Question Number 26770 by julli deswal last updated on 29/Dec/17
$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\left({n}\:!\right)^{\mathrm{1}/{n}} }{{n}}\:= \\ $$
Commented by abdo imad last updated on 29/Dec/17
$${n}!\sim\:{n}^{{n}} \:{e}^{−{n}} \:\sqrt{\mathrm{2}\pi{n}}\:\left({stirling}\:{formula}\right)\Rightarrow\:\left({n}!\right)^{\frac{\mathrm{1}}{{n}}} \sim\:\:{n}\:{e}^{−\mathrm{1}} \left(\mathrm{2}\pi{n}\right)^{\frac{\mathrm{1}}{\mathrm{2}{n}}} \\ $$$$\Rightarrow\:\:\:\frac{\left({n}!\right)^{\frac{\mathrm{1}}{{n}}} }{{n}}\:\:\sim\:\:{e}^{−\mathrm{1}} \:\overset{\:\frac{\mathrm{1}}{\mathrm{2}{n}}\:{ln}\left(\mathrm{2}\pi{n}\right)} {{e}}_{} \:{but}\:\:\:\:{lim}\:_{{n}−>\propto} \:\frac{{ln}\left(\mathrm{2}\pi{n}\right)}{\mathrm{2}{n}}=\mathrm{0} \\ $$$${and}\:{lim}_{{n}−>\propto} {e}^{\frac{{ln}\left(\mathrm{2}\pi{n}\right)}{\mathrm{2}{n}}} \:\:=\mathrm{1}\Rightarrow\:\:\:{lim}_{{n}−>\propto} \:\:\frac{\left({n}!\right)^{\frac{\mathrm{1}}{{n}}} }{{n}}\:\:=\:\frac{\mathrm{1}}{{e}} \\ $$