This is a partial result that I obtained after combining the methods in the following two questions:
Convergence of $\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{\sin(n-m)}{n^2+m^2}$
Does the series $\sum_{n=1}^{\infty}\frac{\sin(\cos(n))}{n}$ converge?
For fixed $n\in\mathbb{N}$, we consider
$$
S_{ n}(N) := \sum_{m=1}^N \sin(\sin(nm)).
$$
Let $f_n (x) = \sin(\sin(nx))$. This is $2\pi$-periodic function with $\int_0^{2\pi} f_n(x) dx = 0$. Let $\mu\leq 7.6063 $ be the irrationality measure of $\pi$. Then by Koksma's inequality and a bound for the discrepancy, we have
$$
|S_{n}(N)|=O(nN^{1-\frac1{\mu-1}+\epsilon}).
$$
Here, the factor $n$ appears due to the variation of $f_n$.
By partial summation, we have
$$
\sum_{m=1}^{\infty} \frac{\sin(\sin(nm))}{n^2+m^2}=\sum_{m=1}^{\infty} \frac{S_{ n}(m)-S_n(m-1)}{n^2+m^2}=\sum_{m=1}^{\infty} S_n(m)\left( \frac1{n^2+m^2} - \frac1{n^2+(m+1)^2}\right)
$$
$$
=O\left(\sum_{m=1}^{\infty} \frac{|S_n(m)|m}{(n^2+m^2)^2} \right)
$$
Thus, we can use $S_n(m)=O(nm^{1-\frac1{\mu-1}+\epsilon})$. Note also that we have a trivial bound $S_n(m)=O(m)$. Now, we compare these bounds and split the sum over $m$ into two parts: $m<n^{\mu-1+\delta}$ and $m\geq n^{\mu-1+\delta}$.
The number $\delta>0$ is chosen so that $1-\frac1{\mu-1}+\frac1{\mu-1+\delta}+\epsilon = 1-\epsilon$.
The sum over $m<n^{\mu-1+\delta}$ is treated by
$$
\sum_{m<n^{\mu-1+\delta}} \frac{|S_n(m)|m}{(n^2+m^2)^2} =O\left( \sum_{m<n^{\mu-1+\delta}} \frac {m^2}{(n^2+m^2)^2}\right) =O\left( \frac1n\right)
$$
which is not any better than the trivial bound $|\sin(\sin(nm))|\leq 1$.
For the sum over $m\geq n^{\mu-1+\delta}$, it follows that
$$
|S_n(m)|=O(m^{1-\epsilon})
$$
We have
$$
\sum_{m\geq n^{\mu-1+\delta}} \frac{|S_n(m)|m}{(n^2+m^2)^2}=O\left( \sum_{m\geq n^{\mu-1+\delta}} \frac{ m^{2-\epsilon}}{(n^2+m^2)^2}\right)=O\left(\frac1{n^{1+\epsilon}}\right)
$$
Therefore, what we obtained is the convergence of
$$
\sum_{n=1}^{\infty} \sum_{m\geq n^{\mu-1+\delta}} \frac{\sin(\sin(nm))}{n^2+m^2}.
$$
which is clearly not enough for proving anything about the original series.
The main difficulty here is the lack of a nontrivial bound for the range $m< n^{\mu-1+\delta}$.
Denote $\gamma = 2 \sin^2(\alpha)$. We have to study the convergence of the series $\sum u_n(\gamma, \beta)$ where $u_n(\gamma, \beta) = \frac{\gamma^n}{n^\beta}$.
Easy case... $\gamma = 0$ or $\alpha = k \pi$ with $k \in \mathbb Z$. The general term of the series is equal to zero, so the series converges.
So let's suppose that $\gamma \neq 0$ and separate the cases:
- $\vert \sin(\alpha)\vert= 1/\sqrt{2}$, then $ \gamma = 1$ and $u_n(\gamma, \beta) = 1/n^\beta$. The series converges for $\beta >1$ and diverges otherwise.
- $\vert \sin(\alpha)\vert \neq 1/\sqrt{2}$, then $\left\vert \frac{u_{n+1}(\gamma, \beta)}{u_n(\gamma, \beta)} \right\vert = \gamma \left(\frac{n+1}{n}\right)^\beta$. According to the ratio test, the series converges for $\gamma <1$ and diverges for $\gamma >1$ whatever the value of $\beta$.
Best Answer
Here is a proof that the answer is (almost certainly) not provable using current techniques. We will prove that the series in fact diverges if $2\pi e$ is a rational number with a prime numerator. We first prove the following claims:
Lemma 1. If $p$ is an odd prime number and $S\subset \mathbb Z$ so that $$\sum_{s\in S}e^{2\pi i s/p}\in\mathbb R,$$ then $\sum_{s\in S}s\equiv 0\bmod p$.
Proof. Let $\zeta=e^{2\pi i/p}$. We have $$\sum_{s\in S}\zeta^s=\sum_{s\in S}\zeta^{-s},$$ since the sum is its own conjugate. As a result, since the minimal polynomial of $\zeta$ is $\frac{\zeta^p-1}{\zeta-1}$, we see $$\frac{x^p-1}{x-1}\bigg|\sum_{s\in S}\left(x^{p+s}-x^{p-s}\right),$$ where we have placed each element of $s$ in $[0,p)$. The polynomial on the left is coprime with $x-1$ and the polynomial on the right has it as a factor, so $$\frac{x^p-1}{x-1}\bigg|\sum_{s\in S}\left(x^{p+s-1}+\cdots+x^{p-s}\right).$$ Now, the quotient of these two polynomials must be an integer polynomial, so in particular the value of the left-side polynomial at $1$ must divide the value of the right-side polynomial at $1$. This gives $p|\sum_{s\in S}2s,$ finishing the proof.
Define $$a_n=\sum_{k=0}^n \frac{n!}{k!}.$$
Lemma 2. If $p$ is a prime number, $$\sum_{n=0}^{p-1}a_n\equiv -1\bmod p.$$ Proof. \begin{align*} \sum_{n=0}^{p-1}a_n &=\sum_{0\leq k\leq n\leq p-1}\frac{n!}{k!}\\ &=\sum_{0\leq n-k\leq n\leq p-1}(n-k)!\binom n{n-k}\\ &=\sum_{j=0}^{p-1}\sum_{n=j}^{p-1}n(n-1)\cdots(n-j+1)\\ &\equiv \sum_{j=0}^{p-1}\sum_{n=0}^{p-1}n(n-1)\cdots(n-j+1)\pmod p, \end{align*} where we have set $j=n-k$. The inside sum is a sum of a polynomial over all elements of $\mathbb Z/p\mathbb Z$, and as a result it is $0$ as long as the polynomial is of degree less than $p-1$ and it is $-1$ for a monic polynomial of degree $p-1$. Since the only term for which this polynomial is of degree $p-1$ is $j=p-1$, we get the result.
Now, let $2\pi e = p/q$. Define $\mathcal E(x)=e^{2\pi i x}$ to map from $\mathbb R/\mathbb Z$, and note that $\mathcal E(x+\epsilon)=\mathcal E(x)+O(\epsilon)$. We have \begin{align*} \sin((n+p)!) &=\operatorname{Im}\mathcal E\left(\frac{(n+p)!}{2\pi}\right)\\ &=\operatorname{Im}\mathcal E\left(\frac{qe(n+p)!}{p}\right). \end{align*} We will investigate $\frac{qe(n+p)!}{p}$ "modulo $1$." We see that \begin{align*} \frac{qe(n+p)!}{p} &=q\sum_{k=0}^\infty \frac{(n+p)!}{pk!}\\ &\equiv q\sum_{k=n+1}^\infty \frac{(n+p)!}{pk!}\pmod 1\\ &=O(1/n)+q\sum_{k=n+1}^{n+p}\frac{(n+p)!}{pk!}\\ &=O(1/n)+\frac qp\left[\sum_{k=n+1}^{n+p}\frac{(n+p)!}{k!}\pmod p\right]. \end{align*} Now, \begin{align*} \sum_{k=n+1}^{n+p}\frac{(n+p)!}{k!}=\sum_{j=0}^{p-1}\frac{(n+p)!}{(n+p-j)!} &=\sum_{j=0}^{p-1}(n+p)(n+p-1)\cdots(n+p-j+1)\\ &\equiv \sum_{j=0}^{p-1}m(m-1)\cdots (m-j+1)\pmod p, \end{align*} where $m$ is the remainder when $n$ is divided by $p$. The terms with $j>m$ in this sum go to $0$, giving us $$\sum_{j=0}^m \frac{m!}{(m-j)!}=a_m.$$ Putting this together, we see that $$\sin((n+p)!)=\operatorname{Im}\mathcal E\left(\frac{qa_{n\bmod p}}p\right)+O\left(\frac 1n\right).$$ In particular, the convergence of our sum would imply, since the $O(1/n)$ terms give a convergent series when multiplied by $O(1/n)$, that $$x_N=\operatorname{Im}\sum_{n=1}^N\frac 1n\mathcal E\left(\frac{qa_{n\bmod p}}p\right)$$ should converge. In particular, $\{x_{pN}\}$ must converge, which implies that $$\sum_{m=0}^{p-1}\mathcal E\left(\frac{qa_m}p\right)$$ must be real (as otherwise the series diverges like the harmonic series). By Lemma 1, this implies that $$\sum_{m=0}^{p-1}a_m=0\bmod p,$$ which contradicts Lemma 2.