# q2s - 1. Thus, by the Alternating Series Test, our series...

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MATH 153 Quiz 2 Solutions Name: Determine whether the following series absolutely converge, conditionally converge or diverge. 1. [8 points] X n =1 ( - 1) n n n 2 + 2 First, does the series converge absolutely? X n =1 ± ± ± ± ( - 1) n n n 2 + 2 ± ± ± ± = X n =1 n n 2 + 2 Let’s compare using 1 n : lim n →∞ n n 2 +2 1 n = lim n →∞ n 2 n 2 + 2 = 1 Since this limit is positive and ﬁnite, and since 1 n diverges ( p -series with p 1), we have by Limit Comparison Test that the series n n 2 +2 diverges, and thus that our series does not converge absolutely. Does the series converge conditionally? It’s an alternating series, so we apply the Alternating Series Test. Let b n = n n 2 ; then lim n →∞ n n 2 + 2 = 1 /n 1 + 2 /n 2 = 0 1 + 0 = 0 , as needed. We also must show that { b n } is a decreasing sequence: n + 1 ( n + 1) 2 + 2 n n 2 + 2 ( n + 1)( n 2 + 2) n [( n + 1) 2 ) + 2] n 3 + n 2 + 2 n + 2 n 3 + 2 n 2 + 3 n 2 n 2 + n, which is true for all n

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Unformatted text preview: 1. Thus, by the Alternating Series Test, our series converges conditionally . 2. [6 points] X n =1 (-3) n n 2 n ! Let us apply the Ratio Test: lim n (-3) n +1 ( n +1) 2 ( n +1)! (-3) n n 2 n ! = lim n 3 n +1 ( n + 1) 2 ( n + 1)! n ! 3 n n 2 = lim n 3( n + 1) 2 ( n + 1) n 2 = lim n 3( n + 1) n 2 = 0 As this limit is less than one, we can say that our series converges abso-lutely . 3. [6 points] X n =1 (2 n ) n 5 n Let us apply the Root Test: lim n n s (2 n ) n 5 n = lim n 2 n 5 = Since this limit is not less than one, we can say that our series diverges ....
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## This note was uploaded on 07/26/2011 for the course MATH 153 taught by Professor Rempe during the Spring '08 term at Ohio State.

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q2s - 1. Thus, by the Alternating Series Test, our series...

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