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

let U_n = (((∫_0 ^n  e^(−x^2 ) dx)^2 )/(∫_0 ^n   e^(−nx^2 ) dx))  1) calculate lim_(n→+∞)   U_n   2) determne nature of Σ  U_n   and Σ U_n ^3  .

$${let}\:{U}_{{n}} =\:\frac{\left(\int_{\mathrm{0}} ^{{n}} \:{e}^{−{x}^{\mathrm{2}} } {dx}\right)^{\mathrm{2}} }{\int_{\mathrm{0}} ^{{n}} \:\:{e}^{−{nx}^{\mathrm{2}} } {dx}} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{lim}_{{n}\rightarrow+\infty} \:\:{U}_{{n}} \\ $$$$\left.\mathrm{2}\right)\:{determne}\:{nature}\:{of}\:\Sigma\:\:{U}_{{n}} \:\:{and}\:\Sigma\:{U}_{{n}} ^{\mathrm{3}} \:. \\ $$

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

we have ∫_0 ^n  e^(−nx^2 ) dx =_((√n)x=t)      ∫_0 ^(n(√n))  e^(−t^2 )  (dt/(√n)) =(1/(√n)) ∫_0 ^(n(√n))  e^(−t^2 ) dt ⇒  U_n =(√n)  (((∫_0 ^n  e^(−x^2 ) dx)^2 )/(∫_0 ^(n(√n)) e^(−t^2 ) dt))  but lim_(n→+∞)      (√n)=+∞  lim_(n→+∞)   (((∫_0 ^n  e^(−x^2 ) dx)^2 )/(∫_0 ^(n(√n)) e^(−t^2 ) dt)) =(((((√π)/2))^2 )/((√π)/2)) =((√π)/2) ⇒lim_(n→+∞)  U_n =+∞  2) we have  U_n  ∼ ((√(πn))/2)  ⇒Σ U_n   diverges also Σ U_n ^3   diverges.

$${we}\:{have}\:\int_{\mathrm{0}} ^{{n}} \:{e}^{−{nx}^{\mathrm{2}} } {dx}\:=_{\sqrt{{n}}{x}={t}} \:\:\:\:\:\int_{\mathrm{0}} ^{{n}\sqrt{{n}}} \:{e}^{−{t}^{\mathrm{2}} } \:\frac{{dt}}{\sqrt{{n}}}\:=\frac{\mathrm{1}}{\sqrt{{n}}}\:\int_{\mathrm{0}} ^{{n}\sqrt{{n}}} \:{e}^{−{t}^{\mathrm{2}} } {dt}\:\Rightarrow \\ $$$${U}_{{n}} =\sqrt{{n}}\:\:\frac{\left(\int_{\mathrm{0}} ^{{n}} \:{e}^{−{x}^{\mathrm{2}} } {dx}\right)^{\mathrm{2}} }{\int_{\mathrm{0}} ^{{n}\sqrt{{n}}} {e}^{−{t}^{\mathrm{2}} } {dt}}\:\:{but}\:{lim}_{{n}\rightarrow+\infty} \:\:\:\:\:\sqrt{{n}}=+\infty \\ $$$${lim}_{{n}\rightarrow+\infty} \:\:\frac{\left(\int_{\mathrm{0}} ^{{n}} \:{e}^{−{x}^{\mathrm{2}} } {dx}\right)^{\mathrm{2}} }{\int_{\mathrm{0}} ^{{n}\sqrt{{n}}} {e}^{−{t}^{\mathrm{2}} } {dt}}\:=\frac{\left(\frac{\sqrt{\pi}}{\mathrm{2}}\right)^{\mathrm{2}} }{\frac{\sqrt{\pi}}{\mathrm{2}}}\:=\frac{\sqrt{\pi}}{\mathrm{2}}\:\Rightarrow{lim}_{{n}\rightarrow+\infty} \:{U}_{{n}} =+\infty \\ $$$$\left.\mathrm{2}\right)\:{we}\:{have}\:\:{U}_{{n}} \:\sim\:\frac{\sqrt{\pi{n}}}{\mathrm{2}}\:\:\Rightarrow\Sigma\:{U}_{{n}} \:\:{diverges}\:{also}\:\Sigma\:{U}_{{n}} ^{\mathrm{3}} \:\:{diverges}. \\ $$

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

I_1 =∫_0 ^∞ e^(−x^2 ) dx  and I_2 =∫_0 ^∞ e^(−nx^2 ) dx  calculating I_2   ∫_0 ^∞ e^(−nx^2 ) dx  t=nx^2   x=((√t)/(√n))   (dx/dt)=(1/(√n))×(1/2)×t^((1/2)−1) =(1/(2(√n) ))×t^((−1)/2)   ∫_0 ^∞ e^(−t) ×(1/(2(√n)))×t^((−1)/2) dt  =(1/(2(√n)))∫_0 ^∞ e^(−t) ×t^((1/2)−1) dt  =(1/(2(√n)))×⌈((1/2))=(1/(2(√n)))×(√π) =(1/2)×(√(π/n)) →value of I_2   I_1 =(1/2)×(√π)   lim_(n→∞) U_n =(I_1 ^2 /I_2 )=((π/4)/((1/2)×(√(π/n))))=((π×2×(√n))/(4×(√π)))=((√(nπ))/2)  sir pls che4k...

$${I}_{\mathrm{1}} =\int_{\mathrm{0}} ^{\infty} {e}^{−{x}^{\mathrm{2}} } {dx}\:\:{and}\:{I}_{\mathrm{2}} =\int_{\mathrm{0}} ^{\infty} {e}^{−{nx}^{\mathrm{2}} } {dx} \\ $$$${calculating}\:{I}_{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\infty} {e}^{−{nx}^{\mathrm{2}} } {dx} \\ $$$${t}={nx}^{\mathrm{2}} \:\:{x}=\frac{\sqrt{{t}}}{\sqrt{{n}}}\:\:\:\frac{{dx}}{{dt}}=\frac{\mathrm{1}}{\sqrt{{n}}}×\frac{\mathrm{1}}{\mathrm{2}}×{t}^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}\sqrt{{n}}\:}×{t}^{\frac{−\mathrm{1}}{\mathrm{2}}} \\ $$$$\int_{\mathrm{0}} ^{\infty} {e}^{−{t}} ×\frac{\mathrm{1}}{\mathrm{2}\sqrt{{n}}}×{t}^{\frac{−\mathrm{1}}{\mathrm{2}}} {dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{n}}}\int_{\mathrm{0}} ^{\infty} {e}^{−{t}} ×{t}^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} {dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{n}}}×\lceil\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{n}}}×\sqrt{\pi}\:=\frac{\mathrm{1}}{\mathrm{2}}×\sqrt{\frac{\pi}{{n}}}\:\rightarrow{value}\:{of}\:{I}_{\mathrm{2}} \\ $$$${I}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}×\sqrt{\pi}\: \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{U}_{{n}} =\frac{{I}_{\mathrm{1}} ^{\mathrm{2}} }{{I}_{\mathrm{2}} }=\frac{\frac{\pi}{\mathrm{4}}}{\frac{\mathrm{1}}{\mathrm{2}}×\sqrt{\frac{\pi}{{n}}}}=\frac{\pi×\mathrm{2}×\sqrt{{n}}}{\mathrm{4}×\sqrt{\pi}}=\frac{\sqrt{{n}\pi}}{\mathrm{2}} \\ $$$${sir}\:{pls}\:{che}\mathrm{4}{k}... \\ $$

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