Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Section 10.7 - Power Series - Exercises 10.7 - Page 613: 51


$2 \lt x \lt 8$

Work Step by Step

We notice that $g(x)=\dfrac{3}{x-2}=\dfrac{3}{3-[-(x-5)]}$ This implies that $(x) =\Sigma_{n=0}^{\infty} (\dfrac{-1}{3})^n (x-5)^n$ Remember that the series will converge when $|g(x)|=|\dfrac{3}{3-[-(x-5)]}| \lt 1$ This implies that $|\dfrac{x-5}{3 }| \lt 1$ or, $|x+5| \lt 3$ or, $-3 \lt x-5 \lt 5$ Thus, $2 \lt x \lt 8$
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