Question Number 163662 by amin96 last updated on 09/Jan/22
Commented by amin96 last updated on 09/Jan/22
$$\boldsymbol{\mathrm{PROVE}}\:\boldsymbol{\mathrm{THAT}} \\ $$
Answered by Ar Brandon last updated on 09/Jan/22
![S=2Σ_(n=0) ^∞ ((n!)/((2n+1)!!))=2Σ_(n=0) ^∞ ((n!×n!×2^n )/((2n+1)!))=Σ_(n=0) ^∞ ((Γ^2 (n+1))/(Γ(2n+2)))×2^(n+1) =Σ_(n=0) ^∞ 2^(n+1) β(n+1, n+1)=Σ_(n=0) ^∞ 2^(n+1) ∫_0 ^1 x^n (1−x)^n dx =2∫_0 ^1 Σ_(n=0) ^∞ (2x−2x^2 )^n dx=2∫_0 ^1 (dx/(2x^2 −2x+1)) =∫_0 ^1 (dx/((x−(1/2))^2 +(1/4)))=2[arctan(2x−1)]_0 ^1 =2((π/4)+(π/4))=2×(π/2)=π](https://www.tinkutara.com/question/Q163694.png)
$${S}=\mathrm{2}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{n}!}{\left(\mathrm{2}{n}+\mathrm{1}\right)!!}=\mathrm{2}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{n}!×{n}!×\mathrm{2}^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\Gamma^{\mathrm{2}} \left({n}+\mathrm{1}\right)}{\Gamma\left(\mathrm{2}{n}+\mathrm{2}\right)}×\mathrm{2}^{{n}+\mathrm{1}} \\ $$$$\:\:\:=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{2}^{{n}+\mathrm{1}} \beta\left({n}+\mathrm{1},\:{n}+\mathrm{1}\right)=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{2}^{{n}+\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} \left(\mathrm{1}−{x}\right)^{{n}} {dx} \\ $$$$\:\:\:=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\mathrm{2}{x}−\mathrm{2}{x}^{\mathrm{2}} \right)^{{n}} {dx}=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}} \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}}=\mathrm{2}\left[\mathrm{arctan}\left(\mathrm{2}{x}−\mathrm{1}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\:\:\:=\mathrm{2}\left(\frac{\pi}{\mathrm{4}}+\frac{\pi}{\mathrm{4}}\right)=\mathrm{2}×\frac{\pi}{\mathrm{2}}=\pi \\ $$
Commented by amin96 last updated on 09/Jan/22
$${yes}\:{sir}.\:{bravoooo} \\ $$