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The following is an interesting problem presented to this site which has yet to bet solved: Does

$$\sum_{n=1}^\infty \frac{\sin(n!)}{n}$$

converge. While attempting this problem, I thought that proving the equidistribution of $n!$ modulo $2\pi$ would be sufficient for the original conjecture. For those who do not know, an sequence $a_n$ is said to be equidistributed on a non-degenerate interval $[a,b]$ if

$$\lim_{n\to \infty}\frac{|\{a_1,a_2,\cdots,a_n\}\cap [c,d]|}{n}=\frac{d-c}{b-a}$$

for all subintervals $[c,d]\subseteq [a,b]$. My thoughts then turned to the more general question: If $a_n$ is any sequence of real numbers such that $\mod(a_n,2\pi)$ is equidistributed over $[0,2\pi]$, does

$$\sum_{n=1}^\infty \frac{\sin(a_n)}{n^\beta}$$

necessarily converge for $\beta>0$. Obviously, if $\beta>1$ then the series converges absolutely, so the interesting cases are $0<\beta<1$ and $\beta=1$ (although they might be the same case overall). One possible way forward is using Weyl's criterion: we know that if $a_n$ is equidistributed over $[0,2\pi]$, then

$$\lim_{n\to\infty} \sum_{j=1}^n\frac{\sin(q a_j)}{n}=0$$

for all $q\in\mathbb{N}$. I'm not sure how this could be useful but it seems pretty close to the original sum. One result in favor of this conjecture is discussed on this mathoverflow post. That is, if $p(n)$ is any polynomial with rational coefficients, then

$$\sum_{n=1}^\infty \frac{\sin(p(n))}{n}$$

converges.

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    $\begingroup$ About $n!\mod{2\pi}$ being equidistributed: I don't think it is a solved problem, but this is interesting: arxiv.org/abs/1505.01198 $\endgroup$ – Maximilian Janisch Dec 26 '19 at 22:48
  • $\begingroup$ I don't think so either, but I realized that even if I did show that it was, then the question of convergence would still stand. $\endgroup$ – QC_QAOA Dec 26 '19 at 22:49
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No.

Suppose we had equidistributed $(a_n)_n$ such that $\sum_{n \le t} \sin(a_n) \sim \frac{t}{\log t}$.

Then, for any $\beta \in (0,1]$, summation by parts gives $$\sum_{n \le N} \frac{\sin(a_n)}{n^\beta} = \frac{\sum_{n \le N} \sin(a_n)}{N^\beta}+\beta \int_1^N \frac{\sum_{n \le t} \sin(a_n)}{t^{1+\beta}}dt,$$ which diverges to $+\infty$ (the integral goes to $+\infty$; if $\beta < 1$, the first term does as well, and if $\beta = 1$, the first term goes to $0$).

We now construct such $(a_n)_n$. Fix $\theta = \sqrt{2}$ (any irrational would do). Let $b_n = n\theta$ mod $1$. Then $(b_n)_n$ is very equidistributed in that, for some absolute $C \ge 1$, $$\left|\sum_{n \le N} \sin(b_n)\right| \le C$$ for each $N \ge 1$ (recall $|\sum_{n \le N} e(n\theta)| \le \min(N,\frac{1}{||\theta||})$). Define $(a_n)_n = (b_1,\frac{\pi}{2},\frac{\pi}{2},b_2,\frac{\pi}{2},b_3,\frac{\pi}{2},b_4,b_5,b_6,\dots)$, where we have inserted $\frac{\pi}{2}$'s into the sequence $(b_n)_n$ at prime indices. Then we have that $(a_n)_n$ is equidistributed, since we just modified an equidistributed sequence on a density $0$ set. And due to the last centered inequality (and the density of the prime numbers), we have $\sum_{n \le t} \sin(a_n) \sim \frac{t}{\log t}$.


The main reason the answer is "no" is that equidistribution is just a sublinear assumption on $\sum_{n \le t} \sin(a_n)$. If you wanted to define a sequence $(x_n)_n$ to be $\beta$-equidistributed if $\sum_{n \le N} \sin(qx_n) = o(N^\beta)$ (rather than $o(N)$) for each $q \in \mathbb{Z}$, then the answer would be "yes", as the summation by parts formula above shows.

One last thing I want to say, at the expense of being annoying. Weyl's criterion is really the converse of what you said. It is completely trivial to show that if a sequence $(x_n)_n$ is equidistributed, then $\sum_{n \le N} \sin(qx_n) = o(N)$ for each $q$. However, the quite remarkable thing is that if $\sum_{n \le N} \sin(qx_n) = o(N)$ and $\sum_{n \le N} \cos(qx_n) = o(N)$, then $(x_n)_n$ is equidstributed. This converse is really at the heart of fourier analysis, and is still quite magical to me.

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  • $\begingroup$ Good answer. If you have the time, I would be very interested in more details of your analysis. I get the general gist of it but always enjoy a deeper dive into the details $\endgroup$ – QC_QAOA Jan 2 at 6:05
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    $\begingroup$ @QC_QAOA Thanks for the response. I think all the details are added now. If you have a specific concern, feel free to let me know. $\endgroup$ – mathworker21 Jan 2 at 6:12
  • $\begingroup$ Very nice trick with the $\pi/2$ at prime indices. Would never have thought of that but it works out to such a great counterexample to my original conjecture. $\endgroup$ – QC_QAOA Jan 2 at 6:16

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