Calculus 8th Edition

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

Chapter 11 - Infinite Sequences and Series - 11.2 Series - 11.2 Exercises - Page 756: 62


The given series converges for all of $x$ and the sum is $\frac{3}{3-sinx}$.

Work Step by Step

Given: $\Sigma^{\infty}_{n=0} \frac{sin^{n}x}{3^{n}}$ $\Sigma^{\infty}_{n=0} \frac{sin^{n}x}{3^{n}} = \Sigma^{\infty}_{n=0} (\frac{sinx}{3})^{n}$ Here, $a=1$ and $r=\frac{sinx}{3}$ The series converges when $|r| \lt 1$ $|\frac{sinx}{3}| \lt 1$ $-3 \lt sinx \lt 3$ Since $|sinx| \leq 1$ for all $x$, we should have $-3 \lt sinx \lt 3$ for all $x$ and so the given series converges for all $x$. Sum can be calculated as follows: $\Sigma^{\infty}_{n=0} \frac{sin^{n}x}{3^{n}} = \frac{a}{1-r}$ $=\frac{1}{1-\frac{sinx}{3}}$ $=\frac{3}{3-sinx}$
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