## Calculus (3rd Edition)

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.