Question Number 182646 by mathlove last updated on 12/Dec/22
$${prove}\:{that} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left[\frac{\sqrt[{{n}+\mathrm{1}}]{\left({n}+\mathrm{1}\right)!\centerdot\left(\mathrm{2}{n}+\mathrm{1}\right)!!}}{{n}+\mathrm{1}}−\frac{\sqrt[{{n}}]{{n}!\centerdot\left(\mathrm{2}{n}−\mathrm{1}\right)!!}}{{n}}\right]=\frac{\mathrm{2}}{{e}^{\mathrm{2}} } \\ $$
Commented by mathlove last updated on 13/Dec/22
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Commented by mathlove last updated on 13/Dec/22
$${prove}\:{that} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left[\frac{\sqrt[{{n}+\mathrm{1}}]{\left({n}+\mathrm{1}\right)!\centerdot\left(\mathrm{2}{n}+\mathrm{1}\right)!!}}{{n}+\mathrm{1}}−\frac{\sqrt[{{n}}]{{n}!\centerdot\left(\mathrm{2}{n}−\mathrm{1}\right)!!}}{{n}}\right]=\frac{\mathrm{2}}{{e}^{\mathrm{2}} } \\ $$