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Question Number 53464 by maxmathsup by imad last updated on 22/Jan/19

$$let{U}_n=\frac{{\left({\int }_0^n{e}^{-x^2}dx\right)}^2}{{\int }_0^n{e}^{-{nx}^2}dx}$$$$1)calculate{lim}_{n\to +\mathrm{\infty }}{U}_n$$$$2)determnenatureof\mathrm{\Sigma }{U}_{n}and\mathrm{\Sigma }{U}_n^3.$$

Commented by maxmathsup by imad last updated on 24/Jan/19

$$wehave{\int }_0^n{e}^{-{nx}^2}dx{=}_{\sqrt{n}x=t}{\int }_0^{n\sqrt{n}}{e}^{-t^2}\frac{dt}{\sqrt{n}}=\frac{1}{\sqrt{n}}{\int }_0^{n\sqrt{n}}{e}^{-t^2}dt\Rightarrow$$$$U_n=\sqrt{n}\frac{{\left({\int }_0^n{e}^{-x^2}dx\right)}^2}{{\int }_0^{n\sqrt{n}}{e}^{-t^2}dt}but{lim}_{n\to +\mathrm{\infty }}\sqrt{n}=+\mathrm{\infty }$$$${lim}_{n\to +\mathrm{\infty }}\frac{{\left({\int }_0^n{e}^{-x^2}dx\right)}^2}{{\int }_0^{n\sqrt{n}}{e}^{-t^2}dt}=\frac{{\left(\frac{\sqrt{\pi }}{2}\right)}^2}{\frac{\sqrt{\pi }}{2}}=\frac{\sqrt{\pi }}{2}\Rightarrow {lim}_{n\to +\mathrm{\infty }}{U}_n=+\mathrm{\infty }$$$$2)wehave{U}_n\sim \frac{\sqrt{\pi n}}{2}\Rightarrow \mathrm{\Sigma }{U}_{n}divergesalso\mathrm{\Sigma }{U}_n^{3}diverges.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 22/Jan/19

$$I_1={\int }_0^{\mathrm{\infty }}{e}^{-x^2}dxand{I}_2={\int }_0^{\mathrm{\infty }}{e}^{-{nx}^2}dx$$$$calculating{I}_2$$$${\int }_0^{\mathrm{\infty }}{e}^{-{nx}^2}dx$$$$t={nx}^{2}x=\frac{\sqrt{t}}{\sqrt{n}}\frac{dx}{dt}=\frac{1}{\sqrt{n}}\times \frac{1}{2}\times t^{\frac{1}{2}-1}=\frac{1}{2\sqrt{n}}\times t^{\frac{-1}{2}}$$$${\int }_0^{\mathrm{\infty }}{e}^{-t}\times \frac{1}{2\sqrt{n}}\times t^{\frac{-1}{2}}dt$$$$=\frac{1}{2\sqrt{n}}{\int }_0^{\mathrm{\infty }}{e}^{-t}\times t^{\frac{1}{2}-1}dt$$$$=\frac{1}{2\sqrt{n}}\times \lceil \left(\frac{1}{2}\right)=\frac{1}{2\sqrt{n}}\times \sqrt{\pi }=\frac{1}{2}\times \sqrt{\frac{\pi }{n}}\to valueof{I}_2$$$$I_1=\frac{1}{2}\times \sqrt{\pi }$$$$\underset{n\to \mathrm{\infty }}{\lim }{U}_n=\frac{I_1^2}{I_2}=\frac{\frac{\pi }{4}}{\frac{1}{2}\times \sqrt{\frac{\pi }{n}}}=\frac{\pi \times 2\times \sqrt{n}}{4\times \sqrt{\pi }}=\frac{\sqrt{n\pi }}{2}$$$$sirplsche4k...$$

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