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Whether the given series ∑ n = 1 ∞ i n n is convergent or divergent. The given series ∑ n = 1 ∞ i n n is divergent _ . Theorem used: Consider a series z 1 + z 2 + ⋅ ⋅ ⋅ with z ≠ 0 is such that lim n → ∞ | z n + 1 z n | = L . Then, If L < 1 the series converges absolutely. If L > 1 , the series diverges. If L = 1 , the series may converge or diverges, so that the test fails and permits no conclusion. Calculation: The given series is ∑ n = 1 ∞ i n n . First separate the even and odd terms. For the even terms, that is n = 2 m , the range of m would be m = 1 to ∞ . Then put n = 2 m and m = 1 to ∞ in the summation ∑ n = 1 ∞ i n n as follows: ∑ m = 1 ∞ ( − i ) 2 m 2 m = ∑ m = 1 ∞ ( − i ) 2 m 2 m = 1 2 ∑ m = 1 ∞ ( − 1 ) m m Since the summation ∑ n = 1 ∞ ( − 1 ) n n converges, the summation ∑ m = 1 ∞ ( i ) 2 m 2 m also converges. Now for the odd terms that is n = 2 m + 1 , the range of m would be m = 0 to ∞ . Then put n = 2 m + 1 and m = 0 to ∞ in the summation ∑ n = 1 ∞ i n n as follows: ∑ m = 0 ∞ ( − i ) 2 m + 1 2 m + 1 = i ∑ m = 0 ∞ ( − 1 ) m 2 m + 1 Then consider the series ∑ m = 0 ∞ ( − 1 ) m 2 m + 1 , for which z m = ( − 1 ) m 2 m + 1 , | z m | = 1 2 m + 1 and | z m + 1 | = 1 2 m + 3 . Applying the ratio test as follows: 1 2 ( 0 ) + 3 > 1 2 ( 0 ) + 1 1 3 > 1 1 | z m + 1 | < | z m | From the above theorem condition L > 1 , series is divergent. Hence, the given series ∑ n = 1 ∞ i n n is divergent _ .

Whether the given series ∑ n = 1 ∞ i n n is convergent or divergent. The given series ∑ n = 1 ∞ i n n is divergent _ . Theorem used: Consider a series z 1 + z 2 + ⋅ ⋅ ⋅ with z ≠ 0 is such that lim n → ∞ | z n + 1 z n | = L . Then, If L < 1 the series converges absolutely. If L > 1 , the series diverges. If L = 1 , the series may converge or diverges, so that the test fails and permits no conclusion. Calculation: The given series is ∑ n = 1 ∞ i n n . First separate the even and odd terms. For the even terms, that is n = 2 m , the range of m would be m = 1 to ∞ . Then put n = 2 m and m = 1 to ∞ in the summation ∑ n = 1 ∞ i n n as follows: ∑ m = 1 ∞ ( − i ) 2 m 2 m = ∑ m = 1 ∞ ( − i ) 2 m 2 m = 1 2 ∑ m = 1 ∞ ( − 1 ) m m Since the summation ∑ n = 1 ∞ ( − 1 ) n n converges, the summation ∑ m = 1 ∞ ( i ) 2 m 2 m also converges. Now for the odd terms that is n = 2 m + 1 , the range of m would be m = 0 to ∞ . Then put n = 2 m + 1 and m = 0 to ∞ in the summation ∑ n = 1 ∞ i n n as follows: ∑ m = 0 ∞ ( − i ) 2 m + 1 2 m + 1 = i ∑ m = 0 ∞ ( − 1 ) m 2 m + 1 Then consider the series ∑ m = 0 ∞ ( − 1 ) m 2 m + 1 , for which z m = ( − 1 ) m 2 m + 1 , | z m | = 1 2 m + 1 and | z m + 1 | = 1 2 m + 3 . Applying the ratio test as follows: 1 2 ( 0 ) + 3 > 1 2 ( 0 ) + 1 1 3 > 1 1 | z m + 1 | < | z m | From the above theorem condition L > 1 , series is divergent. Hence, the given series ∑ n = 1 ∞ i n n is divergent _ .

Advanced Engineering Mathematics

Advanced Engineering Mathematics

10th Edition

ISBN: 9780470458365

Author: Erwin Kreyszig

Publisher: Wiley, John & Sons, Incorporated

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Chapter 15 Solutions

Advanced Engineering Mathematics

Ch. 15.1 - Prob. 1P Ch. 15.1 - Prob. 2P Ch. 15.1 - Prob. 3P Ch. 15.1 - Prob. 4P Ch. 15.1 - Prob. 5P Ch. 15.1 - Prob. 6P Ch. 15.1 - Prob. 7P Ch. 15.1 - Prob. 8P Ch. 15.1 - Prob. 9P Ch. 15.1 - Prob. 10P

Ch. 15.1 - Prob. 12P Ch. 15.1 - Prob. 13P Ch. 15.1 - Prob. 14P Ch. 15.1 - Prob. 15P Ch. 15.1 - Prob. 16P Ch. 15.1 - Prob. 17P Ch. 15.1 - Prob. 18P Ch. 15.1 - Prob. 19P Ch. 15.1 - Prob. 20P Ch. 15.1 - Prob. 21P Ch. 15.1 - Prob. 22P Ch. 15.1 - Prob. 23P Ch. 15.1 - Prob. 24P Ch. 15.1 - Prob. 25P Ch. 15.1 - Prob. 26P Ch. 15.1 - Prob. 27P Ch. 15.1 - Prob. 29P Ch. 15.1 - Prob. 30P Ch. 15.2 - Prob. 1P Ch. 15.2 - Prob. 2P Ch. 15.2 - Prob. 3P Ch. 15.2 - Prob. 4P Ch. 15.2 - Prob. 5P Ch. 15.2 - Prob. 6P Ch. 15.2 - Prob. 7P Ch. 15.2 - Prob. 8P Ch. 15.2 - Prob. 9P Ch. 15.2 - Prob. 10P Ch. 15.2 - Prob. 11P Ch. 15.2 - Prob. 12P Ch. 15.2 - Prob. 13P Ch. 15.2 - Prob. 14P Ch. 15.2 - Prob. 15P Ch. 15.2 - Prob. 16P Ch. 15.2 - Prob. 17P Ch. 15.2 - Prob. 18P Ch. 15.3 - Prob. 1P Ch. 15.3 - Prob. 2P Ch. 15.3 - Prob. 3P Ch. 15.3 - Prob. 4P Ch. 15.3 - Prob. 5P Ch. 15.3 - Prob. 6P Ch. 15.3 - Prob. 7P Ch. 15.3 - Prob. 8P Ch. 15.3 - Prob. 9P Ch. 15.3 - Prob. 10P Ch. 15.3 - Prob. 11P Ch. 15.3 - Prob. 12P Ch. 15.3 - Prob. 13P Ch. 15.3 - Prob. 14P Ch. 15.3 - Prob. 15P Ch. 15.3 - Prob. 16P Ch. 15.3 - Prob. 17P Ch. 15.3 - Prob. 18P Ch. 15.3 - Prob. 19P Ch. 15.4 - Prob. 1P Ch. 15.4 - Prob. 2P Ch. 15.4 - Prob. 3P Ch. 15.4 - Prob. 4P Ch. 15.4 - Prob. 5P Ch. 15.4 - Prob. 6P Ch. 15.4 - Prob. 7P Ch. 15.4 - Prob. 8P Ch. 15.4 - Prob. 9P Ch. 15.4 - Prob. 10P Ch. 15.4 - Prob. 11P Ch. 15.4 - Prob. 12P Ch. 15.4 - Prob. 13P Ch. 15.4 - Prob. 14P Ch. 15.4 - Prob. 16P Ch. 15.4 - Prob. 18P Ch. 15.4 - Prob. 19P Ch. 15.4 - Prob. 20P Ch. 15.4 - Prob. 21P Ch. 15.4 - Prob. 22P Ch. 15.4 - Prob. 23P Ch. 15.4 - Prob. 24P Ch. 15.4 - Prob. 25P Ch. 15.5 - Prob. 2P Ch. 15.5 - Prob. 3P Ch. 15.5 - Prob. 4P Ch. 15.5 - Prob. 5P Ch. 15.5 - Prob. 6P Ch. 15.5 - Prob. 7P Ch. 15.5 - Prob. 8P Ch. 15.5 - Prob. 9P Ch. 15.5 - Prob. 10P Ch. 15.5 - Prob. 11P Ch. 15.5 - Prob. 12P Ch. 15.5 - Prob. 13P Ch. 15.5 - Prob. 14P Ch. 15.5 - Prob. 15P Ch. 15.5 - Prob. 16P Ch. 15.5 - Prob. 17P Ch. 15 - Prob. 1RQ Ch. 15 - Prob. 2RQ Ch. 15 - Prob. 3RQ Ch. 15 - Prob. 4RQ Ch. 15 - Prob. 5RQ Ch. 15 - Prob. 6RQ Ch. 15 - Prob. 7RQ Ch. 15 - Prob. 8RQ Ch. 15 - Prob. 9RQ Ch. 15 - Prob. 10RQ Ch. 15 - Prob. 11RQ Ch. 15 - Prob. 12RQ Ch. 15 - Prob. 13RQ Ch. 15 - Prob. 14RQ Ch. 15 - Prob. 15RQ Ch. 15 - Prob. 16RQ Ch. 15 - Prob. 17RQ Ch. 15 - Prob. 18RQ Ch. 15 - Prob. 19RQ Ch. 15 - Prob. 20RQ Ch. 15 - Prob. 21RQ Ch. 15 - Prob. 22RQ Ch. 15 - Prob. 23RQ Ch. 15 - Prob. 24RQ Ch. 15 - Prob. 25RQ Ch. 15 - Prob. 26RQ Ch. 15 - Prob. 27RQ Ch. 15 - Prob. 28RQ Ch. 15 - Prob. 29RQ Ch. 15 - Prob. 30RQ

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