Convergence of $\sum_{n=1}^{\infty} \frac{\sin(n!)}{n}$

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.