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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? |
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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. |
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