ma241exam4-c-key

ma241exam4-c-key - Spring 2007 Therkelsen MA 241 Exam 4(C...

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Unformatted text preview: Spring 2007 Therkelsen MA 241 - Exam 4 (C) KEY 1. Determine whether each series is convergent or divergent. (a) ∞ ∑ n =1 3+( − 1) n n √ n (b) ∞ ∑ n =1 n − 1 n 5 n Solution: . (a) Note that 0 ≤ 3+( − 1) n n √ n ≤ 4 n √ n = 4 n 3 / 2 . ∞ ∑ n =1 1 n 3 / 2 is a p-series, it converges since p = 3 2 > 1. ∞ ∑ n =1 4 n 3 / 2 therefore converges as well and so ∞ ∑ n =1 3+( − 1) n n √ n is convergent, by the comparison test. (b) Note that 0 ≤ n − 1 n 5 n < n n 5 n = 1 5 n = ( 1 5 ) n . ∞ ∑ n =1 ( 1 5 ) n is a geometric series that converges since r = 1 5 < 1 and so ∞ ∑ n =1 n − 1 n 5 n is convergent, by the comparison test. (The ratio test would also work.) 2. Determine whether each series is convergent or divergent. (a) ∞ ∑ n =1 ( − 1) n cos ( 1 n ) n (b) ∞ ∑ n =0 ( − 1) n n 2 8+6 n 2 Solution: . (a) Since cos ( 1 n ) > 0 for all n ≥ 1, the terms of the series alternate between pos- itive and negative. Since lim n →∞ cos ( 1 n ) n = 0, ∞ ∑ n =1 ( − 1) n cos ( 1 n ) n is convergent, by the alternating series test. 1 (b) Observe that lim n →∞ ( − 1) n n 2 8+6 n 2 does not exist. Therefore, ∞ ∑ n =0 ( − 1) n n 2 8+6 n 2 is divergent, by the divergence test....
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This note was uploaded on 01/25/2011 for the course MA 241 taught by Professor Mccollum during the Spring '08 term at N.C. State.

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ma241exam4-c-key - Spring 2007 Therkelsen MA 241 Exam 4(C...

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