Let $\alpha$ be an irrational number, and $\left\{ \cdot \right\}$ denotes the fractional part function. We have focused on how $\left\{ n! \alpha \right\}$ distributes in this MO question. And now I am interested in the series $\sum_{n = 1}^\infty \left\{ n! \alpha \right\}$.
The above question mentions that, I quote, there is a subset of $\mathbb{R}$ of Hausdorff dimension $1$, such that the corresponding sequence is bounded away from $0$, which means the series $\sum_{n = 1}^\infty \left\{ n! \alpha \right\}$ must diverge.
Though we cannot tell the exact distribution of $\left\{ n! \alpha \right\}$ yet, it might be reasonable to suspect that the convergence of $\sum_{n = 1}^\infty \left\{ n! \alpha \right\}$ have been fully settled, just as the series $\sum_{n = 1}^\infty \sin( n! \pi \alpha )$.
Sadly, I exhaust my scope of search without finding a similar result. As Hanson points out, considering a specific case, such as $\alpha = \pi$, would be unbearably hard. Therefore, it is natural to ask if there exists such a value that the series converges and te4’s comment confirms this as it turns out to be a problem in CIIM 2024. I sketch the solution due to emi3.141592 here for the integrity of this problem.
$\mathrm{e}$ plays well in $\left\{ n! \alpha \right\}$ since $\mathrm{e}=\sum_{k=0}^\infty \frac{1}{k!}$, which cancels with $n!$ leaving a pleasant term $n!\cdot \sum_{k=n+1}^\infty \frac 1{k!}$. However, it diverges as the harmonic series does, since every term is dominated by $\frac{1}{k+1}$. We imitate the structure of the infinite sum of $\mathrm{e}$, deleting most of the terms to make it converge. In particular, it is easy to prove that $\alpha=\sum_{k=1}^\infty \frac{1}{(2^k)!}$ converges. In fact, $$\sum_{n=1}^\infty\left\{ n!\cdot \alpha \right\} \leqslant \sum_{n=1}^\infty \left( \sum_{k=1}^{2^n-1} \frac{k!}{2^n !} \right)< \sum_{n=1}^\infty \left(\sum_{k=1}^{2^n-1} \frac{1}{2^{kn}} \right)=\sum_{n=1}^\infty \left( \frac{1}{2^{n}}-\frac{1}{2^{n(2^n-1)}} \right)<1.$$Consider the set $$\left\{ \left. \sum_{k=1}^\infty \frac{a_k}{(2^k)!} \, \right| a_k \in \{0,1\} \right\}.$$Clearly the series converges at the numbers in this set. Moreover, there are uncountably many elements, hence an irrational one exists.
It is a rigorous proof of existence. But I greedily wonder if there is a constructive example. In spite of the huge gap impeding us to reach a proof, could we heuristically conjecture an explicit set of irrational numbers whose corresponding series converge? For example, we expect $\pi$ not in this set.
Moreover, I still wish to request for references to this series.
Any advice is welcomed.
