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?
Best Answer
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:
$$\sum_{n=1}^\infty \frac{\sin(2^n \pi/7)}{n}=\sum_{k=0}^\infty \left[\sin\left(2\pi \frac{1}{7}\right)\frac{1}{3k+1}+\sin\left(2\pi \frac{2}{7}\right)\frac{1}{3k+2}+\sin\left(2\pi \frac{4}{7}\right)\frac{1}{3k+3}\right]$$
$$>\sum_{k=0}^\infty \left[\frac{3}{4}\cdot\frac{1}{3k+1}+\frac{3}{4}\cdot \frac{1}{3k+2}-\frac{1}{3k+3}\right]=\sum_{k=1}^\infty \frac{18k^2+45k+19}{108 k^3+216k^2+132k+24}$$
$$>\sum_{k=1}^\infty \frac{18k^2+36k+18}{108 k^3+324k^2+324k+108}=\frac{18}{108}\sum_{k=1}^\infty \frac{(k+1)^2}{(k+1)^3}=\frac{18}{108}\sum_{k=1}^\infty \frac{1}{k+1}=\infty$$