(a) Prove that the series in Visual Studio .NET

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11.16. (a) Prove that the series
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(a) The series is 1 1 1 1 1 . 3 5 7 9
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1 n 1 1 1 jun 1 j , then an jun j ,a . 2n 1 2n 1 n 1 2n 1 1 1 1 @ and since lim 0, it follows by Problem 11.5(a) that the series Since n!1 2n 1 2n 1 2n 1 If un
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converges.
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1 1 1 (b) Use the results of Problem 11.15(b). Then the rst 8 terms give 1 1 1 1 1 11 13 15 and the 3 5 7 9 1 error is positive and does not exceed 17. 1 1 1 1 Similarly, the rst 9 terms are 1 1 1 1 1 11 13 15 17 and the error is negative and 3 5 7 9 1 1 greater than or equal to 19, i.e., the error does not exceed 19 in absolute value.
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CHAP. 11]
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INFINITE SERIES
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The absolute value of the error made in stopping after M terms is less than 1= 2M 1 . To obtain the desired accuracy, we must have 1= 2M 1 @ :001, from which M A 499:5. Thus, at least 500 terms are needed.
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ABSOLUTE AND CONDITIONAL CONVERGENCE 11.17. Prove that an absolutely convergent series is convergent.
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Given that jun j converges, we must show that un converges. Let SM u1 u2 uM and TM ju1 j ju2 j juM j. Then SM TM u1 ju1 j u2 ju2 j uM juM j @ 2ju1 j 2ju2 j 2juM j Since jun j converges and since un jun j A 0, for n 1; 2; 3; . . . ; it follows that SM TM is a bounded monotonic increasing sequence, and so lim SM TM exists.
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Also, since lim TM exists (since the series is absolutely convergent by hypothesis),
M!1 M!1
lim SM lim SM TM TM lim SM TM lim TM
M!1 M!1 M!1
must also exist and the result is proved.
11.18. Investigate the convergence of the series
p p p sin 1 sin 2 sin 3 3=2 3=2 . 3 13=2 2
Since each term is in absolute value less than or equal to the corresponding term of the series 1 1 , which converges, it follows that the given series is absolutely convergent and 13=2 23=2 33=2 hence convergent by Problem 11.17.
11.19. Examine for convergence and absolute convergence: a
1 X 1 n 1 n ; n2 1 n 1
1 X 1 n 1 ; 2 n 2 n ln n
1 X n 1
1 X 1 n 1 2n : n2 n 1
(a) The series of absolute values is
n which is divergent by Problem 11.13(b). n2 1
Hence, the given
series is not absolutely convergent. n n 1 However, if an jun j 2 , then an 1 @ an for all n A 1, and and an 1 jun 1 j n 1 n 1 2 1 n also lim an lim 2 0. Hence, by Problem 11.15 the series converges. n!1 n!1 n 1 Since the series converges but is not absolutely convergent, it is conditionally convergent. 1 X 1 . (b) The series of absolute values is 2 M n 2 n ln n dx exists or does not By the integral test, this series converges or diverges according as lim M!1 2 x ln2 x exist. If u ln x; dx du 1 1 c c: 2 u ln x u2 x ln x
M Hence, lim converges. Then
M!1 2
  dx 1 1 1 lim and the integral exists. 2 ln 2 x ln x M!1 ln 2 ln M
Thus, the series
1 X 1 n 1 converges absolutely and thus converges. 2 n 2 n ln n
INFINITE SERIES
[CHAP. 11
Another method: 1 1 1 @ and lim 0, it follows by Problem 11.15(a), that the 2 2 n!1 n ln2 n n 1 ln n 1 n ln n given alternating series converges. To examine its absolute convergence, we must proceed as above. Since (c) 1 n 1 2n , the given series cannot be convergent. To show that n2 2n lim un 6 0, it su ces to show that lim jun j lim 2 6 0. This can be accomplished by L Hospital s n!1 n!1 n!1 n rule or other methods [see Problem 11.21(b)]. Since lim un 6 0 where un
RATIO TEST 11.20. Establish the ratio test for convergence.
Consider rst the series u1 u2 u3 where each term is non-negative. We must prove that if u lim n 1 L < 1, then necessarily un converges. n!1 un By hypothesis, we can choose an integer N so large that for all n A N, un 1 =un < r where L < r < 1. Then uN 1 < r uN uN 2 < r uN 1 < r2 uN uN 3 < r uN 2 < r3 uN etc. By addition, uN 1 uN 2 < uN r r2 r3 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 ju1 j ju2 j ju3 j . Then by the above  u  proof and Problem 11.17, it follows that if lim  n 1  L < 1, then un converges (absolutely). n!1  un      u  u  Similarly, we can prove that if lim  n 1  L > 1 the series un diverges, while if lim  n 1  L 1 n!1  un  n!1  un  the ratio test fails [see Problem 11.21(c)].
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