Page 293 - Schaum's Outline of Theory and Problems of Advanced Calculus

284 INFINITE SERIES [CHAP. 11

Another method:
1 1 1
Since @ and lim ¼ 0, it follows by Problem 11.15(a), that the
2
2 2 n!1 n ln n
ðn þ 1Þ ln ðn þ 1Þ n ln n
given alternating series converges. To examine its absolute convergence, we must proceed as above.
n 1 n
2
(c) Since lim u n 6¼ 0 where u n ¼ ð 1Þ ,the given series cannot be convergent. To show that
n 2 n
n!1 2
lim u n 6¼ 0, it suffices to show that lim ju n j¼ lim 6¼ 0. This can be accomplished by L’Hospital’s
n!1 n 2
n!1 n!1
rule or other methods [see Problem 11.21(b)].
RATIO TEST
11.20. Establish the ratio test for convergence.
Consider first the series u 1 þ u 2 þ u 3 þ where each term is non-negative. We must prove that if
u nþ1
lim ¼ L < 1, then necessarily u n converges.
n!1 u n
By hypothesis, we can choose an integer N so large that for all n A N, ðu nþ1 =u n Þ < r where L < r < 1.
Then
u Nþ1 < ru N
2
u Nþ2 < ru Nþ1 < r u N
3
u Nþ3 < ru Nþ2 < r u N
etc. By addition,
2 3
u Nþ1 þ u Nþ2 þ < u N ðr þ r þ r þ Þ
and so the given series converges by the comparison test, since 0 < r < 1.
In case the series has terms with mixed signs, we consider ju 1 jþju 2 jþju 3 jþ . Then by the above

proof and Problem 11.17, it follows that if lim u nþ1 ¼ L < 1, then u n converges (absolutely).

n!1 u n

Similarly, we can prove that if lim u nþ1 ¼ L > 1the series u n diverges, while if lim u nþ1 ¼ L ¼ 1

n!1 u n n!1 u n
the ratio test fails [see Problem 11.21(c)].
n 1 n n 1
1 1 1
X 4 n 2 X ð 1Þ 2 X ð 1Þ n
11.21. Investigate the convergence of (a) n e ; ðbÞ ; ðcÞ .
2
n 2 n þ 1
n¼1 n¼1 n¼1
2
4 n
(a)Here u n ¼ n e . Then
2 2

4 ðnþ1Þ 4 ðn þ2nþ1Þ
ðn þ 1Þ e ðn þ 1Þ e
lim u nþ1 ¼ lim ¼ lim

4 n 2 4 n 2
n!1 u n n!1 n e n!1 n e
4 4
n þ 1 n þ 1

¼ lim e 2n 1 ¼ lim lim e 2n 1 ¼ 1 0 ¼ 0
n n
n!1 n!1 n!1
Since 0 < 1, the series converges.
n 1 n
2
. Then
ð 1Þ
(b)Here u n ¼ 2
n

n nþ1 2 2

ð 1Þ 2 n 2n
lim u nþ1 ¼ lim ¼ lim ¼ 2

2
2 2
n!1 u n n 1 n
ð 1Þ n!1 ðn þ 1Þ
n!1 ðn þ 1Þ
Since s > 1, the series diverges. Compare Problem 11.19(c).
n 1 n
(c) Here u n ¼ ð 1Þ . Then
2
n þ 1