Question Number 167295 by mnjuly1970 last updated on 12/Mar/22
Answered by qaz last updated on 12/Mar/22
$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\left(\mathrm{2n}\right)!}{\mathrm{n}^{\mathrm{n}} \mathrm{n}!}\right)^{\mathrm{1}/\mathrm{n}} \\ $$$$=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\frac{\left(\mathrm{2n}\right)!}{\mathrm{n}^{\mathrm{n}} \mathrm{n}!}\centerdot\frac{\left(\mathrm{n}−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \left(\mathrm{n}−\mathrm{1}\right)!}{\left(\mathrm{2n}−\mathrm{2}\right)!} \\ $$$$=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{2}\left(\mathrm{2n}−\mathrm{1}\right)}{\mathrm{n}−\mathrm{1}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{n}}\right)^{\mathrm{n}} \\ $$$$=\mathrm{4e}^{−\mathrm{1}} \\ $$
Answered by Mathspace last updated on 13/Mar/22
$${n}!\sim\:{n}^{{n}} {e}^{−{n}} \sqrt{\mathrm{2}\pi{n}} \\ $$$$\left(\mathrm{2}{n}\right)!\sim\left(\mathrm{2}{n}\right)^{\mathrm{2}{n}\:} {e}^{−\mathrm{2}{n}} \sqrt{\mathrm{4}\pi{n}}\:\Rightarrow \\ $$$$\frac{\left(\mathrm{2}{n}\right)!}{{n}^{{n}} .{n}!}=\frac{\mathrm{4}^{{n}} {n}^{\mathrm{2}{n}} \:{e}^{−\mathrm{2}{n}} \mathrm{2}\sqrt{\pi{n}}}{{n}^{{n}} {n}^{{n}} \:{e}^{−{n}} \sqrt{\mathrm{2}\pi{n}}} \\ $$$$=\sqrt{\mathrm{2}}.\mathrm{4}^{{n}} \:.{e}^{−{n}} \:\Rightarrow \\ $$$$\left(\frac{\left(\mathrm{2}{n}\right)!}{{n}^{{n}} {n}!}\right)^{\frac{\mathrm{1}}{{n}}} \sim\left(\sqrt{\mathrm{2}}\right)^{\frac{\mathrm{1}}{{n}}} .\mathrm{4}.{e}^{−\mathrm{1}} \rightarrow\frac{\mathrm{4}}{{e}}\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow+\infty} \left(\frac{\left(\mathrm{2}{n}\right)!}{{n}^{{n}} {n}!}\right)^{\frac{\mathrm{1}}{{n}}} =\frac{\mathrm{4}}{{e}} \\ $$