Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 11 - Infinite Series - 11.5 The Ratio and Root Tests and Strategies for Choosing Tests - Exercises - Page 568: 47


The series converges.

Work Step by Step

Since $\sin n\leq 1$, then $\frac{\sin n}{n^2}\leq \frac{1}{n^2}$. Now, we use the comparison test with the series $\Sigma_{n=1}^{\infty} \frac{1}{n^2}$, which is a convergent p-series with $p=2\gt1$. Hence the series $\Sigma_{n=1}^{\infty} \frac{\sin n}{n^2}$ converges.
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