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Question Number 129047 by Dwaipayan Shikari last updated on 12/Jan/21

1 + \frac{1}{2} {(\frac{1}{2.1!})}^{2} + \frac{1}{2^{2}} {(\frac{1.3}{2^{2} . 2!})}^{2} + \frac{1}{2^{3}} {(\frac{1.3.5}{2^{3} . 3!})}^{2} + \dots = \frac{\sqrt{π}}{Γ^{2} (\frac{3}{4})}

Answered by mindispower last updated on 12/Jan/21

1 + \sum_{n ⩾ 1} \frac{{(\underset{k = 0}{\prod^{n - 1}} (1 + 2 k))}^{2}}{2^{n} {(2^{n} . n!)}^{2}} = S

= 1 + \sum_{n ⩾ 1} \frac{2^{2 n} \underset{k = 0}{\prod^{n - 1}} {(k + \frac{1}{2})}_{n}^{2}}{2^{n} . 2^{2 n} . n!} . \frac{1}{n!} = 1 + \sum_{k ⩾ 1} \frac{{(\frac{1}{2})}_{n} {(\frac{1}{2})}_{n}}{{(1)}_{n}} . \frac{{(\frac{1}{2})}^{n}}{n!}

=_{2} F_{1} (\frac{1}{2}, \frac{1}{2}; 1; [\frac{1}{2}])

{B a i l y}^{'} s t h e o r e m_{2} F_{1} (a, 1 - a; c; \frac{1}{2}) = \frac{Γ (\frac{c}{2}) Γ (\frac{1 + c}{2})}{Γ (\frac{c + a}{2}) Γ (\frac{c + 1 - a}{2})}

w e g e t S = \frac{Γ (\frac{1}{2}) Γ (1)}{Γ (\frac{3}{4}) Γ (\frac{3}{4})} = \frac{\sqrt{π}}{Γ^{2} (\frac{3}{4})}

Commented by Dwaipayan Shikari last updated on 12/Jan/21

I {d i d n}^{'} t k n o w w h a t i s {B a i l e y}^{'} s T h e o r e m . T h a n k i n g y o u

I f o u n d i t u s i n g \frac{Γ (c)}{Γ (c - b) Γ (b)} \int_{0}^{1} t^{b - 1} {(1 - t)}^{c - b - 1} {(1 - \frac{t}{2})}^{- a} d t

= \frac{2^{a} Γ (c)}{Γ (c - b) Γ (b)} \int_{0}^{1} {(1 - t)}^{b - 1} t^{c - b - 1} {(1 + t)}^{- a} d t

Commented by mindispower last updated on 12/Jan/21

y e s s y m e t r i e x, 1 - x,