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Prove that the limit as n tends to infinity from $\{n!\sqrt2\}$ does not exist. where {} denotes fractional part and "!" denotes factorial.

I don't have many ideas. I would try to show that , given an arbitrary term $x_n$ we can find another term $x_m$, m>n such that $|x_n-x_m|>a$ where $a$ is some fixed number. And so the sequence cannot converge.

José Carlos Santos
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Ion
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    Could you please show what you have tried? – LASV Nov 20 '13 at 15:47
  • I suspect that if ${n!\alpha}$ converges, then $\alpha$ is rational or transcendental. Note that the convergence gives us some nice rational approximations for $\alpha$ - but perhaps not nice enough to prove my supicion – Hagen von Eitzen Feb 20 '19 at 04:52
  • @HagenvonEitzen while the conjecture sounds sound, I'm not sure I agree with your remark that the rational approx. are good: even for $\exp(1)$ which has one of the most 'natural' representation for this problem, its approx. is $\sum \limits_{i=1}^N \frac{1}{i!}$, a fraction with denominator $N!$, and the error is $O\big(\frac{1}{N!}\big)$. That does not even reach the Liouville measure of $e$ which is $2$ (we'd need an error around $\frac{1}{(N!)^2}$. So I don't think we could find an absurdity using that $\sqrt{2}$ has Liouville measure $2$ (maybe you meant 'nice approx' in a specific sense) – charmd Jul 16 '21 at 06:42

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