Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Section 11.2 - Series - 11.2 Exercises - Page 716: 62

Answer

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

Work Step by Step

$\Sigma^{\infty}_{n=0} \frac{sin^{n}x}{3^{n}} = \Sigma^{\infty}_{n=0} (\frac{sinx}{3})^{n}$ $a=1$ 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 $x$. $\Sigma^{\infty}_{n=0} \frac{sin^{n}x}{3^{n}} = \frac{a}{1-r}$ $=\frac{1}{1-\frac{sinx}{3}}$ $=\frac{3}{3-sinx}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.