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I need a closed form of the sum $$\sum_{n=1}^{\infty}\frac{((n-1)!)^2(H_{2n}-H_{n-1})}{(2n)!} $$ where $H_n$ denotes the Harmonic number and "!" is the factorial notation.

I tried: Harmonic number has the integral representation see here $$H_n=\int_{0}^{1}\frac{1-x^n}{1-x} dx $$ Denote $$S=\sum_{n=1}^{\infty}\frac{((n-1)!)^2(H_{2n}-H_{n-1})}{(2n)!} $$ So we obtain $$S= \sum_{n=1}^{\infty}\frac{((n-1)!)^2}{(2n)!}\int_{0}^{1}\frac{1-x^{2n}}{1-x}dx-\sum_{n=1}^{\infty}\frac{((n-1)!)^2}{(2n)!}\int_{0}^{1}\frac{1-x^{n-1}}{1-x}dx$$ Interchanging the order of summation and integration (which I am supposing and has to be proved later) we get $$S= \int_{0}^{1}\sum_{n=1}^{\infty}\frac{(\Gamma(n))^2}{\Gamma(2n+1)}\left(\frac{1-x^{2n}}{1-x}\right) dx-\int_{0}^{1}\sum_{n=1}^{\infty}\frac{(\Gamma(n))^2}{\Gamma(2n+1)}\left(\frac{1-x^{n-1}}{1-x}\right) dx$$ Now we have see here $$\sum_{n=1}^{\infty}\frac{(\Gamma(n))^2}{\Gamma(2n+1)}\left(\frac{1-x^{2n}}{1-x}\right)=\frac{\pi^2-36(\arcsin(\frac{x}{2}))^2}{18(1-x)} $$ Also we have see here $$\sum_{n=1}^{\infty}\frac{(\Gamma(n))^2}{\Gamma(2n+1)}\left(\frac{1-x^{n-1}}{1-x}\right)=\frac{\pi^2\ x-36(\arcsin(\frac{\sqrt{x}}{2}))^2}{18x(1-x)} $$ So we have by above two equations $$S=\int_{0}^{1}\frac{\pi^2-36(\arcsin(\frac{x}{2}))^2}{18(1-x)} dx-\int_{0}^{1}\frac{\pi^2\ x-36(\arcsin(\frac{\sqrt{x}}{2}))^2}{18x(1-x)}dx$$ I request some guidance and help for this question. Thank you!

user170231
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Max
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  • Note that your sum is$$\sum_{n\ge1}\frac{H_{2n}-H_{n-1}}{n^2 \binom{2n}n}.$$A similar sum is evaluated here so perhaps the methods demonstrated there may be of some use. – user170231 Mar 23 '23 at 17:48
  • @user170231 Thanks a lot for the link. It would be really helpful if you could write a closed form of the sum as an answer. – Max Mar 23 '23 at 17:58
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    There are a lot of methods (both ad hoc and systematic) to show this sum is $2\zeta(3)/3$. – pisco Mar 23 '23 at 18:22
  • @pisco Thank you. Please write as an answer about any one method. – Max Mar 23 '23 at 18:23
  • It may be a derivative of a hypergeometric function like here. Also, $y=\sin^{-1}\left(\frac{\sqrt x}2\right)$ may help – Тyma Gaidash Mar 25 '23 at 12:57

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