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

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.

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.