Does the series $\sum_{n=1}^{\infty}\frac{\sin(\cos(n))}{n}$ converge?

I was doing some exercises and this one just stunned me. I had to study the convergence of this serie:

$$\sum_{n=1}^{\infty}\frac{\sin(\cos(n))}{n}$$

I tried alot of diffrent things, but I got no where. Can anyone please help me with an idea or a clue just to start ?

Thanks in advance !

Here is an alternate to Jack D'Aurizio's answer, which interestingly also involves the irrationality measure of $\pi$ (call it $\mu$). Suppose $f$ is a $2\pi$-periodic smooth function, such that $\int_0^{2\pi}f(x)\;dx=0$. Let $$ s_n=\frac1n\sum_{k=1}^n f(k). $$ By the Koksma–Hlawka inequality and this bound, $$ |s_n|=O(n^{-1/(\mu-1)+\epsilon}) $$ for any $\epsilon>0$. In particular, choosing $\epsilon<\frac1{2(\mu-1)}$, we have $|s_n|=O(n^{-\epsilon})$. Using summation by parts, $$\begin{eqnarray*} \sum_{n=1}^N\frac{f(n)}n &=&s_N+\sum_{n=1}^{N-1}ns_n\left(\frac1n-\frac1{n+1}\right)\\ &=&s_N+\sum_{n=1}^{N-1}\frac{s_n}{n+1}. \end{eqnarray*}$$ The above implies $s_N\to0$ and the last sum is absolutely convergent, so the LHS converges. In particular setting $f(x)=\sin(\cos(x))$, the series in question converges.

I think it is more practical to expand $\sin\cos(x)$ as a Fourier cosine series. We have

$$\sin\cos(x) = 2\sum_{m\geq 0}(-1)^m J_{2m+1}(1) \cos((2m+1)x) $$ where the coefficients $J_{2m+1}(1)$, depending on a modified Bessel function of the first kind, have an exponential decay. On the other hand $$ \sum_{n\geq 1}\frac{\cos(nx)}{n}=-\log\left|2\sin\frac{x}{2}\right|$$ for any $x\not\in 2\pi\mathbb{Z}$, and $-\log\left|2\sin\frac{x}{2}\right|$ cannot be too large for some small $x\in\mathbb{N}$ since $\pi$ has a finite irrationality measure. By exploiting the exponential decay of the previous coefficients we may conclude that the original series is convergent.