## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 11 - Infinite Series - 11.5 The Ratio and Root Tests and Strategies for Choosing Tests - Exercises - Page 568: 59

Converges

#### Work Step by Step

Given $$\sum_{n=1}^{\infty} \sin\frac{1}{n^2}$$ Since for $n\geq1$ $$\sin\frac{1}{n^2}\leq \frac{1}{n^2}$$ Compare with the convergent series $\sum_{n=1}^{\infty} \frac{1}{n^2}$ ($p-$series, $p>1$), then: $$\sum_{n=1}^{\infty} \sin\frac{1}{n^2}$$ also converges.

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