The following is an interesting problem presented to this site which has yet to bet solved: Does
$$\sum_{n=1}^\infty \frac{\sin(n!)}{n}$$
converge. While attempting this problem, I thought that proving the equidistribution of $n!$ modulo $2\pi$ would be sufficient for the original conjecture. For those who do not know, an sequence $a_n$ is said to be equidistributed on a non-degenerate interval $[a,b]$ if
$$\lim_{n\to \infty}\frac{|\{a_1,a_2,\cdots,a_n\}\cap [c,d]|}{n}=\frac{d-c}{b-a}$$
for all subintervals $[c,d]\subseteq [a,b]$. My thoughts then turned to the more general question: If $a_n$ is any sequence of real numbers such that $\mod(a_n,2\pi)$ is equidistributed over $[0,2\pi]$, does
$$\sum_{n=1}^\infty \frac{\sin(a_n)}{n^\beta}$$
necessarily converge for $\beta>0$. Obviously, if $\beta>1$ then the series converges absolutely, so the interesting cases are $0<\beta<1$ and $\beta=1$ (although they might be the same case overall). One possible way forward is using Weyl's criterion: we know that if $a_n$ is equidistributed over $[0,2\pi]$, then
$$\lim_{n\to\infty} \sum_{j=1}^n\frac{\sin(q a_j)}{n}=0$$
for all $q\in\mathbb{N}$. I'm not sure how this could be useful but it seems pretty close to the original sum. One result in favor of this conjecture is discussed on this mathoverflow post. That is, if $p(n)$ is any polynomial with rational coefficients, then
$$\sum_{n=1}^\infty \frac{\sin(p(n))}{n}$$
converges.