Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - Review - Exercises - Page 825: 27



Work Step by Step

$\Sigma_{n=1}^{\infty}\frac{(-3)^{n-1}}{2^{3n}}=\Sigma_{n=1}^{\infty}\frac{(-3)^{n-1}}{8^{n}}$ $=\Sigma_{n=1}^{\infty}\frac{1}{8}(\frac{-3}{8})^{n-1}$ It appears to be geometric series $\Sigma_{n=1}^{\infty}ar^{n-1}=\frac{a}{1-r}$ where $a$ is common ratio. Here, $a=\frac{1}{8}$ and $r=-\frac{3}{8}$ Thus, $\Sigma_{n=1}^{\infty}\frac{(-3)^{n-1}}{2^{3n}}=\frac{a}{1-r}$ $=\frac{1/8}{1-(-3/8)}$ $=\frac{1}{11}$
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