{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Homework_Set_8_Solution

Homework_Set_8_Solution - 2005 MA 1A HOMEWORK 8 SOLUTIONS(1...

This preview shows pages 1–2. Sign up to view the full content.

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 0 for all n , we get that a n | x | n a 0 | x | n for all n . Therefore we may apply the comparison test, comparing n =0 a n | x | n to n =0 a 0 | 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 = f 2 m ( x ), because a 2 j - 1 x 2 j - 1 a 2 j x 2 j for all j .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}