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.