1-1-2-1-1-2-2-1-3-2-1-3-2-2-2-1-4-3-1-3-5-2-3-2-4-pi-Prove-the-above-Relation-

Question Number 128256 by Dwaipayan Shikari last updated on 05/Jan/21

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

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

Commented by Dwaipayan Shikari last updated on 06/Jan/21

1 + \underset{n ⩾ 1}{\sum^{\infty}} \frac{{(\frac{1}{2})}_{n}^{2}}{{(2)}_{n} n!} =_{2} F_{1} (\frac{1}{2}, \frac{1}{2}; 2; 1) = \frac{Γ (1) Γ (2 - 1)}{Γ^{2} (\frac{3}{2})} = \frac{4}{π}

Commented by A8;15: last updated on 06/Jan/21

Mr . can we prove this relation with

continued fraction ?

Commented by Dwaipayan Shikari last updated on 06/Jan/21

H e y s i r! I a m a s t u d e n t . K i n d l y {d o n}^{'} t c a l l m e S i r .

I a l s o t h i n k t h i s c a n b e p r o v e d b y C o n t i n u e d f r a c t i o n s

Commented by Dwaipayan Shikari last updated on 06/Jan/21

{t a n}^{- 1} x = \frac{x}{1 + \frac{x^{2}}{3 - x^{2} + \frac{9 x^{2}}{5 - 3 x^{2} + \frac{25 x^{2}}{7 - 5 x^{2} + \frac{49 x^{2}}{. .}}}}}

\frac{π}{4} = \frac{1}{1 + \frac{1}{2 + \frac{3^{2}}{2 + \frac{5^{2}}{2 + \frac{7^{2}}{2 + \dots}}}}} \Rightarrow \frac{4}{π} = 1 + \frac{1}{2 + \frac{3^{2}}{2 + \frac{5^{2}}{2 + \frac{7^{2}}{2 + \frac{9^{2}}{2 + \dots}}}}}

Commented by A8;15: last updated on 06/Jan/21

I respect you . May I call you brother ?

(I think I am smaller than you, so you are my elder brother)

Commented by Dwaipayan Shikari last updated on 23/Jan/21

M y a g e e^{\frac{34}{12}}