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.
 A: 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
