[Largedisplaystyle {;}_3F_2left(frac{1}{2},frac{1}{2},frac{1}{2};1,frac{3}{2};1
ight)=frac{4mathbf{G}}{pi}
]
(Largemathbf{Proof:})
Well, after simplifying the pochhammer symbols we get that it is is equal to the following sum:
[S= sum_{n=0}^{infty} inom{2n}{n}^2 frac1{2^{4n} (2n+1)}
]
It is now easy to finish off using the integral (which can be calculated using the beta function and Legendre's duplication formula):
[int_0^1 frac{x^{2 n}}{sqrt{1-x^2}}mathrm{d}x=frac{pi}{2} inom{2n}{n}frac1{2^{2n}}
]
and also the power series expansion of arcsin:
[sin^{-1} x=sum_{n=0}^{infty} inom{2n}{n} frac{x^{2n+1}}{2^{2n}(2n+1)}
]
Thus we get:
[egin{align*}
S&= frac{2}{pi}sum_{n=0}^{infty} inom{2n}{n} frac1{2^{2n}(2n+1)} int_0^1 frac{x^{2 n}}{sqrt{1-x^2}}mathrm{d}x \&= frac{2}{pi} int_0^1 frac1{sqrt{1-x^2}} frac{sin^{-1} x}{x} mathrm{d}x \&=frac{2}{pi} int_0^{largefrac{pi}{2}} frac{x}{sin x}mathrm{ d}x\&=frac{2}{pi }cdot 2mathbf{G}\&=Largeoxed{color{blue}{dfrac{4mathbf{G}}{pi }}}
end{align*}]