For which $p$ does $ \sum_{n=2}^{\infty} \frac{\sin(\frac{\pi}{n})}{n^p}$ converge?

We have $0 < \sin(\frac{\pi}{n}) < \frac{\pi}{n}$ for all $n$. Hence $\sum \frac{\frac{\pi}{n}}{n^p} = \sum \frac{\pi}{n^{p+1}}$ is a majorant series. It is well-known that it converges if $p+1 > 1$, i.e. $p > 0$. This implies that $\sum_{n=2}^{\infty} \frac{\sin(\frac{\pi}{n})}{n^p}$ converges for $p > 0$.

What about $p \le 0$? We have $0 < \frac{2}{n} < \sin(\frac{\pi}{n}) \le \frac{\sin(\frac{\pi}{n})}{n^p}$ for $n \ge 2$, hence the harmonic series $\sum \frac{2}{n}$ is a divergent minorant.