## Calculus (3rd Edition)

Given $$\sum_{n=1}^{\infty}\sin \frac{1}{n^{2}}$$ Since for $n\geq1$ \begin{aligned} \sin \frac{1}{n^{2}} \leq \frac{1}{n^{2}} \end{aligned} Compare with $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{2}}$, a convergent $p-$series $p=2$; then $\displaystyle\sum_{n=1}^{\infty}\sin \frac{1}{n^{2}}$ also converges.