Question Number 65004 by mathmax by abdo last updated on 24/Jul/19
$${let}\:{U}_{{n}} =\:\int_{\frac{\mathrm{1}}{{n}}} ^{\frac{\mathrm{2}}{{n}}} \:\Gamma\left({x}\right)\Gamma\left(\mathrm{1}−{x}\right){dx}\:\:\:\:{with}\:{n}\geqslant\mathrm{3} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{and}\:{determine}\:{lim}_{{n}\rightarrow+\infty} \:{U}_{{n}} \\ $$$$\left.\mathrm{2}\right)\:{study}\:{the}\:{convergence}\:{of}\:\Sigma\:{U}_{{n}} \\ $$
Commented by mathmax by abdo last updated on 24/Jul/19
![1) we have Γ(x).Γ(1−x) =(π/(sin(πx))) ⇒U_n =∫_(1/n) ^(2/n) (π/(sin(πx)))dx =_(πx =t) π ∫_(π/n) ^((2π)/n) (dt/(π sint)) =∫_(π/n) ^((2π)/n) (dt/(sint)) =_(tan((t/2))=u) ∫_(tan((π/(2n)))) ^(tan((π/n))) (1/((2u)/(1+u^2 ))) ((2du)/(1+u^2 )) = ∫_(tan((π/(2n)))) ^(tan((π/n))) (du/u) =[ln∣u∣]_(tan((π/(2n)))) ^(tan((π/n))) =ln∣((tan((π/n)))/(tan((π/(2n)))))∣ ⇒ U_n =ln∣((tan((π/n)))/(tan((π/(2n)))))∣ we have tan((π/n)) ∼(π/n) and tan((π/(2n))) ∼ (π/(2n)) ⇒ ((tan((π/n)))/(tan((π/(2n ))))) ∼(π/n) ((2n)/π) =2 ⇒lim_(n→+∞) U_n =ln(2) 2) lim_(n→+∞) U_n ≠0 ⇒Σ U_n diverges.](https://www.tinkutara.com/question/Q65079.png)
$$\left.\mathrm{1}\right)\:{we}\:{have}\:\Gamma\left({x}\right).\Gamma\left(\mathrm{1}−{x}\right)\:=\frac{\pi}{{sin}\left(\pi{x}\right)}\:\Rightarrow{U}_{{n}} =\int_{\frac{\mathrm{1}}{{n}}} ^{\frac{\mathrm{2}}{{n}}} \:\frac{\pi}{{sin}\left(\pi{x}\right)}{dx} \\ $$$$=_{\pi{x}\:={t}} \:\:\:\:\pi\:\int_{\frac{\pi}{{n}}} ^{\frac{\mathrm{2}\pi}{{n}}} \:\:\:\frac{{dt}}{\pi\:{sint}}\:=\int_{\frac{\pi}{{n}}} ^{\frac{\mathrm{2}\pi}{{n}}} \:\:\frac{{dt}}{{sint}}\:=_{{tan}\left(\frac{{t}}{\mathrm{2}}\right)={u}} \:\:\:\:\int_{{tan}\left(\frac{\pi}{\mathrm{2}{n}}\right)} ^{{tan}\left(\frac{\pi}{{n}}\right)} \:\:\frac{\mathrm{1}}{\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }}\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=\:\int_{{tan}\left(\frac{\pi}{\mathrm{2}{n}}\right)} ^{{tan}\left(\frac{\pi}{{n}}\right)} \:\:\:\frac{{du}}{{u}}\:=\left[{ln}\mid{u}\mid\right]_{{tan}\left(\frac{\pi}{\mathrm{2}{n}}\right)} ^{{tan}\left(\frac{\pi}{{n}}\right)} \:={ln}\mid\frac{{tan}\left(\frac{\pi}{{n}}\right)}{{tan}\left(\frac{\pi}{\mathrm{2}{n}}\right)}\mid\:\Rightarrow\:{U}_{{n}} ={ln}\mid\frac{{tan}\left(\frac{\pi}{{n}}\right)}{{tan}\left(\frac{\pi}{\mathrm{2}{n}}\right)}\mid \\ $$$${we}\:\:{have}\:\:\:{tan}\left(\frac{\pi}{{n}}\right)\:\sim\frac{\pi}{{n}}\:\:\:{and}\:{tan}\left(\frac{\pi}{\mathrm{2}{n}}\right)\:\sim\:\frac{\pi}{\mathrm{2}{n}}\:\:\Rightarrow \\ $$$$\frac{{tan}\left(\frac{\pi}{{n}}\right)}{{tan}\left(\frac{\pi}{\mathrm{2}{n}\:}\right)}\:\sim\frac{\pi}{{n}}\:\frac{\mathrm{2}{n}}{\pi}\:=\mathrm{2}\:\Rightarrow{lim}_{{n}\rightarrow+\infty} \:\:{U}_{{n}} ={ln}\left(\mathrm{2}\right) \\ $$$$\left.\mathrm{2}\right)\:{lim}_{{n}\rightarrow+\infty} \:{U}_{{n}} \:\neq\mathrm{0}\:\:\Rightarrow\Sigma\:{U}_{{n}} \:\:\:{diverges}. \\ $$