Summing three binoharmonic series

The following problem $$\sum_{n=1}^\infty\frac{2^{2n+1}H_n}{n(2n+1)^2\binom{2n}{n}}+\sum_{n=1}^\infty\frac{(H_{n-1})^2}{(2n-1)^2 2^{2n}}\binom{2n}{n}+\sum_{n=1}^\infty\frac{H^{(2)}_{n-1}}{(2n-1)^2 2^{2n}}\binom{2n}{n},$$

was proposed in 2021 by Cornel Valean, and it has not been evaluated yet. He claims it has a nice closed form.

I tried to convert the sum of the second and third series to integral by replacing $$x$$ with $$x^2/4$$ in the generating function

$$\sum_{n=1}^\infty \left((H_{n-1})^2-H_{n-1}^{(2)}\right)x^{n-1}=\frac{\ln^2(1-x)}{1-x},$$

and then integrate both sides from $$x=0$$ to $$x=y$$,

$$\sum_{n=1}^\infty \left((H_{n-1})^2-H_{n-1}^{(2)}\right)\frac{y^{2n-1}}{(2n-1)2^{2n}}=4\int_0^{y}\frac{\ln^2(1-x^2/4)}{1-x^2/4}dx,$$

and not sure how to generate the factor $$\frac{\binom{2n}{n}}{2n-1}$$ on the left side. All methods are appreciated.

Thank you

The following is my approach for $$S=2\sum _{k=1}^{\infty }\frac{4^kH_k}{k\left(2k+1\right)^2\binom{2k}{k}}+\sum _{k=1}^{\infty }\frac{\left(H_{k-1}\right)^2}{4^k\left(2k-1\right)^2}\binom{2k}{k}+\sum _{k=1}^{\infty }\frac{H_{k-1}^{\left(2\right)}}{4^k\left(2k-1\right)^2}\binom{2k}{k}$$ $$=2\sum _{k=1}^{\infty }\frac{4^kH_k}{k\left(2k+1\right)^2\binom{2k}{k}}+\frac{1}{2}\sum _{k=1}^{\infty }\frac{H_k^2+H_k^{\left(2\right)}}{4^k\left(k+1\right)\left(2k+1\right)}\binom{2k}{k}.$$

In order to start, how about we first focus on the second series?

Transforming the series into integrals after applying partial fraction decomposition yields $$\frac{1}{2}\sum _{k=1}^{\infty }\frac{H_k^2+H_k^{\left(2\right)}}{4^k\left(k+1\right)\left(2k+1\right)}\binom{2k}{k}$$ $$=\frac{1}{2}\int _0^1\int _0^1\left(\sum _{k=1}^{\infty }\frac{1}{4^k}\binom{2k}{k}\left(xy\right)^k\right)\frac{\ln ^2\left(1-x\right)}{x}\:dx\:dy$$ $$-\frac{1}{4}\int _0^1\frac{1}{\sqrt{y}}\int _0^1\left(\sum _{k=1}^{\infty }\frac{1}{4^k}\binom{2k}{k}\left(xy\right)^k\right)\frac{\ln ^2\left(1-x\right)}{x}\:dx\:dy$$ $$=-\frac{1}{2}\int _0^1\frac{\ln ^2\left(1-x\right)}{x}\left(\frac{x+2\sqrt{1-x}-2}{x}\right)\:dx-\frac{1}{4}\int _0^1\frac{\ln ^2\left(1-x\right)}{x}\left(\frac{2\arcsin \left(\sqrt{x}\right)}{\sqrt{x}}-2\right)dx$$ $$=-\frac{1}{2}\int _0^1\frac{\ln ^2\left(1-x\right)}{x^2}\left(x+2\sqrt{1-x}-2\right)\:dx-\int _0^1\frac{\arcsin \left(x\right)\ln ^2\left(1-x^2\right)}{x^2}\:dx$$ $$+\int _0^1\frac{\ln ^2\left(1-x^2\right)}{x}\:dx$$ $$=-\frac{1}{2}\int _0^1\frac{\ln ^2\left(1-x\right)}{x^2}\left(x+2\sqrt{1-x}-2\right)\:dx-2\pi \int _0^1\frac{x\ln ^2\left(x\right)}{\left(1-x^2\right)^{\frac{3}{2}}}\:dx$$ $$+4\int _0^1\frac{x\arcsin \left(x\right)\ln ^2\left(x\right)}{\left(1-x^2\right)^{\frac{3}{2}}}\:dx+\int _0^1\frac{\ln ^2\left(1-x^2\right)}{x}\:dx$$ $$=-\frac{2\pi ^2}{3}-\frac{\pi ^3}{6}+2\pi \ln ^2\left(2\right)-8\int _0^1\frac{\ln \left(x\right)\arcsin \left(x\right)}{x\sqrt{1-x^2}}\:dx,$$ which features an integral that requires the use of complex numbers for evaluation.

Wouldn’t it be better to try to relate it to the other series?

The key to accomplishing that is employing $$\arcsin \left(x\right)=\frac{\pi }{2}-\arcsin \left(\sqrt{1-x^2}\right),\:0\le x<1.$$

Therefore $$\int _0^1\frac{\ln \left(x\right)\arcsin \left(x\right)}{x\sqrt{1-x^2}}\:dx=\int _0^1\frac{\ln \left(x\right)\left(\frac{\pi }{2}-\arcsin \left(\sqrt{1-x^2}\right)\right)}{x\sqrt{1-x^2}}\:dx$$ $$=\frac{1}{2}\int _0^1\frac{\ln \left(1-x^2\right)\left(\frac{\pi }{2}-\arcsin \left(x\right)\right)}{1-x^2}\:dx=-\frac{1}{2}\sum _{k=1}^{\infty }H_k\int _0^1x^{2k}\left(\frac{\pi }{2}-\arcsin \left(x\right)\right)\:dx$$ $$-\frac{1}{2}\sum _{k=1}^{\infty }\frac{H_k}{2k+1}\int _0^1\frac{x^{2k+1}}{\sqrt{1-x^2}}\:dx=-\frac{1}{2}\sum _{k=1}^{\infty }\frac{4^kH_k}{\left(2k+1\right)^2\binom{2k}{k}}$$ $$=-\frac{1}{4}\sum _{k=1}^{\infty }\frac{4^kH_k}{k\left(2k+1\right)\binom{2k}{k}}+\frac{1}{4}\sum _{k=1}^{\infty }\frac{4^kH_k}{k\left(2k+1\right)^2\binom{2k}{k}}.$$

This means that $$\frac{1}{2}\sum _{k=1}^{\infty }\frac{H_k^2+H_k^{\left(2\right)}}{4^k\left(k+1\right)\left(2k+1\right)}\binom{2k}{k}=-\frac{2\pi ^2}{3}-\frac{\pi ^3}{6}+2\pi \ln ^2\left(2\right)$$ $$+2\sum _{k=1}^{\infty }\frac{4^kH_k}{k\left(2k+1\right)\binom{2k}{k}}-2\sum _{k=1}^{\infty }\frac{4^kH_k}{k\left(2k+1\right)^2\binom{2k}{k}}$$ $$\require{cancel}S=\cancel{2\sum _{k=1}^{\infty }\frac{4^kH_k}{k\left(2k+1\right)^2\binom{2k}{k}}}-\frac{2\pi ^2}{3}-\frac{\pi ^3}{6}+2\pi \ln ^2\left(2\right)+2\sum _{k=1}^{\infty }\frac{4^kH_k}{k\left(2k+1\right)\binom{2k}{k}}$$ $$-\cancel{2\sum _{k=1}^{\infty }\frac{4^kH_k}{k\left(2k+1\right)^2\binom{2k}{k}}}$$ $$S=-\frac{2\pi ^2}{3}-\frac{\pi ^3}{6}+2\pi \ln ^2\left(2\right)+2\sum _{k=1}^{\infty }\frac{4^kH_k}{k\left(2k+1\right)\binom{2k}{k}},$$ and the first series disappears.

We are left with one last series to calculate, which should be straightforward. Converting it to integrals leaves us with $$\sum _{k=1}^{\infty }\frac{4^kH_k}{k\left(2k+1\right)\binom{2k}{k}}=-2\int _0^1\int _0^1\left(\sum _{k=1}^{\infty }\frac{4^k}{\binom{2k}{k}}\left(xy\right)^{2k}\right)\frac{\ln \left(1-x^2\right)}{x}\:dx\:dy$$ $$=-2\int _0^1\ln \left(1-x^2\right)\int _0^1\frac{y\left(xy\sqrt{1-x^2y^2}+\arcsin \left(xy\right)\right)}{\left(1-x^2y^2\right)^{\frac{3}{2}}}\:dy\:dx$$ $$=-2\int _0^1\frac{\ln \left(1-x^2\right)\arcsin \left(x\right)}{x^2\sqrt{1-x^2}}\:dx+2\int _0^1\frac{\ln \left(1-x^2\right)}{x}\:dx$$ $$=-2\pi \int _0^1\frac{\ln \left(x\right)}{\left(1-x^2\right)^{\frac{3}{2}}}\:dx-2\int _0^1\frac{\ln \left(1-x^2\right)}{x}\:dx-4\int _0^1\frac{\arcsin \left(x\right)}{\sqrt{1-x^2}}\:dx$$ $$+2\int _0^1\frac{\ln \left(1-x^2\right)}{x}\:dx$$ $$=\frac{\pi ^2}{2}.$$

Replacing this in $$S$$ yields $$S=-\frac{2\pi ^2}{3}-\frac{\pi ^3}{6}+2\pi \ln ^2\left(2\right)+2\left(\frac{\pi ^2}{2}\right),$$ and finally $$S=2\sum _{k=1}^{\infty }\frac{4^kH_k}{k\left(2k+1\right)^2\binom{2k}{k}}+\sum _{k=1}^{\infty }\frac{\left(H_{k-1}\right)^2}{4^k\left(2k-1\right)^2}\binom{2k}{k}+\sum _{k=1}^{\infty }\frac{H_{k-1}^{\left(2\right)}}{4^k\left(2k-1\right)^2}\binom{2k}{k}$$ $$=\frac{\pi ^2}{3}-\frac{\pi ^3}{6}+2\pi \ln ^2\left(2\right).$$

• Very nice Jorge (+1) Commented Aug 9 at 19:20
• Thanks Ali. It is a wonderful problem, hopefully we get to see more approaches. Commented Aug 9 at 21:05