Page 294 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 11] INFINITE SERIES 285

n 2 2

ð 1Þ ðn þ 1Þ ðn þ 1Þðn þ 1Þ
lim u nþ1 n þ 1 ¼ lim ¼ 1
2
¼ lim 2 n 1 n!1 ðn þ 2n þ 2Þn
n!1 u n n!1 ðn þ 1Þ þ 1 ð 1Þ n

and the ratio test fails. By using other tests [see Problem 11.19(a)], the series is seen to be convergent.
MISCELLANEOUS TESTS
2
4
5
3
11.22. Test for convergence 1 þ 2r þ r þ 2r þ r þ 2r þ where (a) r ¼ 2=3, (b) r ¼ 2=3,
(c) r ¼ 4=3.

1
Here the ratio test is inapplicable, since u nþ1 ¼ 2jrj or jrj depending on whether n is odd or even.
2
u n
However, using the nth root test, we have
(
n
n
n
p ffiffiffiffiffiffiffiffiffi p ffiffiffi
p ffiffiffiffiffiffiffiffi if n is odd
n 2jr j ¼ 2 jrj
n
ju n j ¼
p ffiffiffiffiffiffiffi
n if n is even
jr j ¼jrj
p ffiffiffiffiffiffiffiffi 1=n
Then lim n ju n j ¼jrj (since lim 2 ¼ 1).
n!1 n!1
Thus, if jrj < 1the series converges, and if jrj > 1the series diverges.
Hence, the series converges for cases (a) and (b), and diverges in case (c).
1 1 4 1 4 7
2 2 2 2
11.23. Test for convergence þ þ þ þ 1 4 7 ... ð3n 2Þ þ .
3 3 6 3 6 9 3 6 9 ... ð3nÞ
2
3n þ 1

The ratio test fails since lim u nþ1 ¼ lim ¼ 1. However, by Raabe’s test,
n!1 3n þ 3
n!1 u n
( )
2
3n þ 1 4

> 1
u nþ1

lim n 1 ¼ lim n 1 ¼
3n þ 3 3
n!1 u n n!1
and so the series converges.
2 2 2 2
1 1 3 1 3 5 1 3 5 ... ð2n 1Þ
11.24. Test for convergence þ þ þ þ þ .
2 2 4 24t 2 4 6 ... ð2nÞ
2

2n þ 1

The ratio test fails since lim u nþ1 ¼ lim ¼ 1. Also, Raabe’s test fails since

n!1 u n n!1 2n þ 2
( )
2
2n þ 1

u nþ1 ¼ 1
lim n 1 ¼ lim n 1
2n þ 2
n!1 u n n!1
However, using long division,
2
2n þ 1 1 5 4=n 1
c n
u nþ1 where jc n j < P
¼ ¼ 1 þ 2 ¼ 1 þ 2
2n þ 2 n 4n þ 8n þ 4 n n
u n
so that the series diverges by Gauss’ test.