In search of $π$ [ $P:2$ ]

Above series is $\sum_{n=0}^∞ \frac{(\frac{4}{5})_n (\frac{1}{5})_n }{(2)_n n!}=_2F_1 (\frac{4}{5} , \frac{1}{5} ; 2;1)$ $(a)_n = \prod_{k=0}^{n-1} (a+k)$ $_2F_1 (a,b;c;1) = \frac{\Gamma{(c)} \Gamma{(c-a-b)}}{\Gamma{(c-a)}\Gamma{(c-b)}}$ Here it is $_2F_1 (\frac{4}{5},\frac{1}{5} ;2;1)=\frac{25}{4\Gamma{(\frac{1}{5})}\Gamma{(\frac{4}{5})}}= \frac{25}{4π} \sin(\frac{π}{5})$ $\sin(\frac{π}{5}) = \sqrt{\frac{5-\sqrt{5}}{8}}$ . and $\Gamma{(a)}\Gamma{(1-a)} = \frac{π}{\sin(πa)}$

It is $\frac{25\sqrt{\frac{5-\sqrt{5}}{2}}}{8π}$ Answer is $\color{#20A900} \boxed{5a-8b=0}$

Proof

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