n=0((12)n)2(2)n.n!=2F1(12,12;2;1)=1Γ2(32)=4π\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} n=0(23)n(13)n(2)n.n!=2F1(23,13;2;1)=934π\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} n=0(34)n(14)n(2)n.n!=2F1(34,14;2;1)=823π\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} n=0(45)n(15)n(2)n.n!=2F1(15,45;2;1)=25sin(π5)4π\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}

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

In all cases I have used 2F1(a,b;c;1)=Γ(c)Γ(cab)Γ(ca)Γ(cb)_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 ππ

When 2F1(11a,1a;2;1)=a2sin(πa)(a1)π\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=97a=97 then it is 940996πsin(π97)\large\frac{9409}{96π} \sin(\frac{π}{97})

Those Expressions can be transformed into infinite sums . For example

2F1(12,12;2;1)=1+12(12.1!)2+13(1.322.2!)2+14(1.3.523.3!)2+_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=1z=1 , then it becomes 2F1(a,b;c,1)=Γ(c)Γ(b)Γ(cb)01tb1(1t)cba1dt_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 =Γ(c)Γ(b)Γ(cb)B(b,cba)=Γ(c)Γ(cab)Γ(ca)Γ(cb)= \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!

Yajat Shamji - 6 months, 1 week ago

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I will add more values later :)

Dwaipayan Shikari - 6 months, 2 weeks ago

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