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 - Page 602: 61


$|x| \displaystyle \lt \frac{1}{3}$ $f(x)=\displaystyle \frac{3x}{1-3x}, \displaystyle \quad|x| \lt \frac{1}{3}$

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, $\displaystyle \sum_{n=1}^{\infty}(3x)^{n}=(3x)^{1}+(3x)^{2}+(3x)^{3}+...$ $=3x[1+(3x)^{1}+(3x)^{2}+(3x)^{3}+...]$ $=(3x)\displaystyle \sum_{n=0}^{\infty}(3x)^{n}$ The series is geometric, $r=3x$, and it converges for $|3x| \lt 1$ $\Rightarrow$ $|x| \displaystyle \lt \frac{1}{3}$ $f(x)=(3x)\displaystyle \sum_{n=0}^{\infty}(3x)^{n}=(3x)\frac{1}{1-3x}=$ $\displaystyle=\frac{3x}{1-3x}, \displaystyle \quad|x| \lt \frac{1}{3}$
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