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Let $z\in\mathbb{R}$; does $\displaystyle\sum_{n=1}^{\infty} \frac{\exp(2\pi i z n!)}{n}$ converge? If $z$ is rational, then surely not because the numerator is eventually just $1$. But what if $z$ is irrational? Beggars and choosers and all that, but a downloadable reference would work too.

Integrand
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    $|\exp(2i\pi zn!)|=1$ for all $n$, if the sum converges we would have $\lim\limits_{n\rightarrow +\infty}\exp(2i\pi zn!)=0$ which is not. – Tuvasbien May 10 '20 at 02:57
  • @Tuvasbien Why are you answering in a comment? – Arthur May 10 '20 at 02:59
  • My answer is « too » simple, I answered in the comments in case the question was not tackling the real issue (for instance if the question was initially meant to be about the convergence of the sum $\sum\sin(2\pi zn!)$ instead, which is not as simple as the above question) – Tuvasbien May 10 '20 at 03:06
  • In the case $z=e$, the Taylor expansion of $e^x$ shows that $\exp(2i\pi en!)=(-1)^{n+1}+\frac{2i\pi (-1)^{n+1}}{n}+\mathcal{O}\left(\frac{1}{n^2}\right)$ and thus the sum converges. In the general case, we can suppose without loss of generality that $z\in[0,1[$, one can show that there exists $(a_k) $ such that $z=\sum_{k=1}^{+\infty}\frac{a_k}{k!}$, $a_k\in[![0,9]!]$ and that this decomposition is unique, then what said above still works and the series diverges (because the $a_k$ are not all $0$ for $k$ large enough because $z$ is not rationnal) – Tuvasbien May 10 '20 at 03:20
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    Actually for $e$ the series diverges since $en!=k(n)+O(1/n), k(n)$ integral, so the series behaves like the harmonic series, but for something like $z=\sum_{n \ge 1} {a_n/n!}, a_n =1/2, n=2k+1, a_n=1, n=2k$ the series behaves like an alternating series ($zn!=k(n)+O(1/n)$ but now $k(n)$ is a half integer for $n=4k+1,2$ and an integer for $n=4k+3,4$, so signs are $-,-,+,+$) hence it converges – Conrad May 10 '20 at 03:36
  • @Conrad you're overloading k. You could rephrase for clarity. – AHusain May 10 '20 at 03:43
  • the point is that if $z=\sum {a_n/n!}$ where $a_n$ alternates between a half-integer like $1/2$ and an integer like $1$ depending on the parity of $n$ (can choose whatever you want as along as they alternate between half integers/integers and they are within a fixed set, starting say at $n=1$ with half integers for definiteness), $zn!$ is a sum of half integers and integers and a remainder so the exponential gives $\pm 1$ multipled by $1+O(1/n)$ and the pattern of signs is $-,-, +,+,-,-,+,+..$ with period $4$ so the series converges – Conrad May 10 '20 at 03:52
  • A simpler example is $z = e/2$. Note that the integer part of $e n!$ is alternately $\pm 1$, so $\exp(i \pi e n!) \sim (-1)^{n+1} + O(1/n)$, and the alternating harmonic series $\sum (-1)^{n+1}/n$ converges. – Robert Israel May 10 '20 at 05:13
  • @Tuvasbien . That is not a valid argument. The convergence of $\sum_n z_n/n$ does not imply $z_n\to 0,$ e.g. if $z_n=(-1)^n.$ – DanielWainfleet May 10 '20 at 08:01
  • Hmm, this question looks a lot like https://math.stackexchange.com/questions/2756038/convergence-of-sum-n-1-infty-frac-sinnn , which is basically an open problem. I feel satisfied with Conrad's answer; thanks for the input, everyone. – Integrand May 10 '20 at 17:28

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Let $a_k=0, k \ne n!$ for any $n \ge 1$ and $a_k=\frac{1}{n} , k=n!$ for some $n \ge 1$

Then $f(x)$~$\sum_{k \ge 1}{a_ke^{2\pi ikx}}$ is a trigonometric series with period $1$ and $\sum_{k \ge 1}|a_k^2| < \infty$ so $f$ is the Fourier series of an $L^2$ function. In particular by the famous theorem of Carleson $\sum_{k \ge 1}{a_ke^{2\pi ikx}}$ converges pointwise a.e on $[0,1]$ hence on $\mathbb R$

This being said, the result can be proven much more elementary here, as by basic Fourier theory, $s_m=\sum_{1 \le k \le m}{a_ke^{2\pi ikx}}$ is summable $C-1$ (Caesaro or by arithmetic means) a.e. on $[0,1]$ (hence on $\mathbb R$) being the Fourier series of an integrable (again on $[0,1]$ here) periodic function.

However for lacunary series $a_{m_k} \ne 0$ only for a sequence $m_k$ s.t $m_{k+1}/m_k \ge q >1, k \ge k_0$ and here we have that and more of course as we can take any $q >1$ as $m_k=k!$), it is not hard to prove that Caesaro summable implies convergence, so the result that the series converges a.e. follows in an elementary way (elementary as Fourier series go of course)

As shown in the comments above, one can easily construct irrational numbers where the series is convergent and divergent, but overall the series converges a.e so the former predominate

Conrad
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