# HW#9 - SOLUTIONS TO HOMEWORK # 9 14.1 (a) Ratio test . lim...

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Unformatted text preview: SOLUTIONS TO HOMEWORK # 9 14.1 (a) Ratio test . lim fl fl fl fl ( n + 1) 4 2 n +1 · 2 n n 4 fl fl fl fl = lim fl fl fl fl 1 2 n + 1 n ¶ 4 fl fl fl fl = 1 2 < 1. Thus ∑ n 4 2 n converges. (b) Ratio test . lim fl fl fl fl 2 n +1 ( n + 1)! · n ! 2 n fl fl fl fl = lim fl fl fl fl 2 n + 1 fl fl fl fl = 0 < 1. Thus ∑ 2 n n converges. (c) Ratio test . lim fl fl fl fl ( n + 1) 4 3 n +1 · 3 n n 2 fl fl fl fl = lim fl fl fl fl 1 3 n + 1 n ¶ 2 fl fl fl fl = 1 3 < 1, so ∑ n 2 3 n converges. (d) n ! > n 4 + 3 for n ≥ 7. Hence n n 4 + 3 does not converge to 0. Thus by Corollary 14.5, ∑ n ! n 4 + 3 diverges. (e) cos 2 n ≥ 0 for all n and cos 2 n n 2 ≤ 1 n 2 . The series ∑ 1 n 2 converges, thus ∑ cos 2 n n 2 converges by the comparison test. (f) log n < n , so 1 log n > 1 n . Since ∑ 1 n diverges our series ∑ 1 log n diverges by the comparison test. 14.2 (a) n- 1 > n 2 for n ≥ 3. Therefore for n ≥ 3 we have n- 1 n 2 ≥ n 2 n 2 = 1 2 1 n ....
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## This note was uploaded on 04/28/2008 for the course MATH 312 taught by Professor Olsavskygregorm during the Spring '08 term at Penn State.

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HW#9 - SOLUTIONS TO HOMEWORK # 9 14.1 (a) Ratio test . lim...

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