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Question Number 128122 by Dwaipayan Shikari last updated on 04/Jan/21

1 + \frac{1}{16} + \frac{5^{2}}{16^{2} . 2!} + \frac{5^{2} . 9^{2}}{16^{3} . 3!} + \frac{5^{2} . 9^{2} . 13^{2}}{16^{4} . 4!} + \dots = \frac{\sqrt{π}}{Γ^{2} (\frac{3}{4})} = F_{1} (\frac{1}{4}, \frac{1}{4}, 1; 1)

P r o v e T h e a b o v e r e l a t i o n

W h e r e

F_{1} (Φ, φ, γ; μ) = \underset{n ⩾ 0}{\sum^{\infty}} \frac{{(Φ)}_{n} {(φ)}_{n}}{n! {(γ)}_{n}} μ^{n}

{(ζ)}_{n} = ζ (ζ + 1) (ζ + 2) \dots (ζ + n - 1)

Answered by mindispower last updated on 05/Jan/21

= 1 + \sum_{n ⩾ 1} \frac{\underset{k = 0}{\prod^{n - 1}} {(4 k + 1)}^{2}}{16^{n} {(n!)}^{2}}

= 1 + \sum_{n ⩾ 1} \frac{4^{2 n} . \underset{k = 0}{\prod^{n - 1}} (k + \frac{1}{4}) . Π (k + \frac{1}{4})}{. 16^{n} . n!} . \frac{1}{n!}

= 1 + \sum_{n ⩾ 1} \frac{{(\frac{1}{4})}_{n} . {(\frac{1}{4})}_{n}}{{(1)}_{n}} . \frac{1^{n}}{n!} =_{2} F_{1} (\frac{1}{4}, \frac{1}{4}; 1; [1])

t h a n u s e r e l a t i o n_{2} F_{1} (a, b, c; 1) w i t h e β,

Γ

Commented by Dwaipayan Shikari last updated on 05/Jan/21

G r e a t s i r!

t h e n_{2} F_{1} (a, b, c; 1) = \frac{Γ (c) Γ (c - a - b)}{Γ (c - a) Γ (c - b)}

_{2} F_{1} (\frac{1}{4}, \frac{1}{4}; 1; 1) = \frac{Γ (\frac{1}{2})}{Γ^{2} (\frac{3}{4})} = \frac{\sqrt{π}}{Γ^{2} (\frac{3}{4})}

I s i t s i r ? ?

Commented by mindispower last updated on 06/Jan/21

y e s s i r g r e a t