Problem Set Nine: Basic Series
Definition: For a given sequence ( a n ) let s n  a 1  a 2  a 3  ...  a n  
n
. The sequence ( s n ) is
a
k 1 k
called the series with nth term a n and nth partial sum s n . The usual notation is to write


k 1
ak
to
denote the series with terms a n . The series converges to s, or the series has sum s, iff s n  s . It is


traditional to also use
k 1
for the sum of a convergent series. A series is divergent iff it does not
ak
converge.
Example: For fixed a and c the geometric series
is


k0
ca
k
 c /(1  a)

Theorem: (a) If
(b) If


k 1
Definition:
k 1
ak

k 1
ak


k 1

n


k 1
ak
ak
k0
k 0
ca
k
converges iff a  1 , in which case the sum
ca
k
 c (1  a
n 1
) /(1  a)
for a  1 .
converges.
is absolutely convergent iff

convergent iff it converges but
Theorem: Let

converges then a n  0 .
converges then
ak


. The partial sums are



k 1
ak
converges.


k 1
ak
is conditionally
diverges.
ak
k 1

be a series with non-negative terms.
(a) The sequence of partial sums ( s n ) is monotone increasing.
(b) The series converges iff sequence of partial sums ( s n ) is bounded above, in which case the sum of
the series is s  lub{ s n : n  1} .
Theorem (Comparison Test): Let ( a n ) and ( b n ) be two non-negative sequences, and suppose that
 c  0  m  N so that n  m  a n  c b n . Then
(a) if


(b) if


k 1
k 1
bk
converges then
ak
diverges then



k 1

k 1
ak
converges, and
diverges.
bk
Theorem: Let f(x) be a non-negative and decreasing function defined for x  m .
Define g n 

n
km
f(k) 

n
f(x) d x
.
m
(a) The sequence ( g n ) in non-negative, monotone decreasing, and thus convergent.
(b, Integral Test) The series


km
n
converges iff lim n  f(x) d x exists, ie, iff
f(k)
m
convergent.
Example: For 0 < p the p-series


k 1
k
p
converges iff p > 1.


f(x) d x
m
is

Definition: The harmonic series is the divergent series
γ  lim
n
γn
of the sequence γ n 

n
k 1
k
1


Example: The alternating harmonic series

n
x
1
dx 
1

k 1
(  1)
k 1

k 1

k
k
1
n
1
k
k 1
1
. Euler’s constant is the limit
 l n (n)
.
is conditionally convergent and has sum
ln(2).
PROBLEMS
Problem 9-1. Give the sum of the geometric series
Problem 9-2. Show that



k2
3
(  1)
4
k 1
.
2k
k

2
k 1
k!
converges by comparison to a geometric series.
Problem 9-3. Do these converge or diverge?

Problem 9-4. For what positive values of p does
 (1 

k 1

k

k2
1
)
k
and
1
k [ l n (k)]
p


k 1
(1 
1 k
)
k
converge? Diverge? For the
divergent series give a closed form lower estimate for the nth partial sum. (“Closed form” means not
involving a summation sign.)
Problem 9-5. Let

constant prove that

k 1

ak

k 1
and

(ak  bk )

k 1
bk
and
be convergent series with sums A and B, respectively. For c a


k 1
cak
have sums A  B and cA , respectively.
Problem 9-6. A permutation of the natural numbers is a function p : N  N that is one-to-one and onto.
For a given permutation p the series


k 1
ak


k 1
a p (k)
is called a rearrangement of


k 1
ak
. Prove that if
is a convergent series with non-negative terms, then every rearrangement converges and has the
same sum.
Problem 9-7. Rearrange the terms of the alternating harmonic series so that the 3nth partial sum is
 1
1 1 1 1 1 1
1
1
1
1 

.
s 3n  1           
   ...  



3 2   5 7 4   9 11 6 

 4 n  3 4 n 1 2 n 
Prove that the series converges to 3 ln(2) / 2 .
Problem 9-8. Rearrange the terms of the alternating harmonic series like this;
1 1 1 1 1 1
1
1  1
1
1
1
1

  



   ...
1           
2
3
5
4
7
9
11
6
13
15
17
19
8

 
 
 
Prove that this series diverges.
Problem 9-9. Suppose that
(an)
is a positive decreasing sequence and


k 1
ak
converges. Show that
n an  0 .
Problem 9-10 (There is no largest convergent series.) If


k 1
ak
is a convergent series with non-
negative terms, prove there is an positive, unbounded sequence ( c k ) so that
Problem 9-11 (There is no smallest divergent series.). If


k 1
bk
terms, prove there is a positive sequence ( c k ) so that c k  0 and


k 1
ck a k
converges.
is a divergent series with non-negative


k 1
ck bk
diverges.
Definition: The Lebesgue outer measure (or simply the measure) of a set A of reals is
meas(A)
 glb {

k 1
len (I k ) } ,
where the glb is taken over all covers C  { I k } k  1 of A by a countable number of open intervals.
Observe that meas(A)  con(A ) .
Problem 9-12. Prove these properties of measure. (a) A  B implies meas(A)  meas(B) .
(b) meas(  n  1 A n )  
n 1
meas(A
n
)
for any countable collection of sets.
(c) meas ( [a, b] )  b  a .
(d) The measure of any countable set is zero.