How to prove that $\sin\left(\dfrac{\pi}{n}\right)$ is decreasing, $\forall n\in\mathbb{N}$.

My original question was to determine the convergence of $\sum (-1)^{n+1}\sin\left(\dfrac{\pi}{n}\right).$

I showed that the absolute value does not converge, so it does not converge absolutely. I now need to check for conditional convergence.

I want to solve the series using the alternating series test.

I already showed that $b_n\to\ 0$, as $n\to\infty$.

Now I need to show $b_n$ decreasing. I found the derivative, which is $-\dfrac{\pi \cos \left(\frac{\pi }{n}\right)}{n^2}.$

The problem is that $\cos$ is sometimes negative, and I have a negative sign in front of the derivative, which means that the derivative is sometimes positive. So it is not decreasing for all $n\in\mathbb{N}$, but the answer says converges conditionally? How?

  • 1
    $\begingroup$ you can use the fact that $cos$ is positive in $[0, \frac \pi 2]$, and therefore for any $n \ge 2$ the derivative is indeed negative and the sequence is decreasing $\endgroup$ – AsafHaas Apr 15 '17 at 20:36
  • $\begingroup$ You have the issue that $\sin\left(\frac\pi{1}\right)=0 \not \gt 1 = \sin\left(\frac\pi{2}\right)$ $\endgroup$ – Henry Apr 15 '17 at 20:43
  • $\begingroup$ @Henry That's not an issue though. Finitely many exceptional terms do not matter for convergence. $\endgroup$ – Hagen von Eitzen Apr 15 '17 at 20:55
  • $\begingroup$ @HagenvonEitzen. You are correct: it does not matter for convergence. But it does mean that "$\sin\left(\dfrac{\pi}{n}\right)$ is decreasing, $\forall n\in\mathbb{N}$" is in fact false $\endgroup$ – Henry Apr 15 '17 at 20:57

One may observe that $$ x \mapsto \sin x \quad \text{is increasing over} \quad \left[0,\frac \pi2\right] $$ and that $$ x \mapsto \frac \pi x \quad \text{is decreasing over} \quad \left[1,\infty\right) $$ giving that $ \sin \circ \:\frac \pi x$ is decreasing over $\left(2,\infty\right)$.

  • $\begingroup$ What about the case $x=1$. $\endgroup$ – hamam_Abdallah Apr 15 '17 at 20:50
  • $\begingroup$ When $x=1$ we are positive, but I get what the question is answering $\endgroup$ – K Split X Apr 15 '17 at 21:08
  • $\begingroup$ $x>2$, we are decreasing* $\endgroup$ – K Split X Apr 15 '17 at 21:09

We have

$$(\forall n\geq 2)\; \;\; \; \frac {\pi}{n}\in (0,\frac {\pi}{2}] $$

on the other hand,

$f:x\mapsto \sin (x) $ is increasing at $(0,\frac{\pi}{2}] $.


$g:n\mapsto \frac{\pi}{n}$ is decreasing from $n=2$.


$$n\mapsto f(g (n))=\sin(\frac {\pi}{n})$$ is decreasing from $n=2$.

  • 1
    $\begingroup$ @K Split X, For proving convergence, all one needs is to prove that $\sin(\pi/n)$ is decreasing for $n >N$ for some $N$. Thus, it's incorrect to rephrase the problem to require that $\sin(\pi/n)$ be decreasing for all $n > 0$. $\endgroup$ – Hoc Ngo Apr 15 '17 at 22:18
  • $\begingroup$ But $x\mapsto \sin (x) $ is not increasing over $(0,\pi] $ so, you need $n>1$. $\endgroup$ – hamam_Abdallah Apr 16 '17 at 11:29
  • $\begingroup$ What I meant was that K Split X stated wrongly: How to prove that $\sin\left(\frac{\pi}{n}\right)$ is decreasing, $\forall n\in\mathbb{N}$, which means to prove for $n = 1,2,3,\ldots$. As you have shown, $n>1$ is what you need. In general, you don't need to pick $N=1$. You can pick $N=10^5$. $\endgroup$ – Hoc Ngo Apr 16 '17 at 12:06

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