Problem Set Fourteen: Partial Summation
Partial Summation Formula: Let ( a n ) and ( b n ) be any sequences, write s n  
n
k 1
bk
for n  1 , and
s 0  0 . For any indices 0  m  n ,

n
k  m 1
a k b k  a n 1 s n  a m 1 s m 

n
k  m 1
( a k  a k 1 ) s k
.
Theorem: If ( a n ) and ( b n ) are sequences such that ( a n ) is monotone decreasing, a n  0 , and the
partial sums s n  
n
k 1
bk
are bounded, then


k 1
ak bk
converges.
Corollary (Alternating Series Test): If ( a n ) is monotone decreasing, a n  0 then


k 1
(  1)
k 1
ak
converges.
Example: Let a k  1/k for k odd and a k  1/k 2 for k even. Then ( a n ) is positive and a n  0 but


k 1
(  1)
k 1
ak
diverges. If fact, the 2nth partial sum satisfies
1
 k 1 2k  1   k 1 (  1)
n
2n
k 1
ak 
1
 k 1 (2k) 2
n
.
If the alternating signs series were to converge, then

1
 k 1 2k  1
would too.
PROBLEMS
Problem 15-1: Why does this series converge?
1

1
1

2
1

3
1

4
1
5

1
6

1
7

1
8

1

9
1

10
1

11
1
12

1
13
1


14
Problem 15-2: If ( a n ) converges monotonically to zero and


k 1
ak bk
1

15

1
 ...
16

k 1
bk
converges then
converges.
Problem 15-3: Prove that a k 
1
(  1)
k 1

is positive, converges to zero and
k


k 1
(  1)
k 1
ak
diverges.
Problem 15-4: (a) For any θ and  , 2 cos (θ ) sin ( )  sin ( θ   )  sin ( θ   ) .
(b) In part (a) use θ  k x,   x/2 , sum and telescope. As a result conclude
2 sin ( x /2)

n
k 1
cos ( k x)  sin  (n  1/2) x   sin  x /2  .
(c) If ( a n ) converges monotonically to zero and 0  x  2 π then


k 1
a k cos ( k x)
converges.