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55 mins ago comment added user65023 Also, by independence, the set of $\alpha$ where it converges will have Lebesgue measure zero. Given the sum up to a fixed $M$, there exists $N>>M$ sufficiently large such that the following sets are almost independent: $\{\alpha>0: \{ M!\alpha\}<\frac{1}{2}\}$ and $\{ \alpha>0:\{N!\alpha\}<\frac{1}{2}\}$. Generate sequences $N_1, N_2, \ldots$ and $\epsilon_1, \epsilon_2,\ldots$ which control the error away from independence. Also, $\epsilon_{i+1} >0$ will depend on $N_i$. The set of $\alpha$ such that $\{N_i \alpha\} \geq \frac{1}{2}$ infinitely often will have measure 1.
1 hour ago comment added user65023 The set of irrationals $\alpha \in [0,1]$ where $\sum_{n=1}^{\infty} \{ n! \alpha\}$ converges is dense, since we can pick a single $\alpha_0$ where this works and add any rational number to it.
6 hours ago history edited GendoTendoLendo CC BY-SA 4.0
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7 hours ago comment added te4 See here: artofproblemsolving.com/community/u1120788h3421811p32954781
8 hours ago history edited GendoTendoLendo CC BY-SA 4.0
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10 hours ago comment added GendoTendoLendo @JamesEHanson Thanks for your comment. I might have greedily expected too much. However I want to know if there is some partial progress. I will edit my question to be more applicable.
10 hours ago history edited GendoTendoLendo CC BY-SA 4.0
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10 hours ago comment added James E Hanson If I recall correctly, distribution of the sequence $\{n!\pi\}$ is closely related to the problem of whether $\pi + e$ is irrational, which is a fairly hard open problem. As such I'm guessing this is not something we understand very well.
10 hours ago history asked GendoTendoLendo CC BY-SA 4.0