Calculus 8th Edition

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

Chapter 11 - Infinite Sequences and Series - 11.9 Representations of Functions as Power Series - 11.9 Exercises - Page 797: 5

Answer

Interval of convergence is $(-3,3)$ Radius of convergence is $3$

Work Step by Step

Sum of geometric series with inityal term $a$ and common ratio $r$ is $S=\Sigma_{n=0}^{\infty}ar^{n}=\frac{a}{1-r}$ Therefore, $f(x)=\Sigma_{n=0}^{\infty}ar^{n}=\Sigma_{n=0}^{\infty}(\frac{2}{3})(\frac{x}{3})^{n}=\Sigma_{n=0}^{\infty}(\frac{2}{3^{n+1}}){x}^{n}$ This is the power series representation of $f(x)$. we know that the power series converges when $r=|\frac{x}{3}|\lt 1$ $|x|\lt 3$ Interval of convergence is $(-3,3)$ Radius of convergence is $3$
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