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Question Number 133928 by mnjuly1970 last updated on 25/Feb/21

Answered by mathmax by abdo last updated on 25/Feb/21

$$2){\mathrm{u}}_{\mathrm{n}}={\left({\mathrm{C}}_{2\mathrm{n}}^{\mathrm{n}}\right)}^{\frac{1}{\mathrm{n}}}\mathrm{we}\mathrm{have}{\mathrm{C}}_{2\mathrm{n}}^{\mathrm{n}}=\frac{\left(2\mathrm{n}\right)!}{{(\mathrm{n}!)}^2}\Rightarrow$$$${\mathrm{u}}_{\mathrm{n}}={\mathrm{e}}^{\frac{1}{\mathrm{n}}\log \left(\frac{\left(2\mathrm{n}\right)!}{{(\mathrm{n}!)}^2}\right)}\mathrm{we}\mathrm{have}\mathrm{n}!\sim {\mathrm{n}}^{\mathrm{n}}{\mathrm{e}}^{-\mathrm{n}}\sqrt{2\pi \mathrm{n}}$$$$\Rightarrow {(\mathrm{n}!)}^2\sim {\mathrm{n}}^{2\mathrm{n}}{\mathrm{e}}^{-2\mathrm{n}}\left(2\pi \mathrm{n}\right)\mathrm{and}\left(2\mathrm{n}\right)!\sim {\left(2\mathrm{n}\right)}^{2\mathrm{n}}{\mathrm{e}}^{-2\mathrm{n}}\sqrt{4\pi \mathrm{n}}$$$$=2^{2\mathrm{n}}.{\mathrm{n}}^{2\mathrm{n}}{\mathrm{e}}^{-2\mathrm{n}}\sqrt{4\pi \mathrm{n}}\Rightarrow \frac{\left(2\mathrm{n}\right)!}{{(\mathrm{n}!)}^2}\sim \frac{2^{2\mathrm{n}}.{\mathrm{n}}^{2\mathrm{n}}.{\mathrm{e}}^{-2\mathrm{n}}}{{\mathrm{n}}^{2\mathrm{n}}{\mathrm{e}}^{-2\mathrm{n}}\left(2\pi \mathrm{n}\right)}\sqrt{4\pi \mathrm{n}}$$$$=2^{2\mathrm{n}}\times \frac{1}{\sqrt{\pi \mathrm{n}}}\Rightarrow \log \left(\frac{\left(2\mathrm{n}\right)!}{{(\mathrm{n}!)}^2}\right)\sim 2\mathrm{nlog}\left(2\right)-\frac{1}{2}\log \left(\mathrm{n}\pi \right)\Rightarrow$$$$\frac{1}{\mathrm{n}}\log (...)\sim 2\log \left(2\right)-\frac{\log \left(\mathrm{n}\pi \right)}{2\mathrm{n}}\to \log \left(4\right)\Rightarrow {\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{u}}_{\mathrm{n}}={\mathrm{e}}^{\log \left(4\right)}=4$$

Commented by mnjuly1970 last updated on 26/Feb/21

$$thanksalotsirmax$$

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