How to prove that $\sin\left(\frac{\pi}{n}\right)$ is decreasing, $\forall n\in\mathbb{N}$ (series test) 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?

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