Sum of $\sum_{n=1}^{\infty }\frac{1}{\pi n }\sin ^k\left(\frac{2\pi n}{k}\right)$ We have: $$S_k=\sum_{n=1}^{\infty }\frac{1}{\pi n }\sin ^k\left(\frac{2\pi n}{k}\right)$$
where $k$ is an odd number greater than $1$.
I was able to find the sum of the series when $k=3,5$ as follows:
$$S_3={\sum_{n=1}^{\infty }\frac{1}{n\pi }\sin ^3\left(\frac{2\pi n}{3}\right)}=\frac{1}{8}$$
$$S_5={\sum_{n=1}^{\infty }\frac{1}{n\pi }\sin ^5\left(\frac{2\pi n}{5}\right )}=\frac{7}{32}$$
Can I get the closed-form of the other values of $S_k$ when $k=7,9,11,13,....$?
 A: We can treat this problem with Fourier series. For instance, since:
$$ \sin^3 x = \frac{1}{4}\left(3 \sin x - \sin(3x)\right) $$
we have:
$$\begin{eqnarray*}S_3 &=& \sum_{n\geq 1}\frac{\sin^3(2\pi n/3)}{\pi n} = \frac{1}{4}\left(3\sum_{n\geq 1}\frac{\sin(2\pi n/3)}{\pi n}-\sum_{n\geq 1}\frac{\sin(2\pi n)}{\pi n}\right)\\&=&\frac{3}{4\pi}\sum_{n\geq 1}\frac{\sin(2\pi n/3)}{n}\tag{1}\end{eqnarray*}$$
and since for any $x\not\in \pi\mathbb{Z}$ we have:
$$ \sum_{n\geq 1}\frac{\sin(nx)}{n}=\frac{1}{2}\left(\pi-(x\operatorname{mod} 2\pi)\right)\tag{2}$$
it follows that:
$$ S_3 = \frac{3}{4\pi}\cdot\frac{\pi}{6} = \color{red}{\frac{1}{8}}.\tag{3}$$
With the same approach we can compute $S_k$: we take the Fourier sine series of $\sin^k x$, then use identity $(2)$ and collect pieces. From:
$$ \sin^5 x = \frac{1}{16}\left(10\sin x-5\sin(3x)+\sin(5x)\right)$$
we get:
$$ S_5 = \frac{1}{16\pi}\left(3\pi+\frac{\pi}{2}\right)=\color{blue}{\frac{7}{32}}.$$
In general, from:
$$ \sin^{2k+1}(x) = \frac{1}{4^k}\sum_{j=0}^{k}(-1)^j\binom{2k+1}{k-j}\sin((2j+1)x)\tag{4}$$
we get:
$$ S_{2k+1}=\frac{1}{4^k \pi}\sum_{j=0}^{k-1}(-1)^j\binom{2k+1}{k-j}\frac{1}{2}\left(\pi-\frac{2(2j+1)\pi}{2k+1}\right)$$
or:
$$ S_{2k+1}=\frac{1}{2^{2k+1}}\sum_{j=1}^{k}(-1)^{k-j}\binom{2k+1}{j}\frac{4j-2k-1}{2k+1}\tag{5}$$
that further simplifies to:

$$ \color{purple}{S_{2k+1}=\sum_{n\geq 1}\frac{\sin^{2k+1}(2\pi n/(2k+1))}{n\pi}=\frac{1}{2^{2k+1}}\left((-1)^k+\binom{2k}{k}\right)}. \tag{6}$$

