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Question Number 29511 by abdo imad last updated on 09/Feb/18

$$find{lim}_{n\to +\mathrm{\infty }}\frac{{(n!)}^2}{\left(2n\right)!}.$$

Commented by prof Abdo imad last updated on 12/Feb/18

$$letusestirlingformulawehave$$$$n!\sim n^n{e}^{-n}\sqrt{2\pi n}\Rightarrow {(n!)}^2\sim n^{2n}{e}^{-2n}\left(2\pi \right)n$$$$\left(2n\right)!\sim {\left(2n\right)}^{2n}{e}^{-2n}\sqrt{4\pi n}=2^{2n}{n}^{2n}{e}^{-2n}2\sqrt{\pi n}\Rightarrow$$$$\frac{{(n!)}^2}{\left(2n\right)!}=\frac{n^{2n}{e}^{-2n}\left(2\pi \right)n}{2^{2n}{n}^{2n}{e}^{-2n}2\sqrt{\pi n}}=\frac{2\pi }{2\sqrt{\pi }}\frac{\sqrt{n}}{4^n}=\sqrt{\pi }\frac{\sqrt{n}}{4^n}=v_n$$$$ln\left(v_n\right)=ln\left(\sqrt{\pi }\right)+\frac{1}{2}ln\left(n\right)+2nln\left(2\right)\to +\mathrm{\infty }\Rightarrow$$$${lim}_{n\to \mathrm{\infty }}{v}_n=\mathrm{\infty }\Rightarrow {lim}_{n\to \mathrm{\infty }}\frac{{(n!)}^2}{\left(2n\right)!}=+\mathrm{\infty }.$$

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