Question Number 30759 by abdo imad last updated on 25/Feb/18
$${find}\:{lim}_{{n}\rightarrow\infty} \:\:\:\:\left(\frac{{n}!}{{n}^{{n}} }\right)^{\frac{\mathrm{1}}{{n}}} \:. \\ $$
Commented by abdo imad last updated on 27/Feb/18
$${let}\:{use}\:{the}\:{stirling}\:{formula}\:{we}\:{have} \\ $$$${n}!\:\:\sim\:{n}^{{n}} \:{e}^{−{n}} \sqrt{\mathrm{2}\pi{n}}\:\:\Rightarrow\frac{{n}!}{{n}^{{n}} }\:\sim\:{e}^{−{n}} \sqrt{\mathrm{2}\pi{n}}\:\Rightarrow \\ $$$$\left(\frac{{n}!}{{n}^{{n}} }\right)^{\frac{\mathrm{1}}{{n}}} =\left({e}^{−{n}} \sqrt{\mathrm{2}\pi{n}}\:\right)^{\frac{\mathrm{1}}{{n}}} =\:{e}^{−\mathrm{1}} \:\left(\mathrm{2}\pi{n}\right)^{\frac{\mathrm{1}}{\mathrm{2}{n}}} \:{but} \\ $$$$\left(\mathrm{2}\pi{n}\right)^{\frac{\mathrm{1}}{\mathrm{2}{n}}} ={e}^{\frac{\mathrm{1}}{\mathrm{2}{n}}{ln}\left(\mathrm{2}\pi{n}\right)} ={e}^{\:\frac{{ln}\left(\mathrm{2}\pi\right)}{\mathrm{2}{n}}\:+\frac{{ln}\left({n}\right)}{\mathrm{2}{n}}} \rightarrow\mathrm{1}\:\:\left({n}\rightarrow\infty\right)\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow\infty} \:\left(\frac{{n}!}{{n}^{{n}} }\right)^{\frac{\mathrm{1}}{{n}}} \:=\:\frac{\mathrm{1}}{{e}}\:. \\ $$