Does $\ \sum_{n=1}^{\infty} \frac{\sin(2^n x)}{n}\ $ converge for all $x\ ?$ Does $\ \displaystyle\sum_{n=1}^{\infty} \frac{\sin(2^n x)}{n}\ $ converge for all $x\ ?$
It obviously converges for any $x\ $ of the form $\ 2^mk \pi\ $ where $\ m,k\in\mathbb{Z},\ $ but for any other values of $\ x\ $ the question is interesting because I don't see how to answer it.
I tried using Cauchy's condensation test with $\ 2^nf(2^n) = \frac{\sin(2^n x)}{n},\ $which implies that $\ f(n) = \frac{\sin(nx)}{n\log_2(n)},\ $ and so $\ \displaystyle\sum_{n=1}^{\infty} \frac{\sin(2^n x)}{n}\ $ converges if and only if $\ \displaystyle\sum_{n=1}^{\infty} \frac{\sin(nx)}{n\log_2(n)}\ $ converges.
I then found out from this video that we can prove using some basic complex analysis that $\ \displaystyle\sum_{k=1}^{\infty} \frac{\sin(n x)}{n}\ $ converges for all $\ x,\ $ and then I thought we were done by the limit comparison test. However, this attempt is wrong because $$\ \frac{\frac{\sin(nx)}{n}}{\frac{\sin(nx)}{n\log_2(n)}}\ \not\to c>0. $$
We also cannot use the integral test for convergence because $\ \displaystyle\sum_{n=1}^{\infty} \frac{\sin(2^n x)}{n}\ $ is not monotone.
Finally, I do not think we can use Abel's test or Dirichlet's test because $\ \displaystyle\sum_{n=1}^{\infty} \sin(2^n x)\ $ probably diverges. Edit: Maybe we can use Dirichlet's test here. I just realised  we only need to show the series $\ \displaystyle\sum_{n=1}^{\infty} \sin(2^n x)\ $ is bounded, not that it converges! But I don't know how to do this... Apparently $\ \displaystyle\sum_{n=1}^{\infty} \sin(kx)\ $ is bounded but $\ \displaystyle\sum_{n=1}^{\infty} \sin(k^2)\ $ isn't, although proving this is difficult! So whether or not $\ \displaystyle\sum_{n=1}^{\infty} \sin(2^n x)\ $ is bounded is probably difficult too. So we should look to use a different test...
Maybe some version of the Alternating Series test or Absolute convergence test ? Although I think that determining convergence of $\ \displaystyle\sum_{n=1}^{\infty} \frac{\lvert \sin(2^n x) \rvert }{n}\ $ is more difficult than the original question here. But Alternating series test might be promising, not sure...
Or maybe there is some other test from complex analysis that is applicable here?
 A: As it has been over a year and no answer/edit has been written to encompass the answer in the comments by @achille_hui, I'll present their answer here for posterity:
At $x=\frac{\pi}{7}$ and $n\geq 1$ we have
$$\sin\left(2^n\frac{\pi}{7}\right)=\sin\left(2\pi\frac{2^{n-1}}{7}\right)=\begin{cases} 
      \sin\left(2\pi\frac{1}{7}\right) & n\equiv 1\ (\text{mod }3) \\
      \sin\left(2\pi\frac{2}{7}\right) & n\equiv 2\ (\text{mod }3) \\
      \sin\left(2\pi\frac{4}{7}\right) & n\equiv 3\ (\text{mod }3) 
   \end{cases}$$
These are bounded by
$$=\begin{cases} 
      \sin\left(2\pi\frac{1}{7}\right)>\frac{3}{4} & n\equiv 1\ (\text{mod }3) \\
      \sin\left(2\pi\frac{2}{7}\right)>\frac{3}{4} & n\equiv 2\ (\text{mod }3) \\
      \sin\left(2\pi\frac{4}{7}\right)>-1 & n\equiv 3\ (\text{mod }3) 
   \end{cases}$$
Taken together, the sum is bounded from below:
$$\sum_{n=1}^\infty \frac{\sin(2^n \pi/7)}{n}=\sum_{k=0}^\infty \left[\sin\left(2\pi \frac{1}{7}\right)\frac{1}{3k+1}+\sin\left(2\pi \frac{2}{7}\right)\frac{1}{3k+2}+\sin\left(2\pi \frac{4}{7}\right)\frac{1}{3k+3}\right]$$
$$>\sum_{k=0}^\infty \left[\frac{3}{4}\cdot\frac{1}{3k+1}+\frac{3}{4}\cdot \frac{1}{3k+2}-\frac{1}{3k+3}\right]=\sum_{k=1}^\infty \frac{18k^2+45k+19}{108 k^3+216k^2+132k+24}$$
$$>\sum_{k=1}^\infty \frac{18k^2+36k+18}{108 k^3+324k^2+324k+108}=\frac{18}{108}\sum_{k=1}^\infty \frac{(k+1)^2}{(k+1)^3}=\frac{18}{108}\sum_{k=1}^\infty \frac{1}{k+1}=\infty$$
