Calculus 8th Edition

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

Chapter 11 - Infinite Sequences and Series - 11.10 Taylor and Maclaurin Series - 11.10 Exercises - Page 811: 22


$-\Sigma_{n=0}^{\infty}\frac{(x+3)^{n}}{3^{n+1}}$, $R=3$.

Work Step by Step

$\frac{1}{x}=-\Sigma_{n=0}^{\infty}\frac{(x+3)^{n}}{3^{n+1}}$ $\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{\frac{(x+3)^{n+1}}{3^{n+2}}}{\frac{(x+3)^{n}}{3^{n+1}}}|$ $=\lim\limits_{n \to\infty}|\frac{x+3}{3}|\lt 1$ $|x+3|\lt 3$ $-3\lt x+3 \lt 3$ $-6 \lt x\lt 0$ The radius of convergence is always half of the width of the interval, so $R=3$.
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