6
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ Try by Abel’s or Dirichlet test for uniform convergence for sereis of functions . $\endgroup$ – neelkanth Feb 8 at 17:39
3
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

Hint $:$ Comparison test.

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.