.... nice mathematics... prove that:: Σ_(n=1) ^∞ (([ (((2n)),(n) )]^2 )/((2n−1)2^(4n) )) =1−(2/π) ... m.n.july. 1970#


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Question Number 114735 by mnjuly1970 last updated on 20/Sep/20

$. . . . n i c e m a t h e m a t i c s . . .$ $p r o v e t h a t ::$ $\underset{n = 1}{\sum^{\infty}} \frac{{[(\begin{matrix} 2 n \\ n \end{matrix})]}^{2}}{(2 n - 1) 2^{4 n}} = 1 - \frac{2}{π}$ $You can't use 'macro parameter character #' in math mode$
	Answered by maths mind last updated on 20/Sep/20

	$\underset{n ⩾ 1}{\sum^{2}} \frac{{[(\begin{matrix} 2 n \\ n \end{matrix})]}^{2}}{(2 n - 1) 2^{4 n}} = Σ \frac{{(\frac{2 n!}{n! . n!})}^{2}}{(2 n - 1) 2^{4 n}} = S$ $= 1 + \sum_{n ⩾ 2} \frac{{(\frac{2 n!}{n! . n!})}^{2}}{(2 n - 1) 2^{4 n}} ∣ 2 n! = 2^{n} . \underset{k = 0}{\prod^{n - 1}} (2 k + 1) = 2^{2 n} n! . \underset{k = 0}{\prod^{n - 1}} (k + \frac{1}{2})$ $\underset{n ⩾ 1}{\sum^{\infty}} \frac{{[(\begin{matrix} 2 n \\ n \end{matrix})]}^{2}}{(2 n - 1) 2^{4 n}} = \sum_{n ⩾ 1} \frac{{(\frac{2 n!}{n! . n!})}^{2}}{(2 n - 1) 2^{4 n}} = \sum_{n ⩾ 1} \frac{\frac{2^{4 n} n!^{2} \underset{k = 0}{\prod^{n - 1}} {(k + \frac{1}{2})}^{2}}{{(n!)}^{4}}}{(2 n - 1) 2^{4 n}}$ $\sum_{n ⩾ 1} \frac{\underset{k = 0}{\prod^{n - 1}} {(k + \frac{1}{2})}^{2}}{(2 n - 1) {(n!)}^{2}} . 1 = \frac{1}{4} + \sum_{n ⩾ 2} \frac{\underset{k = 0}{\prod^{n - 1}} {(k + \frac{1}{2})}^{2}}{(2 n - 1) n!^{2}} . 1$ $= \frac{1}{4} + \sum_{n ⩾ 2} \frac{\underset{k = 0}{\prod^{n - 1}} (k + \frac{1}{2}) \underset{k = 0}{\prod^{n - 2}} (k + \frac{1}{2}) (n - 1 + \frac{1}{2})}{(2 n - 1) n!} . \frac{{(1)}^{n}}{n!}$ $= \frac{1}{4} + \sum_{n ⩾ 2} \frac{\underset{k = 0}{\prod^{n - 1}} (\frac{1}{2} + k) \underset{k = 0}{\prod^{n - 2}} (k + \frac{1}{2}) (2 n - 1)}{2 (2 n - 1) n!} . \frac{1^{n}}{n!}$ $\underset{k = 0}{\prod^{n - 2}} (k + \frac{1}{2}) = - 2 \underset{k = 0}{\prod^{n - 1}} (k - \frac{1}{2}) = - 2 {(- \frac{1}{2})}_{n}$ $n! = \underset{k = 0}{\prod^{n - 1}} (1 + k) = {(1)}_{n}$ $s o W e g e t S = \frac{1}{4} + \sum_{n ⩾ 2} \frac{{(\frac{1}{2})}_{n} . - 2 {(- \frac{1}{2})}_{n}}{2 {(1)}_{n}} . \frac{1^{n}}{n!}$ $r e c a l l t h a t_{2} F_{1} (a, b; c; x) = 1 + \sum_{n ⩾ 1} \frac{a_{n} b_{n}}{c_{n}} \frac{x^{n}}{n!}$ $\Rightarrow S = \frac{1}{4} - \sum_{n ⩾ 2} \frac{{(\frac{1}{2})}_{n} {(- \frac{1}{2})}_{n}}{{(1)}_{n}} = \frac{1}{4} - (_{2} F_{1} (\frac{1}{2}, - \frac{1}{2}; 1; 1) - 1 - \frac{{(\frac{1}{2})}_{1} {(- \frac{1}{2})}_{1}}{{(1)}_{1}} . \frac{1}{1!})$ $= \frac{1}{4} - (_{2} F_{1} (\frac{1}{2}, - \frac{1}{2}; 1; 1) - 1 + \frac{1}{4}) = 1 -_{2} F_{1} (\frac{1}{2}; - \frac{1}{2}; 1; 1)$ $w e u s e g a u s s e H y p e r g e o m e t r i q u e t h e o r e m$ $_{2} F_{1} (a, b; c, 1) = \frac{Γ (c) Γ (c - b - a)}{Γ (c - a) Γ (c - b)} t h i s i s j u s t r e s u l t$ $o f β {(b, c - b)}_{2} F_{1} (a, b; c; z) = \int_{0}^{1} x^{b - 1} {(1 - x)}^{c - b + 1} {(1 - z x)}^{- a} d x$ $s o w e g e t$ $2 F 1 (\frac{1}{2}, - \frac{1}{2}; 1; 1) = \frac{Γ (1) Γ (1 + \frac{1}{2} - \frac{1}{2})}{Γ (\frac{1}{2}) Γ (\frac{3}{2})} = \frac{Γ (1) Γ (1)}{\frac{1}{2} Γ {(\frac{1}{2})}^{2}} = \frac{2}{π}$ $S = 1 -_{2} F_{1} (\frac{1}{2}, - \frac{1}{2}; 1; 1) = 1 - \frac{2}{π}$
	Commented by mnjuly1970 last updated on 21/Sep/20

	$v e r y n i c e t h a n k y o u s i r . .$
	Commented by maths mind last updated on 21/Sep/20

	$w i t h e p l e a s u r s i r$
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