# Does $\sum_{n=1}^{\infty} \frac{\exp(2\pi i z n!)}{n}$ converge for irrational $z$?

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.

• $|\exp(2i\pi zn!)|=1$ 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 at 2:57
• @Tuvasbien Why are you answering in a comment? – Arthur May 10 at 2:59
• 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 at 3:06
• 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 at 3:20
• 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 at 3:36

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)