# sol6 - 7 1 n ² = ln 7 Converges(j Since 0 ≤ cos 2 n ≤...

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MATH 138 Calculus 2 Fall 2010 Solutions #6 (a) lim n →∞ 1 - (0 . 2) n = 1 - 0 = 1 with | r | = | 0 . 2 | < 1; Converges. (b) lim n →∞ 3 n +2 5 n = lim n →∞ 9 3 n 5 n = lim n →∞ 9 ± 3 5 ² n = 9 · 0 = 0 with | r | = 3 5 < 1; Converges. (c) lim n →∞ e 1 n = e 0 = 1; Converges. (d) lim n →∞ (ln n ) 2 n = lim n →∞ 2 1 n ln n 1 = 2 lim n →∞ ln n n = 2 lim n →∞ 1 n 1 = 0 where l’Hˆ opital’s rule was used in the ﬁrst and third equalities; Converges. (e) lim n →∞ n 3 n + 1 = lim n →∞ n 2 1 + 1 n = ; Diverges. (f) lim n →∞ n 3 n 3 + 1 = lim n →∞ 1 1 + 1 n 3 = 1 1 + 0 = 1; Converges. (g) lim n →∞ ( - 1) n n 3 n 3 + 2 n 2 + 1 = lim n →∞ ( - 1) n 1 + 2 n + 1 n 3 ; Diverges. (h) lim n →∞ cos ± 2 n ² = cos(0) = 1; Converges. (i) lim n →∞ (ln(7 n + 1) - ln n ) = lim n →∞ ln 7 n + 1 n = lim n →∞ ln ±
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Unformatted text preview: 7 + 1 n ² = ln 7; Converges. (j) Since 0 ≤ cos 2 n ≤ 1 for all n ≥ 1, we have 0 ≤ cos 2 n n ≤ 1 n for all n ≥ 1. Since lim n →∞ 0 = 0 = lim n →∞ 1 n , we have lim n →∞ cos 2 n n = 0 by the Squeeze Theorem. (k) a n = √ n 2 + 2 n-n = ( √ n 2 +2 n-n )( √ n 2 +2 n + n ) √ n 2 +2 n + n = 2 n √ n 2 +2 n + n = 2 √ 1+ 2 n +1 . So, lim n →∞ a n = lim n →∞ 2 √ 1+ 2 n +1 = 2 √ 1+0+1 = 1; Converges. (l) Diverges because r =-2 ≤ -1. ....
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