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Unformatted text preview: for just “yes” or “no”; do not refer to “behavior”). ∞ X n =1 n 2 √ n 6 + 1 I will use the Limit Comparison Test to compare n 2 √ n 6 +1 to 1 n . (The inspiration to choose 1 n was by thinking about the behavior of n 2 √ n 6 +1 as n gets large, but talk about approximating behaviors does not count as a solution for this class.) So I consider lim n →∞ 1 n n 2 √ n 6 +1 = lim n →∞ √ n 6 + 1 n 3 = p 1 + 1 /n 6 1 = 1 . By the Limit Comparison Test, since ∑ ∞ n =1 1 n diverges, so does ∑ ∞ n =1 n 2 √ n 6 +1 . 3. (2 points) Give an example of (a) an alternating series that converges ∑ ∞ n =1 (1) n n (b) an alternating series that does not converge ∑ ∞ n =1 (1) n...
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 Fall '08
 Reshetiken
 Math, Calculus, Mathematical Series

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