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?

  • $\begingroup$ Is this the sum that comes up in Apery's proof of the irrationality of $\zeta(3)$? $\endgroup$ – Gerry Myerson Feb 13 '13 at 12:09

This series evaluates to the exact same solution as the integrals:

$\displaystyle 4\int_{0}^{\frac{\pi}{2}}x^{2}\csc(x)dx$

and $\displaystyle 16\int_{0}^{1}\frac{(\tan^{-1}(x))^{2}}{x}dx$


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 $\displaystyle\frac{\pi}{2}$


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:


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.

  • $\begingroup$ Thanks! I can solve the integral myself. $\endgroup$ – Shobhit Bhatnagar Feb 17 '13 at 14:18

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