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Unformatted text preview: Homework 2 Key Answers 140B 1. ( 3 points each ) For each of the following power series, find the radius of convergence and the interval of convergence: 1) X nx n , 2) X 1 n n x n , 3) X 3 n n x 2 n +1 . Solution: (a) For the first series, let us compute: = lim sup | a n | 1 /n = lim sup( n ) 1 /n = lim ( n ) 1 /n = 1 , since the limit exists (and n 1 /n = e ln( n ) /n = e ln( n ) / (2 n ) which goes to 1 since lim ln( n ) n = 0). The radius of convergence is then R = 1 = 1. We need now to check the converge at the endpoints x = 1. It is easy to see that: X n, X (- 1) n n, are both divergent (the first one obviously goes to infinity and, for the second one, you can prove that lim (- 1) n n does not exist). The intergral of convergence is then (- 1 , 1). Remark: To compute , we could have used the ratio test: lim a n +1 a n = lim n + 1 n . The limit exists (and is equal to 1). This implies that = lim a n +1 a n (Please, keep in mind that this result is true because the limit of a n +1 a n exists!). (b) Let us compute: = lim sup | a n | 1 /n = lim sup 1 ( n ) n/n = lim 1 ( n ) 1 / n = 1 . Once again, to prove this result, we can use: n 1 / n = e ln( n ) / n , which goes to zero as n . The radius of convergence is R = 1. At the endpoints x = 1, we have: X (- 1) n 1 n n , X 1 n n . 1 The first series is convergent (it is an alternating series and the terms a n are decreas- ing. By the Alternating Series Theorem, the series is convergent). So x =...
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- Spring '09
- Power Series