## Calculus 8th Edition

Published by Cengage

# Chapter 11 - Infinite Sequences and Series - 11.2 Series - 11.2 Exercises - Page 756: 28

#### Answer

Convergent with sum $\frac{5}{8}$

#### Work Step by Step

$\frac{1}{3}+\frac{2}{9}+\frac{1}{27}+\frac{2}{81}+\frac{1}{243}+\frac{2}{729}+...$ $= \Sigma^{\infty}_{n=1} \frac{1}{3^{2n-1}} + 2\Sigma^{\infty}_{n=1} \frac{1}{9^{n}}$ The series $\Sigma^{\infty}_{n=1} \frac{1}{3^{2n-1}}$ is a geometric series with $a=\frac{1}{3}$ and $r=\frac{1}{9}$. $\Sigma^{\infty}_{n=1} \frac{1}{3^{2n-1}}= \frac{\frac{1}{3}}{1-\frac{1}{9}}$ $= \frac{3}{8}$ The series $\Sigma^{\infty}_{n=1} \frac{1}{9^{n}}$ is a geometric series with $a=\frac{1}{9}$ and $r=\frac{1}{9}$ $\Sigma^{\infty}_{n=1} \frac{1}{9^{n}}= \frac{\frac{1}{9}}{1-\frac{1}{9}}$ $=\frac{1}{8}$ Hence the series is convergent $\Sigma^{\infty}_{n=1} \frac{1}{3^{2n-1}} + 2\Sigma^{\infty}_{n=1} \frac{1}{9^{n}}=\frac{3}{8}+2(\frac{1}{8})$ $=\frac{5}{8}$

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