Chapter 1 Infinite Series, Power Series
January 20,22 Geometric series
1.1 The geometric series
1.2 Definitions and notation
1.3 Applications of series
Infinite sequence: An ordered set of infinite number of quantities. Examples:
1) 2, 4, 6, 8,  , 2n, 
1 1 1
1
2) 1, , , ,  , n 1 , 
2 4 8
2
3) 1,  1, 1,  1, 1,  1, 
4) a1 , a2 , a3 , a1 ,  , an , 
Limit of an infinite sequence:
lim an  l means :
n 
For any   0, however small it is, there exists an integer N such that an  l   for all n  N .
1
 0.
n  n 2
Examples p5.1-3; Problems 2.7, 8.
Example : lim
1
Infinite series: The sum of an infinite sequence of numbers. Examples:

1) 2  4  6  8    2n     2n
n 1

1 1 1
1
1
2)1      n 1     n 1
2 4 8
2
n 1 2

3) 1  ( 1)  1  ( 1)     ( 1)n 1
n 1

4) a1  a2  a3    an     an
n 1
Note:
1) The sum of an infinite series may not be finite.
2) Even when the sum is finite, we still cannot do it by adding the terms one by one.
n
Partial sum of an infinite series: S n   ai  a1  a2  a3    an
i 1
Sum of an infinite series is defined as: S  lim S n
n 
2
Geometric series: A geometric series has the general term of an  ar n 1 and

 ar
n 1
 a  ar  ar 2  
n 1
n
Partial sum of a geometric series: Sn   ar
i 1
 a  ar  ar    ar
2
i 1
Sum of a geometric series: S  lim S n 
n 
Example : 1 
n 1
a(1  r n )

1 r
a
, if and only if r  1.
1 r
2 4 8
1
 
 
 3.
2
3 9 27
1
3
Application: Change recurring decimals into fractions. Examples:
abc  103 abc
3
6
0.abcabc   0.abc  abc(10  10  ) 

.
3
1  10
999
185 5
0.185 

999 27
25
0.25 
99
1 3 7
0.583  0.25  0.3   
4 9 12
3
Reading: L’Hospital’s rule:
If lim f ( x)  lim g ( x)  0 (or  ), and lim
x  x0
( or  )
x  x0
( or  )
x  x0
( or  )
f ' ( x)
f ( x)
f ' ( x)
exists, then lim
 lim
.
x  x0 g ( x )
x  x0 g ' ( x )
g ' ( x)
( or  )
( or  )
f ( x)
f ' ( x)
f ' ' ( x)
f ( n ) ( x)
Repeatedly , lim
 lim
 lim
   lim ( n )
until the limit exsists.
x  x0 g ( x )
x  x0 g ' ( x )
x  x0 g ' ' ( x )
x  x0 g
( x)
Proof :
1) If lim f ( x)  lim g ( x)  0, then lim
x  x0
x  x0
x  x0
f ( x0 )  f ' ( x0 )( x  x0 )  
f ( x)
f ' ( x)
 lim
 lim
.
x

x
x

x
0
0
g ( x)
g ( x0 )  g ' ( x0 )( x  x0 )  
g ' ( x)
2) If lim f ( x)  lim g ( x)  , then
x  x0
x  x0
f ( x)
1 / g ( x)
 g ' ( x) / g 2 ( x)
g ' ( x)
f 2 ( x)
f ( x)
f ' ( x)
lim
 lim
 lim

lim
lim

lim

lim
.
x  x0 g ( x )
x  x0 1 / f ( x )
x  x0  f ' ( x ) / f 2 ( x )
x  x0 f ' ( x ) x  x 0 g 2 ( x )
x  x0 g ( x )
x  x0 g ' ( x )
f ( x)
f (1 / t )
 (1 / t 2 ) f ' (1 / t )
3) If lim f ( x)  lim g ( x)  0, then lim
 lim
 lim
x 
x 
x  g ( x )
t  0 g (1 / t )
t  0  (1 / t 2 ) g ' (1 / t )
f ' (1 / t )
f ' ( x)
 lim
 lim
.
t 0 g ' (1 / t )
x  g ' ( x )
4) If lim f ( x)  lim g ( x)  , then
x 
x 
f ( x)
1 / g ( x)
 g ' ( x) / g 2 ( x)
g ' ( x)
f 2 ( x)
f ( x)
f ' ( x)
lim
 lim
 lim

lim
lim

lim

lim
.
x  g ( x )
x  1 / f ( x )
x   f ' ( x ) / f 2 ( x )
x  f ' ( x ) x  g 2 ( x )
x  g ( x )
x  g ' ( x )
4
Read: Chapter 1: 1-3
Homework:
1.1.2,8;
1.2.1,7.
Due: January 29
5
January 25 Convergence of series
1.4 Convergent and divergent series
Convergence of series: If for an infinite series we have lim S n  S , where S is a finite
n 
number, the series is said to be convergent. Otherwise it is divergent.
Note:
1) Whether or not a series is convergent is of essential interest for the series.
2) For a divergent series, Sn either approaches infinity or is oscillatory.
3) Adding or removing a finite number of terms from an infinite series will not affect
whether or not it converges.
Problems 4.3, 6.
More examples :
1 1 1
1 1 1 1 1 1 1 1 1
1
    1                
2 3 4
2  3 4   5 6 7 8   9 10
16 
1
1
1
1
1
 1   2   4   8     2 p  p 1   is divergent.
2
4
8
16
2
1) Harmonic series 1 

2) 1  (1)  1  (1)     (1) n is divergent.
n 1
6
1.5 Testing series for convergence; the preliminary test
Preliminary test:
If lim an  0, then the series is divergent.
n 
Proof :
If the series is convergent , then lim an  lim S n  S n1   S  S  0.
n 
n 
Note:
an  0 is a necessary condition for convergence of a series,
1) The requirement that lim
n 
but is not sufficient. E.g., the harmonic series.
an  0 , then further test is needed.
2) If lim
n 
Problems 5.3, 9.
7
1.6 Convergence test for series of positive terms; absolute convergence
A. The comparison test:
If 0  an  bn for all n, then

1) If

b
n 1
n
converges, then  an converges.
n 1

2) If
a
n 1

n
diverges, then
b
n 1
n
diverges.
Proof :

If
b
n 1
n

converges, let
b
n 1
n
 Sb ,
n
a
i 1
i
 S an , then S an is monotonica lly increasing ,
and S an  Sb , therefore S an has a limit.
Problems 6.4, 5.
More examples :
1
1
1

 n 1 for n  1.
n! 1 2   (n  1)  n 2

1
( 2) is convergent , therefore

n 1
2
n 1

1
is convergent .

n
!
n 1
8
Reading: Monotone convergence theorem:
Let S be a set of real numbers. A real number x is called an upper bound for S if x ≥ s
for all s ∈ S. A real number x is the least upper bound for S if x is an upper bound for S,
and x ≤ y for every upper bound y of S.
Least-upper-bound property: Any set of real numbers that has an upper bound must
have a least upper bound.
Monotone convergence theorem:
A bounded monotonic sequence of real numbers has a finite limit.
Proof :
Suppose an  an 1 and an  c, where c is the least upper bound of an , then for any given
number   0, there exists N so that for all n  N we have an  c   . This is because if this N
does not exist then c   will be an upper bound, which is in contradict ion with t he fact that
c is the least upper bound of an . Now an  c    an  c   , which means lim an  c.
n 
9
Read: Chapter 1: 4-6
Homework:
1.4.7;
1.5.3,9;
1.6.4,5.
Due: February 5
10
January 27, 29 Convergence test
1.6 Convergence test for series of positive terms; absolute convergence
B. The integral test:
If an  f (n), and f ( x) is positive and monotonica lly decreasing , then

 an converges if and only if
n 1


f ( x)dx converges.
Proof : The proof is easy to understand with the following two figures.
Note:
The lower limit in the integral is not necessary. Using x = 0 or x =1 may cause problems.

 1
1
E.g.,  2 dx is infinite, but  2 converges .
0 n
n 1 n
Example p12; Problems 6.12.
11
The p-series:

1
( p  0, p  1)
p
n
n 1
 p 1 b

 dx
b dx
x
b  p 1  1
1

lim

lim

lim

p
1 x p b 1 x p b  p  1 b  p  1 
n
n

1
1

1
diverges

n 1 n
More examples : The p - series : 

converges if p  1,
diverges if p  1. 





1
converges if p  1,
(
p

0
)
diverges if p  1.
p

n
n 1

12
C. The ratio test:
If lim
n 


1) If   1,  an converges,
n 1



  , then 2) If   1,  an diverges.
n 1



3) If   1, further te st is needed.
an 1
an
Proof :
1) If   1, take  so that     1. Then for all n  N ,
an 1
 .
an
Form bn   n  N a N for n  N , then an  bn .

Since
b
n
n 1

converges,
a

n
n 1
(and thus
 a . An absolutely
n
convergent series is convergent .
n 1
To be proved.) must converge.
a
2) If   1, take  so that     1. Then for all n  N , n 1    1, lim an  0.
n 
an

a
n
must diverge.
n 1
Example p14.1,2; Problems 6.19,21.
13
D. Limit comparison test (a special comparison test):
an
 l , then
n  b
n
If an and bn are both positive, and lim
1) If 0  l  , then
a
n
and  bn either both converge or both diverge.
2) If l  0, and  bn converges, then
3) If l  , and  bn diverges,
 a converges.
then  a diverges.
n
n
Proof :
1) If 0  l  , then ther e exist l1 and l2 , which satisfy
0  l1  l  l2 , and


an
  l1bn  an  l2bn . Therefore
l1 
 l2 for large n 
bn

or both diverge according to the comparisio n test.
2)If l  0, for large n we have
a
n
an
 1  an  bn . Therefore if
bn
3)If l  , then for large n we have
and  bn either both converge
b
n
an
 1  bn  an . Therefore if
bn
converges then
b
n
a
diverges then
n
converges.
a
n
diverges.
Example p15.1; Problems 6.31,35.
14
Read: Chapter 1:6
Homework:
1.6.12,13,21,26,28,32,35.
Due: February 5
15
February 1 Alternating series
1.7 Alternating series
1.8 Conditionally convergent series


n 1
n 1
Absolute convergence: A series  an is said to be absolutely convergent if  | an |
is convergent.
Theorem: An absolutely convergent series is convergent.
Proof :

Suppose  | an | converges, let bn  an  | an | . The partial sum of
n 1



n 1
n 1
n 1

b
n 1
n
is non - decreasing
and is bounded from above by  bn   an  | an |   2 | an |.

Therefore
b
n 1
n

converges. Then


 a   b  | a
n 1
n
n 1
n
n 1
n
| also converges.

Conditional convergence: A series  an is said to be conditionally convergent if

a
n 1

n
is convergent but
| a
n 1
n 1
n
| is not convergent.
16
Alternating series: An alternating series is a series whose successive terms alternate in
1 1 1
sign. E.g., 1     
2 3 4
Test for alternating series (Leibnitz’s alternating series theorem):
An alternating series converges if the absolute values of the terms decreases steadily to 0.
an  0.
That is an1  an and lim
n
Proof :
Suppose a1  0, then the sequence
S2 n  ( a1  a2 )  ( a3  a4 )    ( a2 n 1  a2 n )  0 and is non - decreasing .
Also S2 n  a1  ( a2  a3 )  ( a4  a5 )    ( a2 n 2  a2 n 1 )  a2 n  a1 is bounded from above.
Therefore lim S2 n exists. Let lim S2 n  L.
n 
n 
Now lim S2 n 1  S2 n   lim a2 n 1  0, Therefore lim S2 n 1  lim S2 n  L.
n 
n 
n 
n 
These two facts shows that lim Si  L.
i 
Example :
(1) n
1 1 1

1

    converges ( ln2).

n
2
3 4
n 1

Problems 7.4,6,7.
17
1.9 Useful facts about series
1) The convergence or divergence of a series is not affected by multiplying each term by
the same none-zero constant.
2) The sum of two convergent series is convergent.
Problems 9.11,18,19.
18
Read: Chapter 1:7-9
Homework:
1.7.3,4,6,7;
1.9.1,4,9,10,16,20.
Due: February 10
19
February 3, 5 Power series
1.10 Power series; interval of convergence
Power series: A power series is an infinite series of the form

S ( x)   an ( x  a) n  a0  a1 ( x  a )  a2 ( x  a ) 2  
n 0
We say that the power series is in x and is about the point a.

A special case is a=0 and S ( x)   an x n  a0  a1 x  a2 x 2  
n 0
Interval of convergence: The range of x where the power series converges.
Ratio test of the convergence of a power series:
an 1 ( x  a) n 1
an 1
lim

lim
x  a  l x  a  .
n  a ( x  a ) n
n  a
n
n
1
1
1) The series converges when a   x  a  (all x if l  0, no x if l  ).
l
l
1
1
2) The series diverges when x  a  or x  a  .
l
l
1
1
3) Further te st is needed for the end points x  a  and x  a  .
l
l
Example p21.2,3; Problems 10.2,7,17.
20
1.11 Theories about power series
1) A power series may be differentiated or integrated term by term. The resultant series
converges within the original interval of convergence (but not necessarily at the
endpoints).


xn
x n1
Example : f ( x)   2 converges in x [1,1]. f ' ( x)  
converges in x [1,1).
n 1 n
n 1 n
2) Two power series may be added, subtracted, or multiplied. The resultant series
converges at least in the common interval of convergence. Two power series can be
divided if the denominator series is not 0, and the resultant series has some interval of
convergence.

Example :
[ x
n
 (2 x) n ] converges in x (1 / 2,1 / 2).
n 0
3) One series can be substituted into another if the value of the substituted series is in the
interval of convergence of the other series.
4) The power series of a function is unique.
These theories are easy to understand if we treat a power series as a well defined
function.
21
1.12 Expanding functions in power series
Taylor series: A Taylor series expansion of a function f (x) at x=a has the form

f ( x)  a0  a1 ( x  a)  a2 ( x  a)     an ( x  a) n .
2
n 0
We can test t hat f (a)  a0 , f ' (a)  a1 , f " (a)  2a2 , , and f ( n ) (a)  n!an . Therefore

f ( x)  
n 0
f " (a)
f ( n ) (a)
( x  a) 2  .
( x  a) n  f (a)  f ' (a)( x  a) 
2
n!
Maclaurin series: The Taylor series expansion of a function f (x) at x=0:

f ( x)  
n 0
f ( n ) (0) n
f " (0) 2
x  f (0)  f ' (0) x 
x  .
n!
2
22
1.13 Techniques for obtaining power series expansions
Some important Taylor series expansions (print in your head):
(1) n x 2 n 1
x3 x5 x7
sin x  
 x    
3! 5! 7!
n  0 ( 2n  1)!

(1) n x 2 n
x2 x4 x6
cos x  
 1   
(2n)!
2! 4! 6!
n 0


xn
x 2 x3 x 4
e    1 x    
2! 3! 4!
n  0 n!
x
(1) n 1 x n
x 2 x3 x 4
ln( 1  x)  
 x      , x  (1,1]
n
2 3 4
n 1

m
m(m  1) 2 m(m  1)( m  2) 3
(1  x)     x n  1  mx 
x 
x  , x  (1,1)
2!
3!
n 0  n 
(Binomial expansion) , especially

m
1
 1  x  x 2  x 3  , x  (1,1).
1 x
1
 1  x  x 2  x 3   , x  (1,1).
1 x
23
Read: Chapter 1:10-13
Homework:
1.10.2,7,15,23.
Due: February 10
24
February 8 Power series expansion
1.13 Techniques for obtaining power series expansions
A. Multiply a series by a polynomial or by another series
Example p26.1,2; Problem 13.5,9.
B. Division of two series or of a series by a polynomial
Example p27.1; Problem 13.22.
C. Binomial series
Example p29.2.
D. Substitution of a polynomial or a series for a variable in another series
Example p29.1; Problem 13.10,27.
E. Combination of methods
Example p30; Problem 13.13.
F. Taylor series using the basic Maclaurin series
Example p30.1; Problems 13.39.
25
1.14 Accuracy of series approximations
Error in a convergent alternating series:

If S   an is an alternatin g series with an 1  an and lim an  0, then the reminder
n 
n 1
Rn  S  (a1  a2    an )  an 1 .
That is, the error is less than the first neglected term.
Proof :
Suppose an 1  0, [otherwise we consider

 (a ).]
n 1
n
then Rn  (an 1  an  2 )  (an 3  an  4 )    0.
Rn  an 1  (an  2  an 3 )  (an  4  an 5 )    an 1.
Example p34.1; Problem 14.4.
Note: This rule only applies to an alternating series.
26
Theorem

If S   an x n converges for x  1, and an 1  an for n  N , then
n 0
N
S   an x
n

n 0
a N 1 x N 1
1 x
.
Proof :
N
S   an x n

n 0

a x
n  N 1
n
n


a
n  N 1
n
x
n



n  N 1
a N 1 x 
n
a N 1 x N 1
1 x
Problem 14.6.
27
Read: Chapter 1:13-14
Homework:
1.13.5,8,14,20,28,42;
(Find first 3 terms only. No computer work needed.)
1.14.6.
Due: February 19
28