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: 77


Converges to $\dfrac{1}{1-\sin x}$ for $x\ne (2k+1)\pi/2$ (where k is an integer)

Work Step by Step

The sum of a geometric series can be found as: $S=\dfrac{a}{1-r}$ We have a convergent geometric series with first term, $a=1$ and common ratio $r =\sin x$ $S=\dfrac{1}{1-\sin x}$ Hence, the given series converges to $\dfrac{1}{1-\sin x}$ for $x\ne (2k+1)\pi/2$ (where k is an integer)
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.