((2(√2))/(9801))Σ_(n=1) ^∞ (((4n)!(1103+26390n))/((n!)^4 396^(4n) ))=(1/π) (Prove that)


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Question Number 117460 by Dwaipayan Shikari last updated on 11/Oct/20

$\frac{2 \sqrt{2}}{9801} \underset{n = 1}{\sum^{\infty}} \frac{(4 n)! (1103 + 26390 n)}{{(n!)}^{4} 396^{4 n}} = \frac{1}{π} (P r o v e t h a t)$
Commented by Dwaipayan Shikari last updated on 11/Oct/20

$T h i s h a d b e e n d o n e b y R a m a n u j a n . D o y o u h a v e a n y i d e a$ $h o w t o s o l v e i t ?$
Commented by mindispower last updated on 11/Oct/20

$s e e g e n e r a l i s e d h y e r g e o m e t r i c f u n c t i o n$
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