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Question Number 104701 by abony1303 last updated on 23/Jul/20

Commented by abony1303 last updated on 23/Jul/20

$$\mathrm{Pls}\mathrm{help}$$

Answered by mathmax by abdo last updated on 23/Jul/20

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

Commented by abony1303 last updated on 23/Jul/20

$$\mathrm{couldnt}\mathrm{catch}\mathrm{you}\mathrm{ser}:($$

Commented by mathmax by abdo last updated on 23/Jul/20

$$\mathrm{what}\mathrm{do}\mathrm{you}\mathrm{mean}\mathrm{with}\mathrm{this}...$$

Commented by abony1303 last updated on 23/Jul/20

$${\mathrm{didn}}^{\prime }\mathrm{t}\mathrm{understand}\mathrm{at}\mathrm{some}\mathrm{points}\mathrm{from}$$$$\mathrm{beginning}.\mathrm{can}\mathrm{u}\mathrm{pls}\mathrm{write}\mathrm{some}$$$$\mathrm{descriptions}?$$

Commented by abony1303 last updated on 24/Jul/20

$$\mathrm{Thank}\mathrm{you}\mathrm{ser}.\mathrm{It}\mathrm{was}\mathrm{very}\mathrm{helpful}.$$$${\mathrm{Didnt}}^{\prime }\mathrm{t}\mathrm{know}\mathrm{about}{\mathrm{Sterling}}^{\prime }\mathrm{s}$$$$\mathrm{approximation}.$$

Commented by mathmax by abdo last updated on 24/Jul/20

$$\mathrm{here}\mathrm{i}\mathrm{have}\mathrm{used}\mathrm{th}\mathrm{stirling}\mathrm{formula}\mathrm{n}!\sim {\mathrm{n}}^{\mathrm{n}}{\mathrm{e}}^{-\mathrm{n}}\sqrt{2\pi \mathrm{n}}(\mathrm{for}\mathrm{n}\to +\mathrm{\infty })$$$$\mathrm{i}\mathrm{think}\mathrm{the}\mathrm{answer}\mathrm{is}\mathrm{clear}$$$$$$

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