.8.33. Observe that for n > 1,
n
x —k
_ ; Z+ k
a
" = 2o \ 1, 1J1; X + k
1J
Hence
n
^
..
n
^
n
+ fc"
fc=l fc=2 fc=l
If x is a positive integer, then for sufficiently large n, Sn = | . We
now show that for x ^ 1,2,..., lim 5 n = \. Note that for k large
enough,
x—k
x+k
= 1 - ^ £ . Hence, by the result in 3.8.4,
lim [J
n—• o o
•*"*k=l
x—k|
x+k
= 0,
which in turn gives lim Sn = A, as we have claimed.
n—>oo
3.8.34. Assume that the product \\ (14- can) converges for c = Co
n=l
and c = c\, where Co ^ c\. Then the products
(l + cia n ) c i
n=l
also converge. Moreover,
(l+cian)ci
1 -h coan
and
+ coa n
| | ^-yn=l
co(co-ci)
2
2/1
, _ ,
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373
3.8. Infinite Products
where en —• 0 as n —• oo. Thus, by 3.8.3 and 3.8.4, the series ^
a
n
n=l
converges. Next, by 3.8.13, ]T an also converges. Consequently, for
n=l
oo
each c E R, both series Yl (can)2
oo
and J2 can converge. Hence our
n=l
n=l
claim follows from 3.8.7.
oo
n
n=l
k=0
3.8.35. Clearly, the series J2 an Yl {x<2 ~ ^ 2 ) converges to zero if x
is an integer. Assume now that it converges for a noninteger value
XQ. For x e R , consider the sequence whose terms are given by
n(* 2 -* 2 )
,
fc=0
°n = -^
n(*§-
-fc 2 )
fc=0
Then
2
2
A
x -k
fc=0 '
Prom this, we conclude that, beginning with some value of the index
n, the sequence {bn} is monotonic. Moreover, since the product
0 0
2 . 2
FI X2_fe2 converges, the sequence {6 n } is bounded. We have also
oo
n
n=l
k=0
2
2
oo
n
n=l
fc=0
J2*n H(X ~fc )= £ > „ JI(^0 - *%•
Therefore, by the Abel test, the series under consideration converges
for any x £ R.
3.8.36.
(a) We have
Multiplying the first TV equalities, we obtain
" /
n-'
1
w II l -^)
n=l
v
^
n /
1
~, i
>» i
= +EF =E F
fc=l
fc=i
+
~,
E
k=pN+l
i
P
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Solutions. 3: Series of Real Numbers
374
where £}' denotes summation over the integers which in their
prime factorization contain only prime numbers pi,P2> ~,PNHence
0 <T7fi--V
111
n=l
Since
px)
x
lim
^
V
n /
1 ,
- r-= v' - < v
Z ^ kx
fe=l
2^
Z^
kx
fc=P7V
+l
-
k*'
k=pN + l
— = 0, we get
-1
111
nx )
PnJ
Z^
£rn/
.—*
oo
a
i nnX
n=l
(b) By (i) in the solution of part (a),
N
_ 1
PN
nO-r) »£*•
1 \
/
„=1 V
P«/
oo
1
fc=i*
oo
/
Therefore the divergence of J^ ^ implies that J ] (1
n=l
n=l ^
\
—J di-
P n
'
verges to zero, which in turn is equivalent to the divergence of
oo
the series Yl 7T ( s e e 3.8.4).
n=l
3.8.37. [18]
(a) By DeMoivre's law, cos rat + i sin mt = (cost -f isint)171, with
m — 2n+ 1, we get
sin(2n -f l)t = (2n + 1) cos 2n *sin* - (
n
*
J cos 2 n ~ 2 tsin 3 1
+ ... + ( - l ) n s i n 2 n + 1 £ .
So we can write
(1)
sin(2n + l)t = sin t W(sin21),
where W(u) is a polynomial of degree < n. Since the function
on the left-hand side of the equality vanishes at tk = 2^+1 > ^ =
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375
3.8. Infinite Products
1,2, ...,n, which belong to the interval (0, f ) , the polynomial
W(u) vanishes at Uk = sin2 tky k = 1,2, ...,n. Consequently,
tj
sin
kJ[\
Hence, by (1),
(2)
sin(2n + l)t = A sin t J J
sin 2 1
kn
sin2 2n+l
JXLX
1 -
k=l \
The task is now to find A. We have A = lim
/
sin(2n 1)t
|
sint
t_^o
= 2n + 1.
Substituting this value of A into (2) and taking t = 2^+\ > w e &et
(3)
W
2
s i n2n+l
^^
^ kn±
2
^ ,
°sl i Ai n 2n+l
x
A /
sinx = ( 2 n + l ) s i n — ^ — T T 1
V
;
2n + l 1fe=l
1 IV
For fixed x € R and rn € N such that \x\ < (m + l)7r, take n
greater than m. Then, by (3),
Vv
SUl £ =
-Lm,nWm,m
where
sin2
Pm,„ = ( 2 n + l ) s i n ^ - I j i f l
**cm,n — II
I
fc=m+l
V
s i n 2n+l
-^
° xxx
fe=l \
2 x \
sin' 2n+l \
sin" 2
OX
*
2n+l
2
k
^
2n+l/
Letting n —> oo, we obtain
(5)
TT /
lim P m?n = x | ]
fc=l
V
r-2
^2
1 - -2-2
It follows from (4) that for x ^ kir, lim Qm
n—*oo
n
'
= Qm. To esti-
mate Q m , we note that by the above assumptions,
|x|
kn
nir
TT
u<
' ' <
r<~
7 < 7 ^ for fc = r n + l , . . . , n .
2n + 1
2n + 1 ~ 2n + 1
2
' '
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Solutions. 3: Series of Real Numbers
376
Taking into account the inequality ^u < sinu < u, 0 < u < | ,
we see that
00
/
Y\ (1 - f^z)
71=1 ^
(1 — ^W) < Qm,n < 1. Since the product
Y[
fc=m+l
2 \
converges, we have
'
00
/
2 \
n (i-^j<«»<!•
x
/c=ra+l
'
Consequently,
(6)
lim Q m = 1.
m—>oo
Finally, the desired equality follows from (4) , (5) and (6).
(b) Apply (a) and the identity sin2x = 2 sin x cos x.
3.8.38. Substitute x = § in the formula stated in 3.8.37 (a).
3.8.39.
(a) The convergence of the given product is equivalent to the conoo
vergence of the series Yl ( m ( l + n )
—
n)* ^he absolute con-
71=1
vergence of this series follows from the equality
| l n ( l + *) - * |
1
•£_
2
V
lim '
n—•oo
r
— = ~.
(b) We have
So the absolute convergence of the product follows from 3.8.3.
3.8.40. Clearly, the product fj (1 + a n ) , an > — 1, converges if and
71=1
only if the series J2
n=l
m
a
( l + n) converges. Moreover, if P is the
value of the product and 5 is the sum of the series, then P = es.
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377
3.8. Infinite Products
Assume now that the product converges absolutely. Then in view
of the equality
T |ki(l-f-a n )|
hm
:—:
=1
n\
(1)
n—KX>
\CLn\
(because
lim an = 0)
n—>oo
the series ]T] ln(l + a n ) converges absolutely.
Consequently (see
n=l
3.7.8), any of its rearrangements converges to the same sum. Finally,
by the remark at the beginning of this solution, any rearrangement
of the factors of the product does not change its value.
oo
Assume now that the value of the product f ] (1 + an) does not
71=1
depend on the order of its factors. This means that the sum of the
oo
series ^ ln(l 4- an) is also independent of the order of its terms. By
n=l
Riemann's theorem, the series converges absolutely, which in view
oo
of (1) implies the convergence of ^ \an\. Thus the desired result is
71=1
proved.
3.8.41. [20] Set A , = | • J • | •... • 2 g l . * $ ± f . Then
2) \
4/
\
2a
i-iVi_iy_(i*
Z) V
5 / •" V
2/3 + 1 /
Rp'
Hence the (a + /3)nth partial product is equal to ^ m i . By the Wallis
formula (see 3.8.38),
(2n+l)!!
2
r
hm
—
n->oo (2n)\\y/n
y/n'
and therefore
n-^oo Rn/3
y j3
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378
Solutions. 3: Series of Real Numbers
3.8.42. If the product
Yl (1 + an) converges, but not absolutely,
CO
n=l
oo
then the series ]T) ln(l + a n ) converges conditionally (see the solun=l
tion of 3.8.40). On account of the Riemann theorem, its terms can
be rearranged to give either a convergent series whose sum is an arbitrarily preassigned real number 5, or a divergent series (to -f oo or
to — oo). Thus our claim follows from the relation P = es (see the
solution of 3.8.40).
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Bibliography - Books
References
[I]
J. Banas, S. W§drychowicz, Zbior zadari z analizy matematycznej, Wydawnictwa Naukowo-Techniczne, Warszawa, 1994.
[2]
W. I. Bernuk, I. K. Zuk, O.W. Melnikov, Sbornik olimpiadnych zadac
po matematike, Narodnaja Asveta, Minsk, 1980.
[3]
P. Biler, A. Witkowski, Problems in Mathematical Analysis, Marcel
Dekker, Inc, New York and Basel, 1990.
[4]
T. J. Bromwich, An Introduction to the Theory of Infinite
Macmillan and Co., Limited, London, 1949.
[5]
R. B. Burckel, An Introduction to Classical Complex Analysis, Academic Press, New York San Francisco, 1979.
[6]
B. P. Demidovic, Sbornik zadac i upraznenij po matematiceskomu
izu, Nauka, Moskva, 1969.
[7]
A. J. Dorogovcev, Matematiceskij
Skola, Kiev, 1985.
analiz. Spravocnoe posobe, Vyscaja
[8]
A. J. Dorogovcev, Matematiceskij
Kiev, 1987.
analiz. Sbornik zadac, Vyscaja Skola,
[9]
G. M. Fichtenholz, Differential- und Integralrechnung, I, II, III, V.E.B.
Deutscher Verlag Wiss., Berlin, 1966-1968.
Series,
anal-
[10] G. H. Hardy, A Course of Pure Mathematics, Cambridge University
Press, Cambridge, 1946.
[II] G. H. Hardy, J. E. Littlewood, G. Polya, Inequalities, Cambridge University Press, Cambridge, 1967.
379
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380
Bibliography - Books
[12] G. Klambauer, Mathematical Analysis, Marcel Dekker, Inc., New York,
1975.
[13] G. Klambauer, Problems and Propositions in Analysis, Marcel Dekker,
Inc., New York and Basel, 1979.
[14] K. Knopp, Theorie und Anwendung der Unendlichen Reihen, SpringerVerlag, Berlin and Heidelberg, 1947.
[15] L. D. Kudriavtsev, A. D. Kutasov, V. I. Chejlov, M. I. Shabunin,
Problemas de Andlisis Matemdtico. Limite, Continuidad, Derivabilidad, Mir, Moskva, 1989.
[16] K. Kuratowski, Introduction to Calculus, Pergamon Press, OxfordEdinburgh-New York; Polish Scientific Publishers, Warsaw, 1969.
[17] D. S. Mitrinovic, Elementary Inequalities, P. Noordhoff Ltd., Groningen, 1964.
[18] D. S. Mitrinovic, D. D. Adamovic, Nizovi i Redovi. Definicije, stavovi,
zadaci, problemi (Serbo-Croatian), Naucna Knjiga, Belgrade, 1971.
[19] A. Ostrowski, Aufgabensammlung zur Infinitesimalrechnung, Band I:
Funktionen einer Variablen, Birkhauser Verlag, Basel und Stuttgart,
1964.
[20] G. Polya, G. Szego, Problems and theorems in analysis I, SprigerVerlag, Berlin Heidelberg New York, 1978.
[21] Ya. I. Rivkind, Zadaci po matematiceskomu
Minsk, 1973.
analizu, Vysejsaja Skola,
[22] W. I. Rozhkov, G. D. Kurdevanidze, N. G. Panfilov, Sbomik zadac
matematiceskich olimpiad, Izdat. Univ. Druzhby Narodov, Moskva,
1987.
[23] W. Rudin, Principles of Mathematical Analysis, McGraw-Hill Book
Company, New York, 1964.
[24] W. A. Sadownicij, A. S. Podkolzin, Zadaci studenceskich olimpiad po
matematike, Nauka, Moskva, 1978.
[25] W. Sierpinski, Arytmetyka teoretyczna, PWN, Warszawa, 1959.
[26] W. Sierpinski, Dzialania nieskonczone, Czytelnik, Warszawa, 1948.
[27] H. Silverman, Complex variables, Houghton Mifflin Company, Boston,
1975.
[28] G. A. Tonojan, W. N. Sergeev, Studenceskije matematiceskije
dy, Izdatelstwo Erevanskogo Universiteta, Erivan, 1985.
olimpia-
Copyright 2000 American Mathematical Society. Duplication prohibited. Please report unauth
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Copyright 2000 American Mathematical Society. Duplication prohibited. Please report una
Thank You!