Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 11 - Infinite Series - 11.6 Power Series - Exercises - Page 578: 41


$$\sum_{n=0}^{\infty}(-1)^{n+1} \frac{(x-4)^{n}}{3^{n+1}}$$ $ |x-4|\lt 3$

Work Step by Step

Given $$ \frac{1}{1-x}=\frac{1}{-3-(x-4)}=\frac{-\frac{1}{3}}{1+\left(\frac{x-4}{3}\right)}$$ Since $$\frac{1}{1-x}=\sum_{n=0}^{\infty} x^{n},\ \ \ \ |x|\lt 1\tag{1}$$ By using (1), we get \begin{align*} \frac{1}{1-x}&=\frac{-1}{3}\frac{1}{1+\left(\frac{x-4}{3}\right)}\\ &=\frac{-1}{3}\sum_{n=0}^{\infty}\left(-\frac{x-4}{3}\right)^{n}\\ &=\frac{-1}{3}\sum_{n=0}^{\infty}(-1)^{n} \frac{(x-4)^{n}}{3^{n}}\\ &= \sum_{n=0}^{\infty}(-1)^{n+1} \frac{(x-4)^{n}}{3^{n+1}} \end{align*} Such that $ |x-4|\lt 3$
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