Page 292 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 11] INFINITE SERIES 283

(c) The absolute value of the error made in stopping after M terms is less than 1=ð2M þ 1Þ.To obtain the
desired accuracy, we must have 1=ð2M þ 1Þ @ :001, from which M A 499:5. Thus, at least 500 terms
are needed.

ABSOLUTE AND CONDITIONAL CONVERGENCE

11.17. Prove that an absolutely convergent series is convergent.
Given that ju n j converges, we must show that u n converges.
Let S M ¼ u 1 þ u 2 þ þ u M and T M ¼ju 1 jþju 2 jþ þ ju M j. Then

S M þ T M ¼ðu 1 þju 1 jÞ þ ðu 2 þju 2 jÞ þ þ ðu M þju M jÞ
@ 2ju 1 jþ 2ju 2 jþ þ 2ju M j
Since ju n j converges and since u n þju n j A 0, for n ¼ 1; 2; 3; .. . ; it follows that S M þ T M is a bounded
monotonic increasing sequence, and so lim ðS M þ T M Þ exists.
M!1
Also, since lim T M exists (since the series is absolutely convergent by hypothesis),
M!1
lim S M ¼ lim ðS M þ T M T M Þ¼ lim ðS M þ T M Þ lim T M
M!1 M!1 M!1 M!1
must also exist and the result is proved.

ffiffiffi ffiffiffi ffiffiffi
p p p
sin 1 sin 2 sin 3
11.18. Investigate the convergence of the series þ .
1 3=2 2 3=2 3 3=2
Since each term is in absolute value less than or equal to the corresponding term of the series
1 1 1 þ , which converges, it follows that the given series is absolutely convergent and
1 3=2 þ 2 3=2 þ 3 3=2
hence convergent by Problem 11.17.

11.19. Examine for convergence and absolute convergence:
n 1 n 1 n 1 n
1 1 1
X n X X 2
; ; :
ð 1Þ ð 1Þ ð 1Þ
n þ 1 n ln n n
ðaÞ 2 ðbÞ 2 ðcÞ 2
n¼1 n¼2 n¼1
1
X n
(a) The series of absolute values is which is divergent by Problem 11.13(b). Hence, the given
2
n þ 1
n¼1
series is not absolutely convergent.
n n þ 1
n þ 1 ðn þ 1Þ þ 1
However, if a n ¼ju n j¼ 2 and a nþ1 ¼ju nþ1 j¼ 2 ,then a nþ1 @ a n for all n A 1, and
n
also lim a n ¼ lim ¼ 0. Hence, by Problem 11.15 the series converges.
2
n!1 n þ 1
n!1
Since the series converges but is not absolutely convergent, it is conditionally convergent.
1
1
X
(b) The series of absolute values is .
2
n ln n
n¼2 ð M dx
By the integral test, this series converges or diverges according as lim 2 exists or does not
exist. M!1 2 x ln x
ð ð
dx du 1 1
If u ¼ ln x; ¼ ¼ þ c ¼ þ c:
2
x ln x u 2 u ln x
M dx 1 1 1
ð
Hence, lim ¼ lim ¼ and the integral exists. Thus, the series
2
2 x ln x M!1 ln 2 ln M ln 2
M!1
converges.
n 1
1
X
Then ð 1Þ converges absolutely and thus converges.
2
n ln n
n¼2