On the convergence of $\sum_n 1 /(n\sin(2^nx))$ 
Find all values of $x$ such that $\displaystyle\sum_{n=1}^\infty\frac{1}{n\sin(2^nx)}$ converges.

I've been attempting to solve this problem without much success. Firstly, it is defined for all $x\in\mathbb R$ such that $\forall n\in\mathbb N$, $2^nx\not\in\pi\mathbb Z$. Then, if $x$ is such that $\sin(2^nx)$ converges then by comparison to $\sum\frac1n$ the series diverges. If $x\in\pi(\mathbb R\backslash\mathbb Q)$, then $(\sin(2^nx))_{n\in\mathbb N}$ is dense in $[-1,1]$ and so it feels like the series should diverge. On the other hand, I believe that if $x\in\pi\mathbb Q$, then $(\sin(2^nx))_{n\in\mathbb N}$ should be periodic (or not far from) and the series should converge. However, I can't prove any of this. Any help is appreciated!
 A: At least for $\pi$-rationals it's true.
Assume $x = 2\pi k + \frac{p_0}{2q}\pi$ where $q > 1$ is odd (if it can't be represented as such, then either $x$ isn't $\pi$-rational or $2^n x / \pi$ is integer for some $n$).
Wlog, dropping the first term if necessary, assume $p_0$ is even.
By Fermat's little theorem, $2^{q - 1} = q\cdot m + 1$.
Then we have $2^q x = 2 \pi k' + \frac{2 q \cdot m \cdot p_0 + p_0}{2q}\pi = 2\pi k' + m p_0 \pi + \frac{p_0}{2q} = 2\pi k'' + \frac{p_0}{2q}$, so $\sin(2^n x)$ is periodic, and $q$ is period.
Let $p_i = 2^i p_0 \mod 2q$, so argument of $\sin$ of $i$-th element in each period is $2\pi k' + \frac{p_1}{2q}\pi$.
Now, as $q$ is odd, $2$ is invertible in $\mathbb Z_q$, so sequence $p_0, p_1, \ldots, p_{q - 1}$ contains all even numbers from $\overline{0..2q}$.
Then we can split $p_i$ in pairs that sums to $2q$.
Let $\alpha_i = \sin \frac{p_i \pi}{2q}$.
Returning for original series, pair $(p_i, p_j)$ in $t$-th period gives us $\frac{1}{(tq + i) \alpha} - \frac{1}{(tq + j) \alpha} = \frac{c_{ij}}{t^2} + o(\frac{1}{t^2})$.
Summing by all pairs, we get that sum of $t$-th period is $O(\frac{1}{t^2})$, and so the series converges.
Essentially, if we draw points $\exp(2^n \cdot i \cdot x)$ on complex plane, they will (except may be the first one) form set of points symmetric relative to $Ox$. Next, if we multiply each point by $\frac{1}{n} + O(\frac{1}{n^2})$, each symmetric pair will sum to $O(\frac{1}{n^2})$, and as we have constant number of points, so will all the points.
What happens with $\pi$-irrational numbers is more interesting. $\sin(2^n x)$ don't necessary has $0$ as limit point: consider $x = \pi \cdot 0.10\ 11\ 10\ 10\ 11\ 10\ 10\ 10\ 11\ldots_2$ ($k$-th segment is $10$ repeated $k$ times and then $11$) - it's irrational, but $2^n x\pmod{\pi} \in \left[\frac{1}{4}, \frac{15}{16}\right]$.
