Calculus 8th Edition

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

Chapter 11 - Infinite Sequences and Series - 11.2 Series - 11.2 Exercises - Page 756: 59


The series converges if $-1 \lt x \lt 5$ and the sum is $\frac{3}{5-x}$

Work Step by Step

Given: $\Sigma^{\infty}_{n=0} \frac{(x-2)^{n}}{3^{n}}$ The given series can be re-written as $\Sigma^{\infty}_{n=0} (\frac{x-2}{3})^{n}$ Here, $a=1$ and $r=(\frac{x-2}{3})$ Also, $|r| \lt 1$ $|\frac{x-2}{3}| \lt 1$ $|x-2| \lt 3$ $-3 \lt x-2 \lt 3$ $-1 \lt x \lt 5$ Therefore, the series is convergent. The sum can be calculated as follows: $\Sigma^{\infty}_{n=1} \frac{(x-2)^{n}}{3^{n}} = \frac{a}{1-r}$ $=\frac{1}{1-(\frac{x-2}{3})}$ $=\frac{1}{(\frac{3-x+2}{3})}$ $=\frac{3}{5-x}$
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