Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - 11.8 Power Series - 11.8 Exercises - Page 792: 40

Answer

by ratio test the series is divergent for $x\ne a$ which gives us that the radius of the power series is $0=c$

Work Step by Step

If $c=0$ we have that $\lim\limits_{n \to \infty}|\frac{c_{n+1}}{c_{n}}|=+\infty$ Hence, we have that $\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}(|\frac{c_{n+1}}{c_{n}}|x-a|)=0$ if $x=a$ and $\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}(|\frac{c_{n+1}}{c_{n}}|x-a|)=+\infty$ if $x\ne a$. Thus , we have by ratio test the series is divergent for $x\ne a$ which gives us that the radius of the power series is $0=c$
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