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 $$\frac{1}{n^{(p+1)}}$$.

• Try by Abel’s or Dirichlet test for uniform convergence for sereis of functions . – neelkanth Feb 8 at 17:39

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.

There are two steps.

1. 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 $$n$$, $$x = \pi/n > 0$$.

2. 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?

Hint $$:$$ Comparison test.

Observe that $$|\sin x| \leq |x|\ \text{for all}\ x \in \Bbb R$$.