University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.2 - Infinite Series - Exercises - Page 498: 72


Converges to $\dfrac{3+\sin x}{8 + 2\sin x}$ for all $x$

Work Step by Step

The sum of a geometric series can be found as: $S=\dfrac{a}{1-r}$ The given series shows a convergent geometric series with first term, $a=\dfrac{1}{2}$ and common ratio $r =\dfrac{-1}{ 3+\sin x}$ $S=\dfrac{\dfrac{1}{2}}{1-(\dfrac{-1}{ 3+\sin x})}=\dfrac{3+\sin x}{2(4+\sin x)}=\dfrac{3+\sin x}{8 + 2\sin x}$ Hence, the given series converges to $\dfrac{3+\sin x}{8 + 2\sin x}$ for all $x$
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