# When the series does converge it is just the

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When the series does converge, it is just the geometric series X n =0 x - 4 5 n = 1 1 - x - 4 5 = 1 5 - ( x - 4) 5 = 5 9 - x , so when it converges the series converges to the function f ( x ) = 5 9 - x . 17. Does the series X n =1 ( - 1) n ( n 2 + n 3 ) n 4 + 1 converge absolutely, converge conditionally, or diverge? Explain your answer. Answer: The series of absolute values X n =1 ( - 1) n ( n 2 + n 3 ) n 4 + 1 = X n =1 n 2 + n 3 n 4 + 1 should behave similarly to n 3 n 4 , so do a limit comparison to this series: lim n →∞ n 2 + n 3 n 4 n 3 n 4 = lim n →∞ n 2 + n 3 n 4 · n 4 n 3 = lim n →∞ n 2 + n 3 n 3 = 1 . Therefore, since n 3 n 4 = 1 n diverges, the Limit Comparison Test says that the series of absolute values diverges as well. However, the given series satisfies the hypotheses of the Alternating Series Test, so it converges. There- fore, since the series converges but the series of absolute values diverges, we conclude that the series converges conditionally. 18. Does the series X n =1 (2 n )! 2 n ( n !) 2 converge or diverge? Explain your answer. Answer: Use the Ratio Test: lim n →∞ (2( n +1))! 2 n +1 (( n +1)!) 2 (2 n )! 2 n ( n !) 2 = lim n →∞ (2 n + 2)! 2 n +1 (( n + 1)!) 2 · 2 n ( n !) 1 (2 n )! = lim n →∞ (2 n + 2)(2 n + 1) 2 · ( n !) 2 (( n + 1) n !) 2 = lim n →∞ (2 n + 2)(2 n + 1) 2 · 1 ( n + 1) 2 . Since (2 n + 2)(2 n + 1) = 4 n 2 + 6 n + 2 and since ( n + 1) 2 = n 2 + 2 n + 1, the above limit is equal to lim n →∞ 4 n 2 + 6 n + 2 2( n 2 + 2 n + 1) = 4 2 = 2 . Since 2 > 1, the Ratio Test implies that the given series diverges. 7
19. Does the series X n =2 ( - 1) n n ln( n ) converge absolutely, converge conditionally, or diverge? Explain your answer. Answer: The series of absolute values X n =2 ( - 1) n n ln( n ) = X n =2 1 n ln( n ) diverges (see HW #12, Problem 3 for a proof of this). However, the series satisfies the hypotheses of the Alternating Series Test and hence converges, so we see that it converges conditionally. 20. For what values of p does the series X n =0 1 ( n 2 + 1) p converge? Explain your answer. Answer: For large n , the +1 will barely contribute, so do a limit comparison with the series 1 ( n 2 ) p : lim n →∞ 1 ( n 2 +1) p 1 ( n 2 ) p = lim n →∞ 1 ( n 2 + 1) p · ( n 2 ) p 1 = lim n →∞ n 2 n 2 + 1 p = 1 p = 1 , since lim n →∞ n 2 n 2 +1 = 1. Therefore, the given series and the series 1 ( n 2 ) p = 1 n 2 p will either both converge or both diverge. Since 1 n 2 p converges for 2 p > 1 and diverges otherwise, we see that the given series converges when 2 p > 1, which is to say when p > 1 2 . 21. What is the interval of convergence of the following power series? Explain your answer. X n =1 3 n ( x - 2) n n 2 . Answer: Start by applying the Ratio Test to the series of absolute values: lim n →∞ 3 n +1 ( x - 2) n +1 ( n +1) 2 3 n ( x - 2) n n 2 = lim n →∞ 3 n +1 | x - 2 | n +1 ( n + 1) 2 · n 2 3 n | x - 2 | n = lim n →∞ 3 | x - 2 | · n 2 ( n + 1) 2 = 3 | x - 2 | . Therefore, the ratio test implies that the given series converges absolutely when 3 | x - 2 | < 1, meaning when | x - 2 | < 1 3 .
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