Convergence of $\sum_{n=0}^{\infty}\sin(x\pi n!)$

Question

Since this question is closed, I'm asking it myself.

Let $M=\{x\in \mathbb R|\sum_{n=0}^{\infty}\sin(x\pi n!)\text{ converges}\}$

Prove that there is no $a,b\in \mathbb R$ such that $(a,b)\subset M$ (ie $M$ has empty interior).

It has been proved that $e\in M$, and it's easy to prove that $\mathbb Q \subset M$ (essentially because if $n\in \mathbb N$, then $\sin(n\pi)=0$). I haven't managed to prove $M$ has empty interior. The idea would be to point out a dense subset of $M^c$.

Noting that $\sin(x\pi n!)\approx 0 \iff \exists k_n\in \mathbb Z, x\approx \frac{k_n}{n!}$, we're looking for $x$'s that are badly approximated by rationals. A result by Liouville states that if $x$ is an irrational algebraic number of degree $m$, there exists some $c_x$ such that for all $p,q\in \mathbb Z$, $\left|x-\frac pq \right|> \frac{c_x}{q^m}$.

In our case, if $x$ is an irrational algebraic number of degree $m$, then for each $n$ and any $p\in \mathbb Z$, $$\left|xn!\pi -p\pi \right|>\frac{c_x\pi}{(n!)^{m-1}}$$ Hence $|\sin(xn!\pi)|>\left|\sin\left(\frac{c_x\pi}{(n!)^{m-1}}\right) \right|$.

However, nothing may be derived from this lower bound (the RHS goes to $0$ too fast).

Answer 1

Notice that any number $x$ between $[0, 1]$ can be written in the form

$$ x = \sum_{k=2}^{\infty} \frac{a_k}{k!}, \qquad a_k \in \{0, \cdots, k-1\}. $$

Mimicking my previous computation, we have

$$ \sin(x\pi n!) = (-1)^{n a_{n-1} + a_n} \sin\left( \pi \sum_{k=1}^{\infty} \frac{a_{n+k}}{(n+1)\cdots(n+k)} \right). $$

Here an important observation is that

$$ \frac{a_{n+1}}{n+1} \leq \sum_{k=1}^{\infty} \frac{a_{n+k}}{(n+1)\cdots(n+k)} \leq \frac{a_{n+1}+1}{n+1}. \tag{*} $$

Now think of $a_k$ as function of $x$ (as in the usual $n$-ary expansion, ambiguity arises only when $x$ is rational, a Lebesgue-null set). Then each $a_k$ defines a random variable having uniform distribution over the set $\{0,\cdots,k-1\}$ and $\{a_k\}$ are mutually independent. Moreover,

$$ \sum_{k=2}^{\infty} \mathbb{P}\left(\left|\frac{a_k}{k} - \frac{1}{2} \right| < \frac{1}{4}\right) = \infty $$

as $k\to\infty$ and thus by the 2nd Borel-Cantelli theorem, $\frac{1}{4} < \frac{a_k}{k} < \frac{3}{4}$ holds for infinitely many $k$ for almost every $x$ in $[0, 1]$. This, together with $\text{(*)}$, shows that

$$ \liminf_{n\to\infty} | \sin(x\pi n!) | \geq \sin(\pi/4) = \frac{1}{\sqrt{2}} $$

holds for almost every $x$ in $[0, 1]$. Since no $x \in M$ can satisfy this, it follows that $M$ is a Lebesgue-null set and its interior is empty.

Answer 2

A result of Kolmogorov states that if $n_k$ is a Hadamard lacunary sequence, i.e., $\frac{n_k}{n_{k-1}}>\rho$ for all $k\in\mathbb{N}$ and some $\rho>1$, then the trigonometric series \begin{align}\sum_k a_k\cos(n_k x+\theta_k)\tag{1}\label{one}\end{align} converges (diverges) almost everywhere iff $\sum_k a^2_k$ converges (diverges).

This can be seen in Zygmund, A., Trigonometric Series Vol. 1, 3rd edition, Cambridge University Press, pp. 202-204.

The convergence can also be seen in the sense of a Toeplitz summation but that is not too relevant for the purposes of this posting.

The series in the problem is of the form \eqref{one} with $a_k\equiv1$, $\theta_k\equiv\frac{\pi}{2}$, and $n_k=k!$. Then the series in the OP diverges almost surely, and so the set $M$ is of measure zero and has no interior.