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.

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    $\begingroup$ $|\exp(2i\pi zn!)|=1$ for all $n$, if the sum converges we would have $\lim\limits_{n\rightarrow +\infty}\exp(2i\pi zn!)=0$ which is not. $\endgroup$ – Tuvasbien May 10 at 2:57
  • $\begingroup$ @Tuvasbien Why are you answering in a comment? $\endgroup$ – Arthur May 10 at 2:59
  • $\begingroup$ 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) $\endgroup$ – Tuvasbien May 10 at 3:06
  • $\begingroup$ 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) $\endgroup$ – Tuvasbien May 10 at 3:20
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    $\begingroup$ 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 $\endgroup$ – 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)

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

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