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Question Number 131103 by mnjuly1970 last updated on 01/Feb/21
...nice calculus... prove that :: Σ_(n∈N) (((Γ^2 (n))/(2^(−n) (2n−1)!)))=π
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…{nice}\:\:{calculus}… \\ $$$$\:{prove}\:\:{that}\::: \\ $$$$\:\:\underset{{n}\in\mathbb{N}} {\sum}\left(\frac{\Gamma^{\mathrm{2}} \left({n}\right)}{\mathrm{2}^{−{n}} \left(\mathrm{2}{n}−\mathrm{1}\right)!}\right)=\pi \\ $$$$ \\ $$
Answered by Ar Brandon last updated on 01/Feb/21
S=Σ_(n=1) ^∞ ((Γ^2 (n))/(2^(−n) (2n−1)!))=Σ_(n=1) ^∞ ((β(n,n))/2^(−n) ) =Σ_(n=1) ^∞ (1/2^(−n) )∫_0 ^1 x^(n−1) (1−x)^(n−1) dx =∫_0 ^1 {Σ_(n=1) ^∞ 2∙2^(n−1) (x−x^2 )^(n−1) }dx =∫_0 ^1 {2∙(1/(1−(2x−2x^2 )))}dx=∫_0 ^1 ((2dx)/(2x^2 −2x+1)) =∫_0 ^1 (dx/(x^2 −x+(1/2)))=∫_0 ^1 (dx/((x−(1/2))^2 +(1/4))) =2[tan^(−1) (2x−1)]_0 ^1 =2((π/4)+(π/4))=π
$$\mathrm{S}=\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\Gamma^{\mathrm{2}} \left(\mathrm{n}\right)}{\mathrm{2}^{−\mathrm{n}} \left(\mathrm{2n}−\mathrm{1}\right)!}=\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\beta\left(\mathrm{n},\mathrm{n}\right)}{\mathrm{2}^{−\mathrm{n}} } \\ $$$$\:\:\:=\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}^{−\mathrm{n}} }\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}^{\mathrm{n}−\mathrm{1}} \left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{n}−\mathrm{1}} \mathrm{dx} \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \left\{\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{2}\centerdot\mathrm{2}^{\mathrm{n}−\mathrm{1}} \left(\mathrm{x}−\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{n}−\mathrm{1}} \right\}\mathrm{dx} \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \left\{\mathrm{2}\centerdot\frac{\mathrm{1}}{\mathrm{1}−\left(\mathrm{2x}−\mathrm{2x}^{\mathrm{2}} \right)}\right\}\mathrm{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2dx}}{\mathrm{2x}^{\mathrm{2}} −\mathrm{2x}+\mathrm{1}} \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{dx}}{\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$$\:\:\:=\mathrm{2}\left[\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2x}−\mathrm{1}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} =\mathrm{2}\left(\frac{\pi}{\mathrm{4}}+\frac{\pi}{\mathrm{4}}\right)=\pi \\ $$
Commented by mnjuly1970 last updated on 01/Feb/21
excellent mr brandon extraordinary..
$$\:\:\:{excellent}\:{mr}\:{brandon} \\ $$$$\:\:{extraordinary}.. \\ $$
Answered by Dwaipayan Shikari last updated on 01/Feb/21
Σ_(n=1) ^∞ ((Γ^2 (n))/(2^(−n) Γ(2n)))=∫_0 ^1 Σ_(n=1) ^∞ 2^n x^(n−1) (1−x)^(n−1) dx =2∫_0 ^1 (1/(1−2x(1−x)))dx=∫_0 ^1 (1/(x^2 −x+(1/2)))dx=2∫_0 ^1 (1/((x−(1/2))^2 +(1/4)))dx =2((π/2))=π
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\Gamma^{\mathrm{2}} \left({n}\right)}{\mathrm{2}^{−{n}} \Gamma\left(\mathrm{2}{n}\right)}=\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{2}^{{n}} {x}^{{n}−\mathrm{1}} \left(\mathrm{1}−{x}\right)^{{n}−\mathrm{1}} {dx} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}−\mathrm{2}{x}\left(\mathrm{1}−{x}\right)}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{x}^{\mathrm{2}} −{x}+\frac{\mathrm{1}}{\mathrm{2}}}{dx}=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}}{dx} \\ $$$$=\mathrm{2}\left(\frac{\pi}{\mathrm{2}}\right)=\pi \\ $$
Commented by mnjuly1970 last updated on 01/Feb/21
very nice .thank you so much=.
$${very}\:{nice}\:.{thank}\:{you}\:{so}\:{much}=. \\ $$

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