Summation of an Infinite Series: $\sum_{n=1}^\infty \frac{4^{2n}}{n^3 \binom{2n}{n}^2} = 8\pi G-14\zeta(3)$ I am having trouble proving that
$$\sum_{n=1}^\infty \frac{4^{2n}}{n^3 \binom{2n}{n}^2} = 8\pi G-14\zeta(3)$$
I know that
$$\frac{2x \ \arcsin(x)}{\sqrt{1-x^2}} = \sum_{n=1}^\infty \frac{(2x)^{2n}}{n \binom{2n}{n}}$$
What should I do if the binomial coefficient is squared?
 A: This series evaluates to the exact same solution as the integrals:
$\displaystyle 4\int_{0}^{\frac{\pi}{2}}x^{2}\csc(x)dx$
and $\displaystyle 16\int_{0}^{1}\frac{(\tan^{-1}(x))^{2}}{x}dx$
EDIT:
I found a way to evaluate said series using the above csc.
Though, I am going to use some already established identities.
Rewrite the series as $\displaystyle \sum_{n=1}^{\infty}\frac{4^{2n}(n!)^{4}}{n^{3}((2n)!)^{2}}$
Begin with the handy $\displaystyle(\sin^{-1}(t))^{2}=\frac{1}{2}\sum_{n=1}^{\infty}\frac{(n!)^{2}}{n^{2}(2n)!}(2t)^{2n}$
Now, let $\displaystyle x=\sin^{-1}(t)$:
$\displaystyle x^{2}=\frac{1}{2}\sum_{n=1}^{\infty}\frac{(n!)^{2}}{n^{2}(2n)!}2^{2n}\sin^{2n}(x)$
Divide by $\displaystyle\sin(x)$ and integrate from $0$ to $\displaystyle\frac{\pi}{2}$
$\displaystyle\int_{0}^{\frac{\pi}{2}}x^{2}\csc(x)dx=\frac{1}{2}\sum_{n=1}^{\infty}\frac{(n!)^{2}}{n^{2}(2n)!}2^{2n}\int_{0}^{\frac{\pi}{2}}\sin^{2n-1}(x)dx...[1]$
But, $\displaystyle\int_{0}^{\frac{\pi}{2}}\sin^{2n-1}(x)dx=\frac{2^{2n}(n!)^{2}}{2n(2n)!}$
Subbing this into [1] results in:
$\displaystyle\int_{0}^{\frac{\pi}{2}}x^{2}\csc(x)dx=\frac{1}{4}\sum_{n=1}^{\infty}\frac{4^{2n}(n!)^{4}}{n^{3}((2n)!)^{2}}$
Now, multiply it all by 4 and get the final:
$\displaystyle4\int_{0}^{\frac{\pi}{2}}x^{2}\csc(x)dx=\sum_{n=1}^{\infty}\frac{4^{2n}(n!)^{4}}{n^{3}((2n)!)^{2}}$
Since $\displaystyle\int_{0}^{\frac{\pi}{2}}x^{2}\csc(x)dx=2\pi G-7/2\zeta(3)$, 
multiplying by 4 gives the required result.
Of course, the evaluation of the integral can be shown if needed.  But, it is a rather famous one and can be found here and there. 
