Do we have :
$$\sup\{\sqrt {n!} - E(\sqrt {n!}); n\in I\!\!N\}=1?$$
Where $E(\cdot)$ is the integer part function, and $n!=1\times 2...\times n$.
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4$\begingroup$ By $\sqrt{n}!$ do you mean $\Gamma(1+\sqrt{n})$? $\endgroup$– Jeremy RouseCommented Jan 16, 2015 at 22:09
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2$\begingroup$ Or maybe $\sqrt{n!}$? $\endgroup$– Emil JeřábekCommented Jan 16, 2015 at 22:18
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1$\begingroup$ I mean $\sqrt{n!}$ $\endgroup$– MedCommented Jan 17, 2015 at 12:32
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$\begingroup$ Question seems clear now, after the edits. $\endgroup$– Todd TrimbleCommented Jan 17, 2015 at 12:59
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3$\begingroup$ In general we don't have any good techniques for controlling the distribution of an exponentially growing sequence modulo 1. For instance nobody knows how to prove that the sup of $\{ 10^n \pi \}$ is $1$, though everyone believes it is true. I don't see any reason why $\sqrt{n!}$ would be any easier than $10^n \pi$ (or $10^n \sqrt{2}$, for that matter). $\endgroup$– Terry TaoCommented Jan 18, 2015 at 3:29
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