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

Converges

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.