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Question Number 131103 by mnjuly1970 last updated on 01/Feb/21

$. . . n i c e c a l c u l u s . . .$ $p r o v e t h a t ::$ $\sum_{n \in N} (\frac{Γ^{2} (n)}{2^{- n} (2 n - 1)!}) = π$
	Answered by Ar Brandon last updated on 01/Feb/21
	$S=Σ_(n=1) ^∞ ((Γ^2 (n))/(2^(−n) (2n−1)!))=Σ_(n=1) ^∞ ((β(n,n))/2^(−n) ) =Σ_(n=1) ^∞ (1/2^(−n) )∫_0 ^1 x^(n−1) (1−x)^(n−1) dx =∫_0 ^1 {Σ_(n=1) ^∞ 2∙2^(n−1) (x−x^2 )^(n−1) }dx =∫_0 ^1 {2∙(1/(1−(2x−2x^2 )))}dx=∫_0 ^1 ((2dx)/(2x^2 −2x+1)) =∫_0 ^1 (dx/(x^2 −x+(1/2)))=∫_0 ^1 (dx/((x−(1/2))^2 +(1/4))) =2[tan^(−1) (2x−1)]_0 ^1 =2((π/4)+(π/4))=π$
	$S = \underset{n = 1}{\sum^{\infty}} \frac{Γ^{2} (n)}{2^{- n} (2 n - 1)!} = \underset{n = 1}{\sum^{\infty}} \frac{β (n, n)}{2^{- n}}$ $= \underset{n = 1}{\sum^{\infty}} \frac{1}{2^{- n}} \int_{0}^{1} x^{n - 1} {(1 - x)}^{n - 1} dx$ $= \int_{0}^{1} {\underset{n = 1}{\sum^{\infty}} 2 \cdot 2^{n - 1} {(x - x^{2})}^{n - 1}} dx$ $= \int_{0}^{1} {2 \cdot \frac{1}{1 - (2 x - {2 x}^{2})}} dx = \int_{0}^{1} \frac{2 dx}{{2 x}^{2} - 2 x + 1}$ $= \int_{0}^{1} \frac{dx}{x^{2} - x + \frac{1}{2}} = \int_{0}^{1} \frac{dx}{{(x - \frac{1}{2})}^{2} + \frac{1}{4}}$ $= 2 {[\tan^{- 1} (2 x - 1)]}_{0}^{1} = 2 (\frac{π}{4} + \frac{π}{4}) = π$
	Commented by mnjuly1970 last updated on 01/Feb/21

	$e x c e l l e n t m r b r a n d o n$ $e x t r a o r d i n a r y . .$
	Answered by Dwaipayan Shikari last updated on 01/Feb/21

	$\underset{n = 1}{\sum^{\infty}} \frac{Γ^{2} (n)}{2^{- n} Γ (2 n)} = \int_{0}^{1} \underset{n = 1}{\sum^{\infty}} 2^{n} x^{n - 1} {(1 - x)}^{n - 1} d x$ $= 2 \int_{0}^{1} \frac{1}{1 - 2 x (1 - x)} d x = \int_{0}^{1} \frac{1}{x^{2} - x + \frac{1}{2}} d x = 2 \int_{0}^{1} \frac{1}{{(x - \frac{1}{2})}^{2} + \frac{1}{4}} d x$ $= 2 (\frac{π}{2}) = π$
	Commented by mnjuly1970 last updated on 01/Feb/21

	$v e r y n i c e . t h a n k y o u s o m u c h = .$
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