## Calculus (3rd Edition)

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