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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?

Adam Rubinson
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    As for my understanding, this is an extremely hard problem. I wouldn't be surprised if it is still an open problem. Check this past discussion, for instance. – Sangchul Lee Dec 23 '21 at 17:20
  • Is Cauchy Condensation only for nonnegative series? – GEdgar Dec 23 '21 at 17:28
  • Also Cauchy condensation test only works for non-negative, monotone sequences. Your series is neither non-negative nor monotone (except for trivial choices of $x$). – Sangchul Lee Dec 23 '21 at 17:39
  • Thanks for your responses. Those two things: the correction about my mis-application of Cauchy's condensation test, and secondly, the fact that this is an extremely hard and possibly still unsolved problem. fully answers my question. If no one else wants to write that as an answer, then I think I will. – Adam Rubinson Dec 23 '21 at 17:48
  • The series converges ae by general theory (it is the Fourier series of an $L^2$ function) but I agree that it is probably hopeless to characterize the points of divergence – Conrad Dec 23 '21 at 22:27
  • However it is not that hard to show that the series with absolute values converges only when the terms become zero from some point on, so for $x=k\pi/2^m$, so actually that is a fairly easy problem – Conrad Dec 23 '21 at 22:34
  • @Conrad If by "only when" you mean "only if", i.e. $\ \implies,\ $ then the contrapositive of what you say is that: $\ \displaystyle\sum_{n=1}^{\infty} \frac{\lvert \sin(2^n x) \rvert }{n}\ $ diverges for $\ x\neq k\pi / 2^m,\ $ right? But then since absolute convergence implies convergence, that would also answer the original question, that $\ \displaystyle\sum_{n=1}^{\infty} \frac{\sin(2^n x)}{n}\ $ converges for all $\ x.\ $ So I don't think I interpret your comment correctly. Could you please elaborate? – Adam Rubinson Dec 24 '21 at 11:53
  • Nope. hint: find a $x$ such that $\sin(2^n x)$ is periodic in $n$ and sum over all those terms is non-zero. – achille hui Dec 24 '21 at 11:54
  • @achillehui I don't know what you mean. Were you replying to my reply to Conrad, or answering the original question? Please elaborate. – Adam Rubinson Dec 24 '21 at 12:04
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    @AdamRubinson your original question. eg. under modulo $7$ arithmetic, the map $k \to 2k$ sends $1$ to $2$ to $4$ then back to $1$. So $(\sin(2^n\frac{\pi}{7}))_{n=1,2\ldots} = (\sin\frac{2\pi}{7}, \sin\frac{4\pi}{7},\sin\frac{8\pi}{7},\sin\frac{2\pi}{7},\sin\frac{4\pi}{7},\cdots)$ is periodic in $n$ with period $3$. – achille hui Dec 24 '21 at 12:07
  • @achillehui Brilliant for spotting that. Feel free to post that as an answer! – Adam Rubinson Dec 24 '21 at 12:25
  • If a series diverges absolutely one cannot really say much about its conditional convergence or lack of it, so the fact that $\sum |\sin 2^nx|/n$ diverges except at $k\pi/2^m$ says nothing about the series itself which as noted is convergent almost everywhere wr Lebesgue measure – Conrad Dec 24 '21 at 13:12

1 Answers1

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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:

$$\sum_{n=1}^\infty \frac{\sin(2^n \pi/7)}{n}=\sum_{k=0}^\infty \left[\sin\left(2\pi \frac{1}{7}\right)\frac{1}{3k+1}+\sin\left(2\pi \frac{2}{7}\right)\frac{1}{3k+2}+\sin\left(2\pi \frac{4}{7}\right)\frac{1}{3k+3}\right]$$

$$>\sum_{k=0}^\infty \left[\frac{3}{4}\cdot\frac{1}{3k+1}+\frac{3}{4}\cdot \frac{1}{3k+2}-\frac{1}{3k+3}\right]=\sum_{k=1}^\infty \frac{18k^2+45k+19}{108 k^3+216k^2+132k+24}$$

$$>\sum_{k=1}^\infty \frac{18k^2+36k+18}{108 k^3+324k^2+324k+108}=\frac{18}{108}\sum_{k=1}^\infty \frac{(k+1)^2}{(k+1)^3}=\frac{18}{108}\sum_{k=1}^\infty \frac{1}{k+1}=\infty$$

QC_QAOA
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