HW14 - Math 191 HW 14 Solutions 11.6 6 converges by the...

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Unformatted text preview: Math 191 HW 14 Solutions 11.6: 6 converges by the Alternating Series Test since f ( x ) = ln x x ⇒ f ( 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 √...
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This note was uploaded on 06/19/2008 for the course MATH 1910 taught by Professor Berman during the Spring '07 term at Cornell.

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HW14 - Math 191 HW 14 Solutions 11.6 6 converges by the...

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