For which $p$ does $ \sum_{n=2}^{\infty} \frac{\sin(\frac{\pi}{n})}{n^p}$ converge? For which $p$ does $\sum_{n=2}^{\infty} \frac{\sin(\frac{\pi}{n})}{n^p}$ converge?
I tried to use a convergence test and all I got was that it converges for any $p>0$. I am not sure about this, could you please help?
I am using the fact that the series is absolutely convergent and testing with $\frac{1}{n^{(p+1)}}$.
 A: 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.
A: There are two steps.


*

*Using the fact that $x > \sin{(x)} > 0$ for all $x > 0$ (you can prove this inspecting the graph at desmos.com), you can prove that $\pi/n > \sin{(x/n)} > 0$ for any $n$. This is because for any $n$, $x = \pi/n > 0$.

*Therefore $\sum_{n=2}^{\infty} \frac{\sin(\frac{\pi}{n})}{n^p}$ < $\pi\sum_{n=2}^{\infty} \frac{1}{n^{p+1}}$. What do you know about $\sum_{n=2}^{\infty} \frac{1}{n^{k}}$? When does that converge?
A: Hint $:$ Comparison test. 
Observe that $|\sin x| \leq |x|\ \text{for all}\ x \in \Bbb R$.
