quiz05_s107_solns

# quiz05_s107_solns - for just “yes” or “no” do not...

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Math 1B Quiz 5 Section 107 October 13, 2009 1. (5 points) Say whether the following series converge and clearly explain why. (A “yes” or “no” without explanation is worth 0 points, even if correct.) (a) X n =2 1 n 5 / 4 ln n Since 1 n 5 / 4 ln n < 1 n 5 / 4 for all n 2, n =2 1 n 5 / 4 ln n converges if n =2 1 n 5 / 4 does. But n =2 1 n 5 / 4 is a p -series with p > 1, so it converges. Therefore n =2 1 n 5 / 4 ln n converges. (b) X n =2 1 n ln n For n 2, x and ln x are positive and increasing, so 1 x ln x is positive, decreasing, and continuous. Thus by the integral comparison test, n =2 1 n ln n converges if and only if R 2 1 x ln x dx does. But letting u = ln x , we have R 2 1 x ln x dx = R ln 2 1 u du , which diverges. Therefore, n =2 1 n ln n diverges. (c) X n =2 1 n 4 / 5 ln n Because n 4 / 5 < n for all n 2, we have that 1 n 4 / 5 ln n > 1 n ln n for all n 2. By the comparison test, since n =2 1 n ln n diverges, n =2 1 n 4 / 5 ln n diverges also.

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2. (3 points) Does the following series converge? Clearly explain your answer (no points
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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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quiz05_s107_solns - for just “yes” or “no” do not...

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