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Unformatted text preview: 2005 MA 1A HOMEWORK 8 SOLUTIONS (1) Suppose a n > a n +1 ≥ 0 for all n . Define f ( x ) = ∞ X n =0 ( 1) n a n x n . (a) Prove that this series is absolutely convergent for  x  < 1. Proof. ∞ X n =0  ( 1) n a n x n  = ∞ X n =0 a n  x n  = ∞ X n =0 a n  x  n . Since a n ≤ a for all n , we get that a n  x  n ≤ a  x  n for all n . Therefore we may apply the comparison test, comparing ∑ ∞ n =0 a n  x  n to ∑ ∞ n =0 a  x  n . But the latter series is just a geometric series, with ratio  x  . Since  x  < 1, this series is convergent. Thus by the comparison test, we get that ∑ ∞ n =0 a n  x  n is convergent, so ∞ X n =0 ( 1) n a n x n is absolutely convergent. (b) Suppose from now on that a n → 0 as n → ∞ . This means that ∑ ∞ n =0 ( 1) n a n is convergent. Define L = ∑ ∞ n =0 ( 1) n a n . Let f k ( x ) be the partial sum used to define f ( x ) where n runs from 0 to k . Prove that, for 0 < x < 1, f 2 m ( x ) ≥ f ( x ) ≥ f 2 m +1 . Proof. Since the series is absolutely convergent, we get f ( x ) = ∞ X n =0 ( 1) n a n x n = 2 m X n =0 ( 1) n a n x n + ∞ X n =2 m +1 ( 1) n a n x n = 2 m X n =0 ( 1) n a n x n + ∞ X j = m +1 ( a 2 j 1 x 2 j 1 + a 2 j x 2 j ) = 2 m X n =0 ( 1) n a n x n ∞ X j = m +1 ( a 2 j 1 x 2 j 1 a 2 j x 2 j ) ≤ 2 m X n =0 ( 1) n a n x n...
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This note was uploaded on 03/30/2010 for the course MA 1a taught by Professor Borodin,a during the Fall '08 term at Caltech.
 Fall '08
 Borodin,A

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