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: 64

Answer

$-1 \lt x \lt 5$ $ f(x)=\displaystyle \frac{15}{5-x},\quad -1 \lt x \lt 5$

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 $\displaystyle \sum_{n=0}^{\infty}ar^{n}=\frac{a}{1-r},\quad 0 \lt |r| \lt 1$. ------------ The series $\displaystyle \sum_{n=0}^{\infty}5(\frac{x-2}{3})^{n} $ $a=5, r=\dfrac{x-2}{3} $, converges for $|r| \lt 1$ $|\displaystyle \frac{x-2}{3}| \lt 1$ $|x-2| \lt 3$ $-3 \lt x-2 \lt 3\quad/+2$ $-1 \lt x \lt 5$ $f(x)=\displaystyle \sum_{n=0}^{\infty}5(\frac{x-2}{3})^{n} =\frac{5}{1-(\frac{x-2}{3})}$ $=\displaystyle \frac{5}{\frac{3-x+2}{3}} =\frac{15}{5-x},\quad -1 \lt x \lt 5$
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