Page 287 - Schaum's Outline of Theory and Problems of Advanced Calculus

278 INFINITE SERIES [CHAP. 11

6. Summability. Let S 1 ; S 2 ; S 3 ; ... be the partial sums of a divergent series u n .If the sequence
S 1 ; S 2 S 1 þ S 2 þ S 3
S 1 ; ; ; ... (formed by taking arithmetic means of the first n terms of
2 3
S 1 ; S 2 ; S 3 ; .. .)converges to S,we say that the series u n is summable in the Ce ´saro sense,or
C-1 summable to S (see Problem 11.51).
If u n converges to S, the Ce ´ saro method also yields the result S. For this reason the
Ce ´ saro method is said to be a regular method of summability.
In case the Ce ´ saro limit does not exist, we can apply the same technique to the sequence
S 1 þ S 2 S 1 þ S 2 þ S 3
S 1 ; ; ; ... : If the C-1 limit for this sequence exists and equals S,we say
3 3
that u k converges to S in the C-2 sense. The process can be continued indefinitely.

Solved Problems

CONVERGENCE AND DIVERGENCE OF SERIES OF CONSTANTS
1 1 1 X 1
1
11.1. (a)Prove that þ þ þ ¼ converges and (b) find its sum.
1 3 3 5 5 7
n¼1 ð2n 1Þð2n þ 1Þ
1 1 1 1
: Then
2 2n 1 2n þ 1
u n ¼ ¼
ð2n 1Þð2n þ 1Þ

1 1 1 1 1 1 1 1 1
2 1 3 2 3 5 2 2n 1 2n þ 1
S n ¼ u 1 þ u 2 þ þ u n ¼ þ þ þ
1 1 1 1 1 1 1 1 1 1

¼ þ þ þ ¼ 1
2 1 3 3 5 5 2n 1 2n þ 1 2 2n þ 1
1 1 1
Since lim S n ¼ lim ¼ ; the series converges and its sum is 1 :
n!1 2 1 2n þ 1 2 2
n!1
The series is sometimes called a telescoping series, since the terms of S n ,other than the first and last,
cancel out in pairs.
1
X
2 2 2 2 3 2 n
3
3
3
3
11.2. (a)Prove that þð Þ þð Þ þ ¼ ð Þ converges and (b) find its sum.
n¼1 n
. Since jrj < 1
að1 r Þ
This is a geometric series; therefore, the partial sums are of the form S n ¼ 1 r
a
2
2
and in particular with r ¼ and a ¼ ,we obtain S ¼ 2.
1 r
S ¼ lim S n ¼ 3 3
n!1
1
X n
1 2 3 4
2 3 4 5 n þ 1
11.3. Prove that the series þ þ þ þ ¼ diverges.
n¼1
n
lim u n ¼ lim ¼ 1. Hence by Problem 2.26, Chapter 2, the series is divergent.
n!1 n þ 1
n!1
p ffiffiffiffiffiffiffiffiffiffiffi
p ffiffiffi
n diverges although lim u n ¼ 0.
11.4. Show that the series whose nth term is u n ¼ n þ 1
n!1
The fact that lim u n ¼ 0follows from Problem 2.14(c), Chapter 2.
n!1 ffiffiffi ffiffiffi ffiffiffi ffiffiffi ffiffiffi
p p p p p ffiffiffiffiffiffiffiffiffiffiffi ffiffiffi p ffiffiffiffiffiffiffiffiffiffiffi p
p
1.
Now S n ¼ u 1 þ u 2 þ þ u n ¼ð 2 1Þþð 3 2Þþ þ ð n þ 1 nÞ¼ n þ 1
Then S n increases without bound and the series diverges.
This problem shows that lim ¼ 0isa necessary but not sufficient condition for the convergence of u n .
See also Problem 11.6. n!1