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: 37


$$\sum_{n=0}^{\infty}\frac{x^n}{3^{n+1}}$$ $|x|<3$

Work Step by Step

Given $$ f(x)=\frac{1}{3- x}$$ Since $$\frac{1}{1-x}=\sum_{n=0}^{\infty} x^{n},\ \ \ \ |x|<1\tag{1}$$ By using (1), we get: \begin{align*} \frac{1}{3- x}&=\frac{1}{3(1-(x/3))}\\ &=\frac{1/3}{(1-(x/3))}\\ &=\frac{1}{3} \sum_{n=0}^{\infty}( x/3)^{n}\\ &= \frac{1}{3} \sum_{n=0}^{\infty}\frac{x^n}{3^n}\\ &= \sum_{n=0}^{\infty}\frac{x^n}{3^{n+1}}\\ \end{align*} Such that $ |x/3|<1$, or $|x|<3$
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