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# HW14 - Math 191 11.6 6 converges by the Alternating Series...

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Math 191 HW 14 Solutions 11.6: 6 converges by the Alternating Series Test since f ( x ) = ln x x f 0 ( x ) = 1 - ln x x 2 < 0 when x > e f ( x ) is decreasing u n u n +1 ; also u n 0 for n 1 and lim n →∞ u n = lim n →∞ ln n n = lim n →∞ 1 n 1 = 0 11.6: 10 diverges by the nth-Term Test since lim n →∞ 3 n + 1 n + 1 = lim n →∞ 3 q 1 + 1 n 1 + 1 n = 3 6 = 0 11.6: 18 converges absolutely because the series X n =1 sin n n 2 converges by the Direct Comparison Test since sin n n 2 1 n 2 11.6: 39 converges conditionally since n +1 - n 1 · n +1+ n n +1+ n = 1 n +1+ n and n 1 n +1+ n o is a decreasing sequence of positive terms which converges to 0 X n =1 ( - 1) n n + 1 + n converges; but X n =1 | a n | = X n =1 1 n + 1 + n diverges by the Limit Comparison Test (part 1) with 1 n ; a divergent p -series; lim n →∞ 1 n +1+ n 1 n = lim n →∞ n n + 1 + n = lim n →∞ = 1 q 1 + 1 n + 1 = 1 2 11.6: 40 diverges by the nth-Term Test since lim n →∞ ( p n 2 + n - n ) = lim

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HW14 - Math 191 11.6 6 converges by the Alternating Series...

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