Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 9 - Infinite Series - 9.2 Exercises: 65

Answer

$-1 \lt x \lt 1$ $f(x)=\displaystyle \frac{1}{1+x},\ \quad -1 \lt x \lt 1$

Work Step by Step

We aim for terms of Th.9.6 (Convergence of a Geometric Series). A geometric series with ratio $r$ diverges when $|r| \geq 1$. If $0 \lt |r| \lt 1$, then the series converges to the sum $\displaystyle \sum_{n=0}^{\infty}ar^{n}=\frac{a}{1-r},\quad 0 \lt |r| \lt 1$. ------------ $\displaystyle \sum_{n=0}^{\infty}(-1)^{n}x^{n}=\sum_{n=0}^{\infty}(-x)^{n}$ Geometric series, $a=1, r=-x$, converges for $|-x| \lt 1$ $|x| \lt 1$ $-1 \lt x \lt 1$ $f(x)=\displaystyle \sum_{n=0}^{\infty}(-x)^{n}=\frac{1}{1+x},\ -1 \lt x \lt 1$
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