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Unformatted text preview: MAC 2312 Quiz 6 February 28, 2012 SOLUTIONS 1. Determine the nonzero values of x for which the series n =1 x 5 x 2 n 1 converges. Solution: First, observe that the numerator of the summand is independent of n , so summationdisplay n =1 x 5 x 2 n 1 = x 5 summationdisplay n =1 1 x 2 n 1 . Therefore, we only need to find the values of x for which the sum on the righthand side converges. To achieve this, consider two cases separately. Case 1:  x  < 1 In this case, lim n x 2 n = 0, so the limit of the summand is lim n 1 x 2 n 1 = 1 negationslash = 0 . Hence, by the test for divergence of series, n =1 1 x 2 n 1 is divergent. Case 2:  x  > 1 In this case, lim n x 2 n = + , so lim n 1 x 2 n 1 x 2 n 1 = lim n x 2 n 1 x 2 n = 1 . Therefore, by the limit comparison test, n =1 1 x 2 n 1 converges if and only if n =1 1 x 2 n converges. But since  x  > 1, 1 x 2 < 1. Therefore in view of the geometric series, summationdisplay n =1 1 x 2 n = summationdisplay n =1 parenleftbigg 1 x 2 parenrightbigg n < , so n 1 1 x 2 n 1 converges for  x  > 1. 2. Determine the positive values of k for which the series n =2 n n n k converges. Solution: Here is the intuition behind finding the solution (this is not a solution; it is just a way to come up with a candidate for a solution). If k > 1, then as n , n k dominates n , so the summand is approximately n n n k n n k = 1 n k 1 / 2 . Therefore, by the pseries test, we suspect that the given sum will converge if and only if k > 3 / 2 (because this makes k 1 2 > 1). With a candidate for a solution in hand (i.e. k > 3 2 ), we are ready to check whether our proposed solution is an actual solution. This amounts to performing two comour proposed solution is an actual solution....
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 Spring '08
 Bonner
 Calculus

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