Interesting trigonometric sum: why is $\sum_{k\ge 1} (-1)^k \frac{\sin(k)^3}{k} = 0?$

Question

Here's an esasy one for Christmas:

The nice problem Calculating $\sum\limits_{n=1}^{\infty}\frac{(-1)^n \sin(n)}{n}$ led me to consider the sums

$$s_m = \sum_{k\ge 1} (-1)^k \frac{\sin(k)^m}{k}$$

for $m=1,2,...$

It turns out that $s_{3} = 0$.

I wonder what might be the shortest proof to make this result obvious.

Answer 1

We have $\sin^3(x)=\frac{1}{4}\left[3\sin(x)-\sin(3x)\right]$ and $$\frac{(1+e^i)^3}{(1+e^{3i})}=\frac{(1+e^i)^2}{1-e^{i}+e^{2i}}=\frac{2\cos 1+2}{2\cos 1-1}\in\mathbb{R}^+,$$ hence $\sum_{m\geq 1}\frac{(-1)^m \sin^3(m)}{m}=0$.

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$\sum_{m\geq 1}\frac{(-1)^m \sin(m)^{2t+1}}{m}$ can be computed through the same principle. For any $k\in\mathbb{N}$, be denoting through $[k]$ the element of $(-\pi,\pi)$ fulfilling $k-[k]\in 2\pi\mathbb{Z}$, we have $\sum_{m\geq 1}\frac{(-1)^m \sin(km)}{m}=-\frac{[k]}{2}$, and $\sin(m)^{2t+1}$ can always be expanded as a Fourier sine series.

Answer 2

This is not the sought most simple proof but let us write down the explicit expression for the general $s_m$ and study some of its properties.

Letting $\sin(k) = \frac{1}{2 i} \left(e^{i k}-e^{-i k}\right)$ we find

$$\boxed{s_m=\frac{-1}{(2 i)^m}\sum _{j=0}^m (-1)^{m-j} \binom{m}{j} \log \left(1+e^{i (2 j-m)}\right)}\tag{1}$$

The expressions are simple for odd $m$ (as was noticed by Claude Leibovici) but more complicated for even $m$.

The first few expressions in the format $\{m, s_m\}$ are

$$\begin{array}{l} \left\{2,\frac{1}{2} \log (\cos (1))\right\} \\ \left\{4,\frac{1}{16} \left(\log \left(\sec ^2(2)\right)+8 \log (\cos (1))\right)\right\} \\ \left\{6,\frac{1}{64} \left(-6 \log \left(\cos ^2(2)\right)+\log \left(\cos ^2(3)\right)+30 \log (\cos (1))\right)\right\} \\ \end{array}\tag{2a}$$

$$\left(\begin{array}{cc} 1 & -\frac{1}{2} \\ 3 & 0 \\ 5 & \frac{\pi }{16} \\ 7 & \frac{3 \pi }{32} \\ 9 & \frac{7 \pi }{64} \\ 11 & \frac{119 \pi }{1024} \\ 13 & \frac{483 \pi }{4096} \\ 15 & \frac{1911 \pi }{16384} \\ 17 & \frac{7449 \pi }{65536} \\ 19 & \frac{14391 \pi }{131072} \\ \end{array} \right) \tag{2b}$$

Letting for odd $m$

$$s_{2k-1} =\pi \frac{a_k}{2^{b_k}}\tag{3}$$

the two series appearing here start as follows

$$a_k=\{-1,0,1,3,7,119,483,1911,7449,14391,1729,212173,812877,1557335,373255,22930815,88182135,1358725815,5242980555,20266509315\}$$

$$b_k=\{1,0,4,5,6,10,12,14,16,17,14,21,23,24,22,28,30,34,36,38\}$$

Both do not seem to obey simple rules. Also they are not (yet) contained in https://oeis.org/