For which positive integer $k$ does the series $\sum_{n=1}^\infty\frac{\sin(n\pi/k)}{n}$ converge?

Question

For which positive integer $k$ does the series $$\sum_{n=1}^\infty\frac{\sin(n\pi/k)}{n}$$ converge?

The cases when $k=1$ or $k=2$ are trivial. For $k>2$, I don't see how to approach it. If one looks at the partial sums: $$ s_k=\frac{\sin(\frac{\pi}{k})}{1}+\frac{\sin(2\cdot\frac{\pi}{k})}{2}+\cdots+\frac{\sin((k-1)\cdot\frac{\pi}{k})}{k-1} $$ $$ s_{2k}-s_k= (-1)\frac{\sin(\frac{\pi}{k})}{k+1}+(-1)\frac{\sin(2\cdot\frac{\pi}{k})}{k+2}+\cdots+(-1)\frac{\sin((k-1)\cdot\frac{\pi}{k})}{k+k-1} $$ $$ s_{3k}-s_{2k}=\frac{\sin(\frac{\pi}{k})}{2k+1}+\frac{\sin(2\cdot\frac{\pi}{k})}{2k+2}+\cdots+\frac{\sin((k-1)\cdot\frac{\pi}{k})}{2k+k-1}\\ \vdots $$ one might tend to sum the terms vertically so that the alternating test can be used. But unfortunately this series does not converge absolutely.

Answer 1

Inspired by Salahaman_Fatima's comment.

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Apply the Dirichlet test to see that:

$$\begin{cases}a_{n+1}\le a_n\forall n\in\mathbb N\\\lim_{n\to\infty}a_n=0\\\left|\sum_{n=1}^mb_n\right|<M\in\mathbb R^+\forall m\in\mathbb N\end{cases}~\implies\left|\sum_{n=1}^\infty a_nb_n\right|<M\in\mathbb R^+$$

It is trivial to see that

$$\frac1{n+1}\le\frac1n\\\lim_{n\to\infty}\frac1n=0$$

So all we need to do is show that

$$\left|\sum_{n=1}^m\sin(n\pi/k)\right|<M\in\mathbb R^+\forall m\in\mathbb N$$

Here, the sum may be evaluated directly, yielding

$$\sum_{n=1}^{m-1}\sin(n\pi/k)=\frac{\sin(m\pi/2k)}{\sin(\pi/2k)}\cos((m-1)\pi/2k),\quad\frac1{2k}\notin\mathbb Z$$

And under absolute value bars:

$$\left|\sum_{n=1}^m\sin(n\pi/k)\right|<|\csc(\pi/2k)|\in\mathbb R^+\forall m\in\mathbb N$$

The other cases may each be handled individually (and notably, they are very simple.)

Answer 2

$$\sum_{n=1}^\infty\frac{\sin (n\pi/k)}{n} = \sum_{n=1}^\infty\frac{\pi}{n\pi}\sin2\pi \frac{n}{2}\frac1{k}$$

The right hand side is a sawtooth wave of period $2$, at $t = 1/k$:

$$f(t) = \frac{\pi}{2}-\frac{\pi}2t, \quad 0<t<2\\ f(t+2) = f(t)\\ f(2n) = 0, \quad n\in\mathbb Z$$

So specifically,

$$\sum_{n=1}^\infty\frac{\sin (n\pi/k)}{n} = f\left(\frac1k\right) = \frac{\pi}{2} - \frac{\pi}{2k}$$

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The Fourier series coefficients of $f(t)$ is given by

$$\begin{align*} b_n &= \frac22\int_{0}^2\frac{\pi}{2}(1-t)\sin\frac{2\pi n t}{2}dt\\ &= \frac\pi2\int_0^2\sin\frac{2\pi n t}{2}dt - \frac\pi2\int_0^2t\sin\frac{2\pi n t}2dt\\ &= 0 + \frac1{2n}\int_0^2 t\ d\cos\frac{2\pi n t}{2}\\ &= \frac1{2n}\left[t\cos\frac{2\pi n t}2\right]_0^2 + \int_0^2\cos\frac{2\pi nt}2 dt\\ &= \left(\frac1{2n}\cdot2\cos2\pi n-0 \right) + 0\\ &= \frac1n \end{align*}$$