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Question Number 131328 by rs4089 last updated on 03/Feb/21

Commented by MJS_new last updated on 04/Feb/21

$$\mathrm{really}?$$$$\mathrm{just}\mathrm{try}\mathrm{some}\mathrm{values}\mathrm{with}n=k^2\wedge k\in \mathbb{N}$$$$y_k=\frac{\left(2k^2\right)!}{2^k}$$$$y_1=1$$$$y_2=10080$$$$y_3=800296713216000$$$$...$$$$y_{10}\approx 7.70\times {10}^{371}$$$$...$$$$\Rightarrow \mathrm{the}\mathrm{answer}\mathrm{is}+\mathrm{\infty }$$

Commented by liberty last updated on 04/Feb/21

$$\mathrm{wrong}\mathrm{answer}0$$

Answered by mathmax by abdo last updated on 03/Feb/21

$${\mathrm{u}}_{\mathrm{n}}=\frac{\left(2\mathrm{n}\right)!}{2^{\sqrt{\mathrm{n}}}}\mathrm{we}\mathrm{have}\mathrm{n}!\sim {\mathrm{n}}^{\mathrm{n}}{\mathrm{e}}^{-\mathrm{n}}\sqrt{2\pi \mathrm{n}}\left(\mathrm{stirling}\mathrm{formula}\right)\Rightarrow$$$$\left(2\mathrm{n}\right)!\sim {\left(2\mathrm{n}\right)}^{2\mathrm{n}}{\mathrm{e}}^{-2\mathrm{n}}\sqrt{4\pi \mathrm{n}}\Rightarrow {\mathrm{u}}_{\mathrm{n}}\sim \frac{2^{2\mathrm{n}}{\mathrm{n}}^{2\mathrm{n}}{\mathrm{e}}^{-2\mathrm{n}}2\sqrt{\pi \mathrm{n}}}{2^{\sqrt{\mathrm{n}}}}\Rightarrow$$$${\mathrm{u}}_{\mathrm{n}}\sim {\mathrm{n}}^{2\mathrm{n}}.{\mathrm{e}}^{-2\mathrm{n}}.2^{2\mathrm{n}-\sqrt{\mathrm{n}}}\sqrt{\pi \mathrm{n}}\Rightarrow$$$$\ln \left({\mathrm{u}}_{\mathrm{n}}\right)=2\mathrm{nln}\left(\mathrm{n}\right)-2\mathrm{n}+((\mathrm{n}-\sqrt{\mathrm{n}})\ln \left(2\right)+\frac{1}{2}\ln \left(\mathrm{n}\pi \right)\Rightarrow$$$$\ln \left({\mathrm{u}}_{\mathrm{n}}\right)=2\mathrm{n}\{\frac{\ln \left(\mathrm{n}\right)}{2\mathrm{n}}-1\frac{(\mathrm{n}-\sqrt{\mathrm{n}})\ln 2}{2\mathrm{n}}+\frac{\ln \left(\mathrm{n}\pi \right)}{4\mathrm{n}}\}\sim 2\mathrm{n}\{-1+\frac{1}{2}\}\sim -\mathrm{n}\Rightarrow$$$${\lim }_{\mathrm{n}\to +\mathrm{\infty }}\ln \left({\mathrm{u}}_{\mathrm{n}}\right)=-\mathrm{\infty }\Rightarrow {\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{u}}_{\mathrm{n}}=0$$

Answered by liberty last updated on 04/Feb/21

$$\frac{\left(2\mathrm{n}\right)!}{2^{\sqrt{\mathrm{n}}}}⩾\frac{2^{\mathrm{n}}}{2^{\sqrt{\mathrm{n}}}}=2^{\mathrm{n}-\sqrt{\mathrm{n}}}$$$$\mathrm{this}\mathrm{follows}\mathrm{from}\left(2\mathrm{n}\right)!\mathrm{having}\mathrm{factors}2,4,6,8,...,2\mathrm{n}$$$$\mathrm{because}\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }(\mathrm{n}-\sqrt{\mathrm{n}})=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\sqrt{\mathrm{n}}(\sqrt{\mathrm{n}}-1)=\mathrm{\infty }$$$$\mathrm{we}\mathrm{have}{\underset{\mathrm{n}\to \mathrm{\infty }}{\lim 2}}^{\mathrm{n}-\sqrt{\mathrm{n}}}=\mathrm{\infty }.$$$$\mathrm{so}\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{\left(2\mathrm{n}\right)!}{2^{\sqrt{\mathrm{n}}}}=\mathrm{\infty }$$

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