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Question Number 111772 by mathmax by abdo last updated on 04/Sep/20

$$\mathrm{calculate}{\lim }_{\mathrm{n}\to +\mathrm{\infty }}\prod _{\mathrm{k}=1}^{\mathrm{n}}(1+\frac{\sqrt{\mathrm{k}(\mathrm{n}-\mathrm{k})}}{{\mathrm{n}}^2})$$

Answered by mathmax by abdo last updated on 07/Sep/20

$$\mathrm{let}{\mathrm{A}}_{\mathrm{n}}=\prod _{\mathrm{k}=1}^{\mathrm{n}}(1+\frac{\sqrt{\mathrm{k}(\mathrm{n}-\mathrm{k})}}{{\mathrm{n}}^2})\Rightarrow \ln \left({\mathrm{A}}_{\mathrm{n}}\right)=\sum _{\mathrm{k}=1}^{\mathrm{n}}\ln (1+\frac{\sqrt{\mathrm{k}(\mathrm{n}-\mathrm{k})}}{{\mathrm{n}}^2})$$$$\mathrm{we}\mathrm{have}\frac{\mathrm{d}}{\mathrm{dx}}\ln (1+\mathrm{x})=\frac{1}{1+\mathrm{x}}=1-\mathrm{x}+{\mathrm{x}}^2-...\mathrm{for}\mid \mathrm{x}\mid <1\Rightarrow$$$$\ln (1+\mathrm{x})=\mathrm{x}-\frac{{\mathrm{x}}^2}{2}+\frac{{\mathrm{x}}^3}{3}-...\Rightarrow \mathrm{x}-\frac{{\mathrm{x}}^2}{2}⩽\ln (1+\mathrm{x})⩽\mathrm{x}\Rightarrow$$$$\frac{\sqrt{\mathrm{k}(\mathrm{n}-\mathrm{k})}}{{\mathrm{n}}^2}-\frac{\mathrm{k}(\mathrm{n}-\mathrm{k})}{{\mathrm{n}}^4}⩽\ln (1+\frac{\sqrt{\mathrm{k}(\mathrm{n}-\mathrm{k})}}{{\mathrm{n}}^2})⩽\frac{\sqrt{\mathrm{k}(\mathrm{n}-\mathrm{k})}}{{\mathrm{n}}^2}\Rightarrow$$$$\frac{1}{{\mathrm{n}}^2}\sum _{\mathrm{k}=1}^{\mathrm{n}}\sqrt{\mathrm{k}(\mathrm{n}-\mathrm{k})}-\frac{1}{{\mathrm{n}}^4}\sum _{\mathrm{k}=1}^{\mathrm{n}}\mathrm{k}(\mathrm{n}-\mathrm{k})⩽\ln \left({\mathrm{A}}_{\mathrm{n}}\right)⩽\frac{1}{{\mathrm{n}}^2}\sum _{\mathrm{k}=1}^{\mathrm{n}}\sqrt{\mathrm{k}(\mathrm{n}-\mathrm{k})}$$$$\frac{1}{{\mathrm{n}}^4}\sum _{\mathrm{k}=1}^{\mathrm{n}}(\mathrm{nk}-{\mathrm{k}}^2)=\frac{1}{{\mathrm{n}}^3}\sum _{\mathrm{k}=1}^{\mathrm{n}}\mathrm{k}-\frac{1}{{\mathrm{n}}^4}\sum _{\mathrm{k}=1}^{\mathrm{n}}{\mathrm{k}}^2$$$$=\frac{1}{{\mathrm{n}}^3}\times \frac{\mathrm{n}(\mathrm{n}+1)}{2}-\frac{1}{{\mathrm{n}}^4}\frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{6}\sim \frac{1}{2\mathrm{n}}-\frac{2}{\mathrm{n}}\to 0(\mathrm{n}\to \mathrm{\infty })$$$${\lim }_{\mathrm{n}\to +\mathrm{\infty }}\frac{1}{{\mathrm{n}}^2}\sum _{\mathrm{k}=1}^{\mathrm{n}}\sqrt{\mathrm{k}(\mathrm{n}-\mathrm{k})}={\lim }_{\mathrm{n}\to +\mathrm{\infty }}\frac{1}{\mathrm{n}}\sum _{\mathrm{k}=1}^{\mathrm{n}}\sqrt{\frac{\mathrm{k}(\mathrm{n}-\mathrm{k})}{{\mathrm{n}}^2}}$$$$={\lim }_{\mathrm{n}\to +\mathrm{\infty }}\frac{1}{\mathrm{n}}\sum _{\mathrm{k}=1}^{\mathrm{n}}\sqrt{\frac{\mathrm{k}}{\mathrm{n}}-{\left(\frac{\mathrm{k}}{\mathrm{n}}\right)}^2}={\int }_0^1\sqrt{\mathrm{x}-{\mathrm{x}}^2}\mathrm{dx}$$$$={\int }_0^1\sqrt{\mathrm{x}}\times \sqrt{1-\mathrm{x}}\mathrm{dx}{=}_{\mathrm{x}={\sin }^2\theta }{\int }_0^{\frac{\pi }{2}}\sin \theta \times \cos \theta \left(2\sin \theta \cos \theta \right)\mathrm{d}\theta$$$$=2{\int }_0^{\frac{\pi }{2}}{\left(\sin \theta \cos \theta \right)}^2\mathrm{d}\theta =2{\int }_0^{\frac{\pi }{2}}{\left(\frac{1}{2}\sin \left(2\theta \right)\right)}^2\mathrm{d}\theta$$$$=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\frac{1-\cos \left(4\theta \right)}{2}\mathrm{d}\theta =\frac{1}{4}\times \frac{\pi }{2}-\frac{1}{16}{\left[\sin \left(4\theta \right)\right]}_0^{\frac{\pi }{2}}=\frac{\pi }{8}\Rightarrow$$$${\lim }_{\mathrm{n}\to +\mathrm{\infty }}\ln \left({\mathrm{A}}_{\mathrm{n}}\right)=\frac{\pi }{8}\Rightarrow {\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{A}}_{\mathrm{n}}={\mathrm{e}}^{\frac{\pi }{8}}$$$$$$

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