Calculus (3rd Edition)

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

Chapter 11 - Infinite Series - 11.4 Absolute and Conditional Convergence - Exercises - Page 563: 6


The series $\sum_{n=1}^{\infty} \frac{\sin(\pi n/4)}{n^2}$ converges absolutely.

Work Step by Step

Since $\sin(\pi n/4)\leq 1$, then $$\frac{\sin(\pi n/4)}{n^2}\leq \frac{1}{n^2}.$$ Also, the series $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges as it is a p-series with $p=2\gt 1$. Then the series $\sum_{n=1}^{\infty} \frac{\sin(\pi n/4)}{n^2}$ converges by the comparison test. Now, the postitive series $\sum_{n=1}^{\infty} |\frac{\sin(\pi n/4)}{n^2}|$ converges and hence the series $\sum_{n=1}^{\infty} \frac{\sin(\pi n/4)}{n^2}$ converges absolutely.
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