7
$\begingroup$

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 $\Gamma(1+\sqrt{n})$? $\endgroup$ Jan 16, 2015 at 22:09
  • 2
    $\begingroup$ Or maybe $\sqrt{n!}$? $\endgroup$ Jan 16, 2015 at 22:18
  • 1
    $\begingroup$ I mean $\sqrt{n!}$ $\endgroup$
    – Med
    Jan 17, 2015 at 12:32
  • $\begingroup$ Question seems clear now, after the edits. $\endgroup$
    – Todd Trimble
    Jan 17, 2015 at 12:59
  • 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 Tao
    Jan 18, 2015 at 3:29

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