# $π$

$\sum_{n=0}^∞ \frac{((\frac{1}{2})_n)^2 }{(2)_n .n!} = _2F_1 (\frac{1}{2},\frac{1}{2};2;1)= \frac{1}{\Gamma^2{(\frac{3}{2})}}= \frac{4}{π}$ $\sum_{n=0}^∞ \frac{(\frac{2}{3})_n (\frac{1}{3})_n }{(2)_n .n!}= _2F_1 (\frac{2}{3},\frac{1}{3};2;1) =\frac{9\sqrt{3}}{4π}$ $\sum_{n=0}^∞ \frac{(\frac{3}{4})_n (\frac{1}{4})_n}{(2)_n .n!}= _2F_1(\frac{3}{4} , \frac{1}{4} ;2;1)= \frac{8\sqrt{2}}{3π}$ $\sum_{n=0}^∞ \frac{(\frac{4}{5})_n (\frac{1}{5})_n }{(2)_n .n!} = _2F_1(\frac{1}{5},\frac{4}{5} ;2;1) = \frac{25\sin{(\frac{π}{5})}}{4π}$

$\textrm{And many more...}$

In all cases I have used $_2F_1 (a,b;c;1)= \frac{\Gamma{(c)}\Gamma{(c-a-b)}}{\Gamma{(c-a)}\Gamma{(c-b)}}$

I have noticed that , we can easily found out an Infinite sum generator for $π$

When $\color{#20A900}_2F_1 (1-\frac{1}{a} , \frac{1}{a} ;2;1) = \frac{a^2 \sin(\frac{π}{a})}{(a-1)π}$ When $a=97$ then it is $\large\frac{9409}{96π} \sin(\frac{π}{97})$

Those Expressions can be transformed into infinite sums . For example

$_2F_1(\frac{1}{2},\frac{1}{2};2;1)= 1+\frac{1}{2} (\frac{1}{2.1!})^2+\frac{1}{3} (\frac{1.3}{2^2.2!})^2 +\frac{1}{4}(\frac{1.3.5}{2^3.3!})^2 +\cdots$

Proof of Above Identity Take $z=1$ , then it becomes $_2F_1 (a,b;c,1)= \frac{\Gamma{(c)}}{\Gamma{(b)}\Gamma{(c-b)} } \int_0^1 t^{b-1} (1-t)^{c-b-a-1}dt$ $= \frac{\Gamma{(c)}}{\Gamma{(b)}\Gamma{(c-b)}} \Beta{(b,c-b-a)} = \frac{\Gamma{(c)} \Gamma{(c-a-b)}}{\Gamma{(c-a)}\Gamma{(c-b)}}$

Note by Dwaipayan Shikari
6 months, 2 weeks ago

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Very nice! Put in it Percy's Tau discussion!

- 6 months, 1 week ago

I will add more values later :)

- 6 months, 2 weeks ago