Question Number 195259 by mathlove last updated on 28/Jul/23
$${prove}\:{that} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{{e}^{{n}} \centerdot\left({n}!\right)}{{n}^{{n}} \:\sqrt{{n}}}=\sqrt{\mathrm{2}\pi} \\ $$
Commented by Frix last updated on 28/Jul/23
$$\mathrm{Easy}\:\mathrm{because}\:\mathrm{for}\:{n}\rightarrow\infty:\:{n}!\rightarrow\frac{{n}^{{n}} }{\mathrm{e}^{{n}} }\sqrt{\mathrm{2}\pi{n}} \\ $$