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: 58


convergent and sum is $-(\frac{x+2}{x+1})$.

Work Step by Step

Given: $\Sigma^{\infty}_{n=1} (x+2)^{n}$ $\Sigma^{\infty}_{n=1} (x+2)^{n}=(x+2)^{1}+(x+2)^{2}+(x+2)^{3}+...$ Here, $a=x+2$ and $r=x+2$ Also, $|r| \lt 1$ $|x+2| \lt 1$ $-1 \lt x+2 \lt 1$ Thus, $-3 \lt x \lt -1$ Therefore , the series is convergent. The sum is $S_{\infty} = \frac{x+2}{1-(x+2)}$ $=\frac{x+2}{-1-x}$ $=-(\frac{x+2}{x+1})$
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