Sum the infinite series How to solve this:
\begin{equation*}
\sum_{n=1}^{\infty }\left[ \frac{1\cdot 3\cdot 5\cdots \left( 2n-1\right) }{
2\cdot 4\cdot 6\cdots 2n}\right] ^{3}
\end{equation*}
I can make the bracket thing, $\left[ C(2n,n)/4^{n}\right] ^{3}$, but how to proceed now.
 A: We have:
$$ \sum_{n\geq 0}\frac{1}{4^n}\binom{2n}{n}\,x^n = \frac{1}{\sqrt{1-x}},\tag{1} $$
$$\frac{1}{4^n}\binom{2n}{n}=\frac{1}{2\pi}\int_{-\pi}^{\pi}\left(\sin x\right)^{2n}\,dx \tag{2}$$
$$ \sum_{n\geq 0}\left(\frac{1}{4^n}\binom{2n}{n}\right)^2 x^n = \frac{2}{\pi}\,K(x)=\frac{2}{\pi}\int_{0}^{\pi/2}\frac{d\theta}{\sqrt{1-x\sin^2\theta}}\tag{3} $$
hence by $(2)$ and $(3)$ it follows that:
$$\begin{eqnarray*}\sum_{n\geq 0}\left(\frac{1}{4^n}\binom{2n}{n}\right)^3 &=& \frac{1}{\pi^2}\int_{-\pi}^{\pi}K(\sin^2 x)\,dx\\&=&\frac{4}{\pi^2}\int_{0}^{\pi/2}\int_{0}^{\pi/2}\frac{1}{\sqrt{1-\sin^2\varphi\sin^2\theta}}\,d\theta\,d\varphi\\&=&\frac{4}{\pi^2}\,K\left(\frac{1}{2}\right)^2=\color{red}{\frac{\pi}{\Gamma\left(\frac{3}{4}\right)^4}},\tag{4}\end{eqnarray*}$$
so the original series equals $\displaystyle\color{purple}{-1+\frac{\pi}{\Gamma\left(\frac{3}{4}\right)^4}=0.39320392968567685918424626\ldots}.$
Footnote: this is just a very special case of the identity $(6)$ for the square of the complete elliptic integral of the first kind, plus the fact that $K(1/2)$ can be computed through the reflection and multiplication formulas for the $\Gamma$ function.
