Page 289 - Schaum's Outline of Theory and Problems of Advanced Calculus

280 INFINITE SERIES [CHAP. 11

1 2 1 p ffiffiffi 1
X 4n n þ 3 X n X ln n
11.9. Test for convergence; (a) ; ðbÞ n þ ; ðcÞ .
3
3
2
n þ 2n 2n 1 n þ 3
n¼1 n¼1 n¼1
2
2
4n n þ 3 4n 2 4 4n n þ 3 4
(a) For large n, is approximately ¼ . Taking u n ¼ and v n ¼ ,wehave
3
3
n þ 2n n 3 n n þ 2n n
u n
lim ¼ 1.
n!1 v n
Since v n ¼ 4 1=n diverges, u n also diverges by Problem 11.8.
Note that the purpose of considering the behavior of u n for large n is to obtain an appropriate
comparison series v n . Inthe above we could just as well have taken v n ¼ 1=n.
2
4n n þ 3 !
Another method: lim n ¼ 4. Then by Theorem 1, Page 267, the series converges.
3
n þ 2n
n!1
n n 1
p ffiffiffi
n þ
.
(b) For large n, u n ¼ 3 is approximately v n ¼ 3 ¼ 2
2n 1 2n 2n
u n X 1 X 1
Since lim ¼ 1 and v n ¼ converges ( p series with p ¼ 2), the given series converges.
2 n 2
n!1 v n
n 1
p ffiffiffi
Another method: lim n 2 n þ ¼ . Then by Theorem 1, Page 267, the series converges.
3
2n 1 2
n!1
ln n 3=2 ln n ln n

(c) lim n 3=2 @ lim n ¼ lim p ¼ 0(by L’Hospital’s rule or otherwise). Then by
2
n þ 3 n 2 n
ffiffiffi
n!1 n!1 n!1
Theorem 1 with p ¼ 3=2, the series converges.
ln n n 1
Note that the method of Problem 11.6(a)yields 2 < 2 ¼ , but nothing can be deduced since
1=n diverges. n þ 3 n n

1 1
X n 2 X 3 1
11.10. Examine for convergence: (a) e ; ðbÞ sin .
n
n¼1 n¼1
2
2 n
(a) lim n e ¼ 0(by L’Hospital’s rule or otherwise). Then by Theorem 1 with p ¼ 2, the series con-
n!1
verges.
(b) For large n, sinð1=nÞ is approximately 1=n. This leads to consideration of
3
3
lim n sin 3 1 ¼ lim sinð1=nÞ ¼ 1
n 1=n
n!1 n!1
from which we deduce, by Theorem 1 with p ¼ 3, that the given series converges.
INTEGRAL TEST
11.11. Establish the integral test (see Page 267).
We perform the proof taking N ¼ 1. Modifications are easily made if N > 1.
From the monotonicity of f ðxÞ,we have
n ¼ 1; 2; 3; .. .
u nþ1 ¼ f ðn þ 1Þ @ f ðxÞ @ f ðnÞ¼ u n
Integrating from x ¼ n to x ¼ n þ 1, using Property 7, Page 92,
nþ1
ð
u nþ1 @ f ðxÞ dx @ u n n ¼ 1; 2; 3 ...
n
Summing from n ¼ 1to M 1,
ð M
u 2 þ u 3 þ þ u M @ f ðxÞ dx @ u 1 þ u 2 þ þ u M 1 ð1Þ
1
If f ðxÞ is strictly decreasing, the equality signs in (1)can be omitted.