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!