Question Number 206845 by hardmath last updated on 27/Apr/24
$$\mathrm{Find}:\:\:\:\frac{\infty!}{\infty^{\infty} }\:=\:? \\ $$
Commented by A5T last updated on 27/Apr/24
$${Do}\:{you}\:{wish}\:{to}\:{find}\:{this}:\:\underset{{n}\rightarrow\infty} {{lim}}\frac{{n}!}{{n}^{{n}} }\:? \\ $$
Commented by hardmath last updated on 27/Apr/24
$$\mathrm{Yes}\:\mathrm{Ser} \\ $$
Commented by A5T last updated on 27/Apr/24
$$\mathrm{0} \\ $$
Answered by Frix last updated on 27/Apr/24
![lim_(n→∞) ((n!)/n^n ) =_(Formula]) ^([Sterling′s) lim_(n→∞) (((n^n /e^n )(√(2πn)))/n^n ) =lim_(n→∞) ((√(2πn))/e^n ) =0](https://www.tinkutara.com/question/Q206855.png)
$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{{n}!}{{n}^{{n}} }\:\:\underset{\left.\mathrm{Formula}\right]} {\overset{\left[\mathrm{Sterling}'\mathrm{s}\right.} {=}}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\frac{{n}^{{n}} }{\mathrm{e}^{{n}} }\sqrt{\mathrm{2}\pi{n}}}{{n}^{{n}} }\:=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\sqrt{\mathrm{2}\pi{n}}}{\mathrm{e}^{{n}} }\:=\mathrm{0} \\ $$
Commented by hardmath last updated on 28/Apr/24
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{Ser} \\ $$
Commented by MM42 last updated on 30/Apr/24
$$\forall{n}>\mathrm{2}\:\rightarrow\mathrm{0}<\frac{{n}!}{{n}^{{n}} }<\frac{{n}!}{\left({n}+\mathrm{1}\right)!}=\frac{\mathrm{1}}{{n}+\mathrm{1}} \\ $$$$\Rightarrow{lim}_{{n}\rightarrow\infty} \:\frac{{n}!}{{n}^{{n}} }=\mathrm{0}\:\checkmark \\ $$