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 !
 A: 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.
A: 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.
