一个超几何函数等式
\[\Large\displaystyle {\;}_3F_2\left(\frac{1}{2},\frac{1}{2},\frac{1}{2};1,\frac{3}{2};1 \right)=\frac{4\mathbf{G}}{\pi}
\]
\(\Large\mathbf{Proof:}\)
Well, after simplifying the pochhammer symbols we get that it is is equal to the following sum:
\[S= \sum_{n=0}^{\infty} \binom{2n}{n}^2 \frac1{2^{4n} (2n+1)}
\]
It is now easy to finish off using the integral (which can be calculated using the beta function and Legendre's duplication formula):
\[\int_0^1 \frac{x^{2 n}}{\sqrt{1-x^2}}\mathrm{d}x=\frac{\pi}{2} \binom{2n}{n}\frac1{2^{2n}}
\]
and also the power series expansion of arcsin:
\[\sin^{-1} x=\sum_{n=0}^{\infty} \binom{2n}{n} \frac{x^{2n+1}}{2^{2n}(2n+1)}
\]
Thus we get:
\[\begin{align*}
S&= \frac{2}{\pi}\sum_{n=0}^{\infty} \binom{2n}{n} \frac1{2^{2n}(2n+1)} \int_0^1 \frac{x^{2 n}}{\sqrt{1-x^2}}\mathrm{d}x \\&= \frac{2}{\pi} \int_0^1 \frac1{\sqrt{1-x^2}} \frac{\sin^{-1} x}{x} \mathrm{d}x \\&=\frac{2}{\pi} \int_0^{\large\frac{\pi}{2}} \frac{x}{\sin x}\mathrm{ d}x\\&=\frac{2}{\pi }\cdot 2\mathbf{G}\\&=\Large\boxed{\color{blue}{\dfrac{4\mathbf{G}}{\pi }}}
\end{align*}\]
