real analysis - Does $sum frac{sin^k j}{j}$ converge or diverge?

Does $\sum \frac{\sin^k j}{j}$ converge or diverge where $k$ is a positive integer? I know that for $k=1$ the series converges and for $k=2$ the series diverges. What can we say about its convergence when $k>2$? The usual proof for $k=2$ is to reduce the summand to $\frac{1-\cos\left( 2j \right)}{2j}$ using a trig identity, but it's not clear to me how this can be generalized to $k>1$.

Answer

It converges for odd values of $k$ and it diverges for even values of $k$.

Indeed, if $k$ is even then $\sin^k(n)$ has a positive mean value and the divergence is a consequence of Kronecker's lemma.

If $k$ is odd then $\sin^k(n)$ can be written as a linear combination of $\sin(n),\sin(3n),\sin(5n)$ etcetera, and
$$ \sum_{n\geq 1}\frac{\sin(nx)}{n} $$
equals the $2\pi$-periodic extension of the function which equals $\frac{\pi-x}{2}$ on $(0,2\pi)$.

Examples:

$$\sum_{n\geq 1}\frac{\sin^3(n)}{n}=\frac{3}{4}\sum_{n\geq 1}\frac{\sin(n)}{n}-\frac{1}{4}\sum_{n\geq 1}\frac{\sin(3n)}{n}=\frac{3}{4}\cdot\frac{\pi-1}{2}-\frac{1}{4}\cdot\frac{\pi-3}{2}=\color{red}{\frac{\pi}{4}}.$$
Since $\sin^4(x)=\frac{3}{8}-\frac{1}{2}\cos(2x)+\frac{1}{8}\cos(4x)$,

$$ \sum_{n=1}^{N}\frac{\sin^4(n)}{n} \geq \frac{3}{8}H_n - C.$$