## Calculus 10th Edition

Published by Brooks Cole

# Chapter 9 - Infinite Series - 9.2 Exercises - Page 602: 63

#### Answer

$0 \lt x \lt 2$ $f(x)=\displaystyle \frac{x-1}{2-x},\ \quad 0 \lt x \lt 2$

#### 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 $\displaystyle \sum_{n=0}^{\infty}ar^{n}=\frac{a}{1-r},\quad 0 \lt |r| \lt 1$. ------------ Adjusting for the subscript, $\displaystyle \sum_{n=1}^{\infty}(x-1)^{n}=(x-1)^{1}+(x-1)^{2}+(x-1)^{3}+...$ $=(x-1)[1+(x-1)^{1}+(x-1)^{2}+...]$ $=(x-1)\displaystyle \sum_{n=0}^{\infty}(x-1)^{n}$ The series converges for $|x-1| \lt 1 \Rightarrow$ $-1 \lt x-1 \lt 1\qquad/+1$ $0 \lt x \lt 2$ $f(x)=(x-1)\displaystyle \sum_{n=0}^{\infty}(x-1)^{n}$ $=(x-1)\displaystyle \frac{1}{1-(x-1)}=\frac{x-1}{2-x},\ 0 \lt x \lt 2$

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