A Challenging Euler Sum $\sum\limits_{n=1}^\infty \frac{H_n}{\tbinom{2n}{n}}$

Question

Recently, I encountered a strange series involving Harmonic Numbers and Binomial Coefficients both.

According to Mathematica:

$$\displaystyle \sum_{n=1}^\infty \frac{H_n}{\binom{2n}{n}} = -\frac{2\sqrt{3} \pi}{27}(\log (3)-2)+\frac{2}{27} \left( \psi_1 \left( \frac{1}{3}\right)-\psi_1 \left(\frac{2}{3} \right)\right)$$

Here $\psi_n(z)$ denotes the Polygamma Function. Can anybody provide a nice proof of the above statement?

My Failed Attempt

Using the Beta-function identity, $$\frac{1}{\binom{2n}{n}}=(2n+1)\int_0^1 y^n(1-y)^n \ dy$$

$$\displaystyle \begin{aligned} \sum_{n=1}^\infty \frac{H_n}{\binom{2n}{n}} &= \sum_{n=1}^\infty (2n+1)H_n \int_0^1 (y-y^2)^n dy \\ &= \int_0^1 \sum_{n=1}^\infty (2n+1)H_n (y-y^2)^n \ dy \end{aligned}$$

Here, I used the identity

$$\sum_{n=1}^\infty (2n+1)H_n t^n=\frac{2t-(1+t)\log(1-t)}{(t-1)^2}\quad |t|<1$$

and got

$$\sum_{n=1}^\infty \frac{H_n}{\binom{2n}{n}}=\int_0^1 \frac{2y-2y^2-(1+y-y^2)\log(y^2-y+1)}{(y^2-y+1)^2}dy$$

How should I continue from here? I tried making some substitutions but nothing worked. Am I going in the right direction?

Please help.

Answer 1

Note: there is a minus sign missing in front of the 1st term on the left side of your evaluation.

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Put the integral into the form $$I=\int_0^{1}\frac{\left(y^2-y+1\right)\ln\left(y^2-y+1\right)-2\ln\left(y^2-y+1\right)-2\left(y^2-y+1\right)+2}{\left(y^2-y+1\right)^2}dy.$$ Making the change of variable $y=\frac12+\frac{\sqrt{3}}{2}\tan\varphi$ (so that $y^2-y+1=\frac{3}{4\cos^2\varphi}$) and simplifying, this reduces to $$I=\frac{8}{3\sqrt{3}}\int_{-\pi/6}^{\pi/6}\left\{\Bigl(\frac34-2\cos^2 \varphi\right)\left(\ln 3-2-2\ln (2\cos \varphi)\Bigr)-2\cos^2 \varphi\right\}d\varphi.\tag{1}$$ The only nontrivial integrals here are of the form $$\int_{-\pi/6}^{\pi/6}\ln (2\cos \varphi)\,d\varphi,\qquad \int_{-\pi/6}^{\pi/6}\left(2\cos^2\varphi-1\right)\ln (2\cos \varphi)\,d\varphi.$$ The second integral can be easily done by parts - it is equal to \begin{align}\int_{-\pi/6}^{\pi/6}\left(2\cos^2\varphi-1\right)\ln (2\cos \varphi)\,d\varphi&=\Bigl[\sin\varphi\cos\varphi\ln(2\cos \varphi)\Bigr]^{\pi/6}_{-\pi/6}+\int_{-\pi/6}^{\pi/6}\sin^2\varphi\,d\varphi=\\&=\frac{\pi}{6}+\frac{\sqrt{3}}{4}\ln3-\frac{\sqrt{3}}{4}. \end{align} Using this in (1), we reduce it to $$I=\frac{8}{3\sqrt{3}}\left[\frac{\pi\left(2-\ln3\right)}{12}+\frac12\int_{-\pi/6}^{\pi/6}\ln(2\cos\varphi)\,d\varphi\right].$$ Therefore, the proof of your identity reduces to showing that $$\int_{0}^{\pi/6}\ln(2\cos\varphi)\,d\varphi=\frac{\psi_1\left(\frac13\right)-\psi_1\left(\frac23\right)}{12\sqrt{3}},\tag{2}$$ However, the left side is clearly expressible in terms of polylogarithms, so (2) should follow from their known special values.

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Added:

Indeed, as Raymond Manzoni pointed out, the difference of the formulas (5) and (7) here gives $$\psi_1\left(\frac13\right)-\psi_1\left(\frac23\right)=6\sqrt{3}\,\mathrm{Cl}_2\left(\frac{2\pi}{3}\right)\tag{3}$$ Clausen function $\mathrm{Cl}_2\left(x\right)$ is basically the imaginary part of dilogarithm function, characterized by the integral representation $$\mathrm{Cl}_2\left(x\right)=-\int_0^x\ln\left(2\sin\frac{t}{2}\right)dt.\tag{4}$$ Using (3), (4) and the fact that $\mathrm{Cl}_2(\pi)=0$, we deduce from (2) the necessary statement.

Answer 2

Let us denote the sum as $S$ Then by the beta function identity, we have \begin{align*} S &=\sum_{n=1}^{\infty} \sum_{m=1}^{n} \frac{n\beta(n, n+1)}{m} =\sum_{n=1}^{\infty} \sum_{m=1}^{n} \frac{n}{m} \int_{0}^{1} x^{n-1} (1-x)^{n} \, dx. \end{align*} Now switching the order of summation and utilizing some power series identities, we obtain \begin{align*} S &=\sum_{m=1}^{\infty} \sum_{n=m}^{\infty} \frac{n}{m} \int_{0}^{1} x^{n} (1-x)^{n} \, \frac{dx}{x} \\ &=\sum_{m=1}^{\infty} \sum_{n=0}^{\infty} \frac{n+m}{m} \int_{0}^{1} x^{m}(1-x)^{m} x^{n} (1-x)^{n} \, \frac{dx}{x} \\ &=\sum_{m=1}^{\infty} \frac{1}{m} \int_{0}^{1} x^{m}(1-x)^{m} \left( \frac{x(1-x)}{(1 - x + x^2)^{2}} + \frac{m}{1 - x + x^2} \right) \, \frac{dx}{x} \\ &= \int_{0}^{1} \left( - \frac{x(1-x)}{(1 - x + x^2)^{2}} \log(1 - x + x^2) + \frac{x(1-x)}{(1 - x + x^2)^{2}} \right) \, \frac{dx}{x} \\ &= \int_{0}^{1} \frac{1-x}{(1 - x + x^2)^{2}} \left\{ 1 - \log (1 - x + x^2) \right\} \, dx. \end{align*} Splitting the integrand into the symmetric part and the anti-symmetric part with respect to the transform $\displaystyle x \mapsto 1-x$, we find that \begin{align*} S &= \frac{1}{2} \int_{0}^{1} \frac{1 - \log (1 - x + x^2)}{(1 - x + x^2)^{2}} \, dx. \end{align*} Now, we use the substitution $\displaystyle x - \frac{1}{2} = \frac{\sqrt{3}}{2} \tan \theta$. Then $S$ reduces to \begin{align*} S &= \frac{8}{3\sqrt{3}} \int_{0}^{\frac{\pi}{6}} \left\{ 1 + 2 \log \left( \frac{2\cos\theta}{\sqrt{3}} \right) \right\} \cos^2 \theta \, d\theta. \end{align*} By some tedious calculation (integration by parts and cosine double angle formulas are sufficient), it easily follows that \begin{align*} &\int \left\{ 1 + 2 \log \left( \frac{2\cos\theta}{\sqrt{3}} \right) \right\} \cos^2 \theta \, d\theta \\ &= \theta (1 - \log 2 \cos \theta) + \left( \theta + \frac{\sin 2\theta}{2} \right) \log \left( \frac{2\cos\theta}{\sqrt{3}} \right) + \int \log (2 \cos \theta) \, d\theta, \end{align*} yielding \begin{align*} S &= \frac{8}{3\sqrt{3}} \left[ \frac{\pi}{6}\left( 1 - \frac{1}{2}\log 3 \right) + \int_{0}^{\frac{\pi}{6}} \log (2 \cos \theta) \, d\theta \right]. \end{align*} Now, from the identity $$\log(2\cos\theta) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \cos 2n \theta,$$ we have $$\int_{0}^{\frac{\pi}{6}} \log(2\cos\theta) \, d\theta = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{2n^{2}} \sin \left( \frac{\pi n}{3} \right) = \frac{1}{12\sqrt{3}} \left( \psi_{1}\left( \frac{1}{3} \right) - \psi_{1}\left( \frac{2}{3} \right) \right).$$ Putting together, we obtain $$S = \frac{2}{27}\left\{ \sqrt{3}\pi (2 - \log 3) + \psi_{1}\left( \frac{1}{3} \right) - \psi_{1}\left( \frac{2}{3} \right) \right\}.$$

Answer 3

I'll leave it advanced, I need to figure out the last two integrals $$\displaystyle{\sum\limits_{n=1}^{+\infty }{\frac{H_{n}}{\left( \begin{matrix} 2n \\ n \\ \end{matrix} \right)}}=\sum\limits_{n=1}^{+\infty }{H_{n}\cdot \frac{\left( n! \right)^{2}}{\left( 2n \right)!}}=\sum\limits_{n=1}^{+\infty }{\left( 2n+1 \right)H_{n}\cdot \frac{\Gamma \left( n+1 \right)\Gamma \left( n+1 \right)}{\Gamma \left( 2n+2 \right)}}=\sum\limits_{n=1}^{+\infty }{\left( 2n+1 \right)H_{n}\cdot \beta \left( n+1,n+1 \right)}}$$

$$\displaystyle{=\sum\limits_{n=1}^{+\infty }{\left( 2n+1 \right)H_{n}\cdot \int\limits_{0}^{1}{t^{n}\left( 1-t \right)^{n}dt}}=\int\limits_{0}^{1}{\sum\limits_{n=1}^{+\infty }{\left( 2n+1 \right)H_{n}\left( t-t^{2} \right)^{n}}dt}}$$

$$\displaystyle{=\int\limits_{0}^{1}{\left( -2x\cdot \frac{d}{dx}\left( \frac{\log \left( 1-x \right)}{1-x} \right)-\frac{\log \left( 1-x \right)}{1-x} \right)\left| _{x=t-t^{2}} \right.dt}=\int\limits_{0}^{1}{\left( 2x\left( \frac{1-\log \left( 1-x \right)}{\left( 1-x \right)^{2}} \right)-\frac{\log \left( 1-x \right)}{1-x} \right)\left| _{x=t-t^{2}} \right.dt}}$$

$$\displaystyle{=\int\limits_{0}^{1}{\left( \frac{2\left( t-t^{2} \right)-\left( t-t^{2} \right)\log \left( 1-t+t^{2} \right)-\log \left( 1-t+t^{2} \right)}{\left( 1-t+t^{2} \right)^{2}} \right)dt}}$$

$$\displaystyle{=\int\limits_{0}^{1}{\left( \frac{2-2\left( 1-t+t^{2} \right)+\left( 1-t+t^{2} \right)\log \left( 1-t+t^{2} \right)-2\log \left( 1-t+t^{2} \right)}{\left( 1-t+t^{2} \right)^{2}} \right)dt}}$$

$$\displaystyle{=2\int\limits_{0}^{1}{\frac{dt}{\left( 1-t+t^{2} \right)^{2}}}-2\int\limits_{0}^{1}{\frac{dt}{1-t+t^{2}}}+\int\limits_{0}^{1}{\frac{\log \left( 1-t+t^{2} \right)}{1-t+t^{2}}dt}-2\int\limits_{0}^{1}{\frac{\log \left( 1-t+t^{2} \right)}{\left( 1-t+t^{2} \right)^{2}}dt}}$$

$$\displaystyle{=2\left( \frac{2t-1}{3}\cdot \frac{1}{1-t+t^{2}}+\frac{4}{3\sqrt{3}}\cdot \arctan \frac{2t-1}{\sqrt{3}} \right)\left| _{0}^{1} \right.-2\left( \frac{2}{\sqrt{3}}\cdot \arctan \frac{2t-1}{\sqrt{3}} \right)\left| _{0}^{1} \right.+\int\limits_{0}^{1}{\frac{\log \left( 1-t+t^{2} \right)}{1-t+t^{2}}dt}-2\int\limits_{0}^{1}{\frac{\log \left( 1-t+t^{2} \right)}{\left( 1-t+t^{2} \right)^{2}}dt}}$$

$$\displaystyle{=\frac{4}{3}-\frac{4\sqrt{3}\pi }{27}+\underbrace{\int\limits_{0}^{1}{\frac{\log \left( 1-t+t^{2} \right)}{1-t+t^{2}}dt}}_{I_{1}}-2\underbrace{\int\limits_{0}^{1}{\frac{\log \left( 1-t+t^{2} \right)}{\left( 1-t+t^{2} \right)^{2}}dt}}_{I_{2}}}$$

$$\displaystyle{I_{1}=\int\limits_{0}^{1}{\frac{\log \left( 1-t+t^{2} \right)}{1-t+t^{2}}dt}\underbrace{=}_{\theta =\arctan \left( \frac{2}{\sqrt{3}}\left( t-\frac{1}{2} \right) \right)}\int\limits_{\arctan \left( -\frac{1}{\sqrt{3}} \right)}^{\arctan \left( \frac{1}{\sqrt{3}} \right)}{\frac{\log \left( \frac{3}{4}\sec ^{2}\theta \right)}{\frac{3}{4}\sec ^{2}\theta }\frac{\sqrt{3}}{2}\sec ^{2}\theta d\theta }}$$

$$\displaystyle{=\frac{2\sqrt{3}}{3}\int\limits_{-\frac{\pi }{6}}^{\frac{\pi }{6}}{\log \left( \frac{3}{4}\sec ^{2}\theta \right)d\theta }=\frac{2\sqrt{3}}{3}\cdot \frac{\pi }{3}\cdot \log \frac{3}{4}+\frac{4\sqrt{3}}{3}\cdot \frac{\pi }{3}\cdot \log 2-\frac{4\sqrt{3}}{3}\int\limits_{-\frac{\pi }{6}}^{\frac{\pi }{6}}{\log \left( 2\cos \theta \right)d\theta }}$$

$$\displaystyle{=\frac{2\sqrt{3}\pi }{9}\log \frac{3}{4}+\frac{4\sqrt{3}\pi }{9}\log 2-\frac{8\sqrt{3}}{3}\int\limits_{0}^{\frac{\pi }{6}}{\log \left( 2\cos \theta \right)d\theta }=\frac{2\sqrt{3}\pi }{9}\log 3-\frac{8\sqrt{3}}{3}\int\limits_{0}^{\frac{\pi }{6}}{\log \left( 2\cos \theta \right)d\theta }}$$

$$\displaystyle{=\frac{2\sqrt{3}\pi }{9}\log 3-\frac{8\sqrt{3}}{3}\sum\limits_{n=1}^{+\infty }{\frac{\left( -1 \right)^{n-1}}{n}\int\limits_{0}^{\frac{\pi }{6}}{\cos 2n\theta d\theta }}=\frac{2\sqrt{3}\pi }{9}\log 3-\frac{4\sqrt{3}}{3}\sum\limits_{n=1}^{+\infty }{\frac{\left( -1 \right)^{n-1}\sin \frac{n\pi }{3}}{n^{2}}}}$$

$$\displaystyle{=\frac{2\sqrt{3}\pi }{9}\log 3-\frac{4\sqrt{3}}{3}\left( \frac{\sin \frac{\pi }{3}}{1^{2}}-\frac{\sin \frac{2\pi }{3}}{2^{2}}+\frac{\sin \frac{3\pi }{3}}{3^{2}}-\frac{\sin \frac{4\pi }{3}}{4^{2}}+\frac{\sin \frac{5\pi }{3}}{5^{2}}-\frac{\sin \frac{6\pi }{3}}{6^{2}}+\frac{\sin \frac{7\pi }{3}}{7^{2}}+... \right)}$$

$$\displaystyle{=\frac{2\sqrt{3}\pi }{9}\log 3-\frac{4\sqrt{3}}{3}\left( \frac{\frac{\sqrt{3}}{2}}{1^{2}}-\frac{\frac{\sqrt{3}}{2}}{2^{2}}+0-\frac{-\frac{\sqrt{3}}{2}}{4^{2}}+\frac{-\frac{\sqrt{3}}{2}}{5^{2}}-0+\frac{\frac{\sqrt{3}}{2}}{7^{2}}+... \right)}$$

$$\displaystyle{=\frac{2\sqrt{3}\pi }{9}\log 3-\frac{2}{9}\cdot 9\left( 1-\frac{1}{2^{2}}+\frac{1}{4^{2}}-\frac{1}{5^{2}}+\frac{1}{7^{2}}+... \right)=\frac{2\sqrt{3}\pi }{9}\log 3-\frac{2}{9}\cdot \left( 9\sum\limits_{n=0}^{+\infty }{\frac{1}{\left( 3n+1 \right)^{2}}}-9\sum\limits_{n=0}^{+\infty }{\frac{1}{\left( 3n+2 \right)^{2}}} \right)}$$

$$\displaystyle{=\frac{2\sqrt{3}\pi }{9}\log 3+\frac{2}{9}\cdot \left( \sum\limits_{n=0}^{+\infty }{\frac{1}{\left( \frac{2}{3}+n \right)^{2}}}-\sum\limits_{n=0}^{+\infty }{\frac{1}{\left( \frac{1}{3}+n \right)^{2}}} \right)=\frac{2\sqrt{3}\pi }{9}\log 3+\frac{2}{9}\cdot \left( \psi _{1}\left( \frac{2}{3} \right)-\psi _{1}\left( \frac{1}{3} \right) \right)}$$

$$\displaystyle{I_{2}=\int\limits_{0}^{1}{\frac{\log \left( 1-t+t^{2} \right)}{\left( 1-t+t^{2} \right)^{2}}dt}\underbrace{=}_{\theta =\arctan \left( \frac{2}{\sqrt{3}}\left( t-\frac{1}{2} \right) \right)}\int\limits_{\arctan \left( -\frac{1}{\sqrt{3}} \right)}^{\arctan \left( \frac{1}{\sqrt{3}} \right)}{\frac{\log \left( \frac{3}{4}\sec ^{2}\theta \right)}{\left( \frac{3}{4}\sec ^{2}\theta \right)^{2}}\frac{\sqrt{3}}{2}\sec ^{2}\theta d\theta }}$$

$$\displaystyle{=\frac{8\sqrt{3}}{9}\int\limits_{-\frac{\pi }{6}}^{\frac{\pi }{6}}{\cos ^{2}\theta \log \left( \frac{3}{4}\sec ^{2}\theta \right)d\theta }=\frac{4\sqrt{3}}{9}\int\limits_{-\frac{\pi }{6}}^{\frac{\pi }{6}}{\left( 2\cos ^{2}\theta -1+1 \right)\left( \log \frac{3}{4}+2\log 2-2\log \left( 2\cos \theta \right) \right)d\theta }}$$

$$\displaystyle{=\frac{4\sqrt{3}}{9}\left( \frac{\sqrt{3}}{2}\log 3+\frac{\pi }{3}\log 3-2\int\limits_{-\frac{\pi }{6}}^{\frac{\pi }{6}}{\left( 2\cos ^{2}\theta -1 \right)\log \left( 2\cos \theta \right)d\theta }-2\int\limits_{-\frac{\pi }{6}}^{\frac{\pi }{6}}{\log \left( 2\cos \theta \right)d\theta } \right)}$$

$$\displaystyle{=\frac{4\sqrt{3}}{9}\left( \frac{\sqrt{3}}{2}\log 3+\frac{\pi }{3}\log 3-2\left( \sin \theta \cos \theta \log \left( 2\cos \theta \right)\left| _{-\frac{\pi }{6}}^{\frac{\pi }{6}} \right.+\int\limits_{-\frac{\pi }{6}}^{\frac{\pi }{6}}{\sin ^{2}\theta d\theta } \right)-2\int\limits_{-\frac{\pi }{6}}^{\frac{\pi }{6}}{\log \left( 2\cos \theta \right)d\theta } \right)}$$

$$\displaystyle{=\frac{4\sqrt{3}}{9}\left( \frac{\sqrt{3}}{2}\log 3+\frac{\pi }{3}\log 3-2\left( \frac{\pi }{6}-\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}\log 3 \right)-2\int\limits_{-\frac{\pi }{6}}^{\frac{\pi }{6}}{\log \left( 2\cos \theta \right)d\theta } \right)}$$

$$\displaystyle{=\frac{4\sqrt{3}}{9}\left( \frac{\pi }{3}\log 3-\frac{\pi }{3}+\frac{\sqrt{3}}{2}-2\sum\limits_{n=1}^{+\infty }{\frac{\left( -1 \right)^{n-1}\sin \frac{n\pi }{3}}{n^{2}}} \right)=\frac{4\sqrt{3}}{9}\left( \frac{\pi }{3}\log 3-\frac{\pi }{3}+\frac{\sqrt{3}}{2}+\frac{\psi _{1}\left( \frac{2}{3} \right)-\psi _{1}\left( \frac{1}{3} \right)}{3\sqrt{3}} \right)}$$

$$\displaystyle{\sum\limits_{n=1}^{+\infty }{\frac{H_{n}}{\left( \begin{matrix} 2n \\ n \\ \end{matrix} \right)}}=\frac{4}{3}-\frac{4\sqrt{3}\pi }{27}+\frac{2\sqrt{3}\pi }{9}\log 3+\frac{2}{9}\cdot \left( \psi _{1}\left( \frac{2}{3} \right)-\psi _{1}\left( \frac{1}{3} \right) \right)-\frac{8\sqrt{3}}{9}\left( \frac{\pi }{3}\log 3-\frac{\pi }{3}+\frac{\sqrt{3}}{2}+\frac{\psi _{1}\left( \frac{2}{3} \right)-\psi _{1}\left( \frac{1}{3} \right)}{3\sqrt{3}} \right)}$$

$$\displaystyle{=\frac{2}{27}\left( \psi _{1}\left( \frac{1}{3} \right)-\psi _{1}\left( \frac{2}{3} \right)+\sqrt{3}\pi \left( 2-\log 3 \right) \right)}$$

$$\displaystyle{\left. {\underline {\, \sum\limits_{n=1}^{+\infty }{\frac{H_{n}}{\left( \begin{matrix} 2n \\ n \\ \end{matrix} \right)}}=\frac{2}{27}\left( \psi _{1}\left( \frac{1}{3} \right)-\psi _{1}\left( \frac{2}{3} \right)+\sqrt{3}\pi \left( 2-\log 3 \right) \right) \,}}\! \right|}$$

Answer 4

Otherwise.. without Polygamma function..

Lemma 1: $$\displaystyle{g\left( x \right) = \sum\limits_{n = 1}^\infty {\frac{{{{\left( {n!} \right)}^2}}}{{\left( {2n} \right)!}}{x^n}} = \frac{{z\sqrt {4 - z} + 4 \cdot \sqrt z \cdot \arcsin \left( {\frac{{\sqrt z }}{2}} \right)}}{{\left( {4 - z} \right)\sqrt {4 - z} }}}$$ .. It follows elementarily from the series $$\displaystyle{\sum\limits_{n = 0}^\infty {\frac{{{2^{2n}}{{\left( {n!} \right)}^2}{z^{2n + 2}}}}{{\left( {n + 1} \right)\left( {2n + 1} \right)!}}} = {\arcsin ^2}z}$$ https://en.wikipedia.org/wiki/List_of_m ... cal_series with two productions, namely:.

$$\displaystyle{\sum\limits_{n = 0}^\infty {\frac{{{2^{2n}}{{\left( {n!} \right)}^2}{z^{2n + 2}}}}{{\left( {n + 1} \right)\left( {2n + 1} \right)!}}} = {\arcsin ^2}z \Rightarrow \sum\limits_{n = 0}^\infty {\frac{{{2^{2n}}{{\left( {n!} \right)}^2}{z^{2n + 1}}}}{{\left( {2n + 1} \right)!}}} }$$ $$\displaystyle{ = \frac{{\arcsin z}}{{\sqrt {1 - {z^2}} }} \Rightarrow \sum\limits_{n = 0}^\infty {\frac{{{2^{2n}}{{\left( {n!} \right)}^2}\left( {2n + 1} \right){z^{2n}}}}{{\left( {2n + 1} \right)!}}} = }$$

$$\displaystyle{\frac{1}{{1 - {z^2}}} + \frac{{z \cdot \arcsin \left( z \right)}}{{\left( {1 - {z^2}} \right)\sqrt {1 - {z^2}} }} \Rightarrow \sum\limits_{n = 0}^\infty {\frac{{{{\left( {n!} \right)}^2}{{\left( {2z} \right)}^{2n}}}}{{\left( {2n} \right)!}}} = }$$ $$\displaystyle{\frac{1}{{1 - {z^2}}} + \frac{{z \cdot \arcsin \left( z \right)}}{{\left( {1 - {z^2}} \right)\sqrt {1 - {z^2}} }} \Rightarrow \sum\limits_{n = 0}^\infty {\frac{{{{\left( {n!} \right)}^2}{z^{2n}}}}{{\left( {2n} \right)!}}} = }$$

$$\displaystyle{ = \frac{4}{{4 - {z^2}}} + \frac{{4 \cdot z \cdot \arcsin \left( {\frac{z}{2}} \right)}}{{\left( {4 - {z^2}} \right)\sqrt {4 - {z^2}} }}} and finally \displaystyle{\sum\limits_{n = 1}^\infty {\frac{{{{\left( {n!} \right)}^2}{z^n}}}{{\left( {2n} \right)!}}} = \frac{{z\sqrt {4 - z} + 4 \cdot \sqrt z \cdot \arcsin \left( {\frac{{\sqrt z }}{2}} \right)}}{{\left( {4 - z} \right)\sqrt {4 - z} }}}$$

Lemma 2: $$\displaystyle{\log \left( {\cos y} \right) = - \log 2 - \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}\cos \left( {2ny} \right)}}{n}} } .$$ Derived from the series $$\displaystyle{ - \frac{1}{2}\log \left( {2 - 2\cos z} \right) = \sum\limits_{n = 1}^\infty {\frac{{\cos \left( {n \cdot z} \right)}}{n}} } $$ https://en.wikipedia.org/wiki/List_of_m ... cal_series

On to our topic..

Let $$\displaystyle{{a_n} = \frac{{{H_n}}}{{\left( {\begin{array}{*{20}{c}} {2n}\\ n \end{array}} \right)}} = \frac{{{{\left( {n!} \right)}^2}}}{{\left( {2n} \right)!}}{H_n}} . We consider the function \displaystyle{f\left( x \right) = \sum\limits_{n = 1}^\infty {{a_n}{x^n}} } with \displaystyle{\left| x \right| \le 1} . Then \displaystyle{S = \sum\limits_{n = 1}^\infty {\frac{{{H_n}}}{{\left( {\begin{array}{*{20}{c}} {2n}\\ n \end{array}} \right)}}} = f\left( 1 \right)}.$$

However $$\displaystyle{{a_{n + 1}} = \frac{{\left( {n + 1} \right) \cdot {{\left( {n!} \right)}^2}}}{{2 \cdot \left( {2n + 1} \right) \cdot \left( {2n} \right)!}}\left( {{H_n} + \frac{1}{{n + 1}}} \right) = \frac{{\left( {n + 1} \right) \cdot {{\left( {n!} \right)}^2}}}{{2 \cdot \left( {2n + 1} \right) \cdot \left( {2n} \right)!}}{H_n} + \frac{{{{\left( {n!} \right)}^2}}}{{2 \cdot \left( {2n + 1} \right) \cdot \left( {2n} \right)!}}}$$

Ultimately $$\displaystyle{2 \cdot \left( {2n + 1} \right){a_{n + 1}} = \left( {n + 1} \right) \cdot {a_n} + \frac{{{{\left( {n!} \right)}^2}}}{{\left( {2n} \right)!}} \Rightarrow 4\left( {n + 1} \right) \cdot {a_{n + 1}} - 2{a_{n + 1}} = n \cdot {a_n} + {a_n} + \frac{{{{\left( {n!} \right)}^2}}}{{\left( {2n} \right)!}}}$$, therefore:

$$\displaystyle{4 \cdot \sum\limits_{n = 1}^\infty {\left( {n + 1} \right) \cdot {a_{n + 1}}{x^n}} - 2\sum\limits_{n = 1}^\infty {{a_{n + 1}}{x^n}} = \sum\limits_{n = 1}^\infty {n \cdot {a_n}{x^n}} }$$ $$\displaystyle{ + \sum\limits_{n = 1}^\infty {{a_n}{x^n}} + \sum\limits_{n = 1}^\infty {\frac{{{{\left( {n!} \right)}^2}}}{{\left( {2n} \right)!}}{x^n}} \Rightarrow 4 \cdot \sum\limits_{n = 2}^\infty {n \cdot {a_n}{x^{n - 1}}} - \frac{2}{x}\sum\limits_{n = 2}^\infty {{a_n}{x^n}} = }$$

$$\displaystyle{ = x\sum\limits_{n = 1}^\infty {n \cdot {a_n}{x^{n - 1}}} + \sum\limits_{n = 1}^\infty {{a_n}{x^n}} + g\left( x \right) \Rightarrow 4 \cdot {\left( {f\left( x \right) - {a_1} \cdot x} \right){'}}}$$ $$\displaystyle{ - \frac{2}{x}\left( {f\left( x \right) - {a_1} \cdot x} \right) = x \cdot f'\left( x \right) + f\left( x \right) + g\left( x \right) \Rightarrow }$$

$$\displaystyle{ \Rightarrow 4 \cdot \left( {f'\left( x \right) - \frac{1}{2}} \right) - \frac{2}{x}\left( {f\left( x \right) - \frac{x}{2}} \right) = x \cdot f'\left( x \right) + f\left( x \right) + g\left( x \right)} \displaystyle{ \Rightarrow f'\left( x \right) + \frac{{2 + x}}{{x\left( {x - 4} \right)}}f\left( x \right) = \frac{1}{{4 - x}}\left( {1 + g\left( x \right)} \right)}$$

With a classical procedure for solving first-order differential equations and given that $\displaystyle{f\left( 0 \right) = 0}$ we find

$$\displaystyle{f'\left( x \right) + \left( { - \frac{1}{2} \cdot \frac{1}{x} - \frac{3}{2} \cdot \frac{1}{{4 - x}}} \right)f\left( x \right) = \frac{1}{{4 - x}}\left( {1 + g\left( x \right)} \right)}$$ $$\displaystyle{ \Rightarrow {\left( {\frac{{\left( {4 - x} \right)\sqrt {4 - x} }}{{\sqrt x }}f\left( x \right)} \right){'}} = \frac{{\sqrt {4 - x} }}{{\sqrt x }}\left( {1 + g\left( x \right)} \right) \Rightarrow }$$

$$\displaystyle{ \Rightarrow \int\limits_0^1 {{{\left( {\frac{{\left( {4 - x} \right)\sqrt {4 - x} }}{{\sqrt x }}f\left( x \right)} \right)}{'}}} dx = \int\limits_0^1 {\frac{{\sqrt {4 - x} }}{{\sqrt x }}\left( {1 + g\left( x \right)} \right)dx} \Rightarrow }$$ $$\displaystyle{\left[ {\frac{{\left( {4 - x} \right)\sqrt {4 - x} }}{{\sqrt x }}f\left( x \right)} \right]_0^1 = \int\limits_0^1 {\frac{{\sqrt {4 - x} }}{{\sqrt x }}dx} + \int\limits_0^1 {\frac{{\sqrt {4 - x} }}{{\sqrt x }}g\left( x \right)dx} \Rightarrow }$$

$$\displaystyle{ \Rightarrow 3\sqrt 3 \cdot f\left( 1 \right) = \sqrt 3 + \frac{{2\pi }}{3} + \int\limits_0^1 {\left( {\frac{{\sqrt x \cdot \sqrt {4 - x} + 4 \cdot \arcsin \left( {\frac{{\sqrt x }}{2}} \right)}}{{\left( {4 - x} \right)}}} \right)dx} }$$ $$\displaystyle{ = \sqrt 3 + \frac{{2\pi }}{3} + \int\limits_0^1 {\left( {\frac{{\sqrt x }}{{\sqrt {4 - x} }}} \right)dx} + \;4 \cdot \int\limits_0^1 {\left( {\frac{{\arcsin \left( {\frac{{\sqrt x }}{2}} \right)}}{{\left( {4 - x} \right)}}} \right)dx} }$$

Then $$\displaystyle{f\left( 1 \right) = \frac{{4\pi }}{{9\sqrt 3 }} + \frac{4}{{3\sqrt 3 }} \cdot \int\limits_0^1 {\left( {\frac{{\arcsin \left( {\frac{{\sqrt x }}{2}} \right)}}{{\left( {4 - x} \right)}}} \right)dx} \mathop { = = = }\limits^{\sqrt x = w} \frac{{4\pi }}{{9\sqrt 3 }} + \frac{8}{{3\sqrt 3 }} \cdot \int\limits_0^1 {\left( {\frac{{w \cdot \arcsin \left( {\frac{w}{2}} \right)}}{{\left( {4 - {w^2}} \right)}}} \right)dw} \mathop { = = = }\limits^{w = 2y} \frac{{4\pi }}{{9\sqrt 3 }} + }$$

$$\displaystyle{ + \frac{8}{{3\sqrt 3 }} \cdot \int\limits_0^{1/2} {\frac{{y \cdot \arcsin \left( y \right)}}{{1 - {y^2}}}dy} \mathop { = = = }\limits^{y = \sin z} \frac{{4\pi }}{{9\sqrt 3 }} + }$$ $$\displaystyle{\frac{8}{{3\sqrt 3 }} \cdot \int\limits_0^{\pi /6} {\frac{{z \cdot \sin z}}{{\cos z}}dz} = \frac{{4\pi }}{{9\sqrt 3 }} - \frac{8}{{3\sqrt 3 }} \cdot \int\limits_0^{\pi /6} {z\left( {\log \left( {\cos z} \right)} \right)'dz} = }$$

$$\displaystyle{ = \frac{{4\pi }}{{9\sqrt 3 }} - \frac{8}{{3\sqrt 3 }} \cdot \left( {\left[ {z\log \left( {\cos z} \right)} \right]_0^{\pi /6} - \int\limits_0^{\pi /6} {\log \left( {\cos z} \right)dz} } \right)}$$ $$\displaystyle{ = \frac{{2\pi \left( {2 - \log 3 + 2\log 2} \right)}}{{9\sqrt 3 }} + \frac{8}{{3\sqrt 3 }} \cdot \int\limits_0^{\pi /6} {\log \left( {\cos z} \right)dz} = }$$

$$\displaystyle{ = \frac{{2\pi \left( {2 - \log 3 + 2\log 2} \right)}}{{9\sqrt 3 }} - \frac{8}{{3\sqrt 3 }} \cdot \int\limits_0^{\pi /6} {\left( {\log 2 + \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}\cos \left( {2nz} \right)}}{n}} } \right)dz} }$$ $$\displaystyle{ = \frac{{2\pi \left( {2 - \log 3} \right)}}{{9\sqrt 3 }} - \frac{8}{{3\sqrt 3 }} \cdot \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{n}\int\limits_0^{\pi /6} {\cos \left( {2nz} \right)dz} } = }$$

$$\displaystyle{ = \frac{{2\pi \left( {2 - \log 3} \right)}}{{9\sqrt 3 }} - \frac{8}{{3\sqrt 3 }} \cdot \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{n}\int\limits_0^{\pi /6} {\cos \left( {2nz} \right)dz} } = }$$ $$\displaystyle{\frac{{2\pi \left( {2 - \log 3} \right)}}{{9\sqrt 3 }} - \frac{8}{{3\sqrt 3 }} \cdot \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}\sin \left( {\frac{{2n\pi }}{6}} \right)}}{{2{n^2}}}} = }$$

$$\displaystyle{ = \frac{{2\pi \left( {2 - \log 3} \right)}}{{9\sqrt 3 }} - \frac{4}{{3\sqrt 3 }} \cdot \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}\sin \left( {\frac{{n\pi }}{3}} \right)}}{{{n^2}}}} = }$$ $$\displaystyle{\frac{{2\pi \left( {2 - \log 3} \right)}}{{9\sqrt 3 }} - \frac{4}{{3\sqrt 3 }} \cdot \sum\limits_{n = 1}^\infty {\frac{{\sin \left( {n\pi + \frac{{n\pi }}{3}} \right)}}{{{n^2}}}} = }$$

$$\displaystyle{ = \frac{{2\pi \left( {2 - \log 3} \right)}}{{9\sqrt 3 }} + \frac{{2 \cdot i}}{{3\sqrt 3 }}\left( {L{i_2}\left( {{e^{i\frac{{4\pi }}{3}}}} \right) - L{i_2}\left( {{e^{ - i\frac{{4\pi }}{3}}}} \right)} \right) = }$$ $$\displaystyle{\frac{{2\pi \left( {2 - \log 3} \right)}}{{9\sqrt 3 }} + \frac{{2 \cdot i}}{{3\sqrt 3 }}\left( {L{i_2}\left( { - \frac{{1 + i\sqrt 3 }}{2}} \right) - L{i_2}\left( { - \frac{{1 - i\sqrt 3 }}{2}} \right)} \right)}$$

Eventually $$\displaystyle{S = \sum\limits_{n = 1}^\infty {\frac{{{H_n}}}{{\left( {\begin{array}{*{20}{c}} {2n}\\ n \end{array}} \right)}}} = \frac{{2\pi \left( {2 - \log 3} \right)}}{{9\sqrt 3 }} - \frac{4}{{3\sqrt 3 }}{\mathop{\rm Im}\nolimits} \left( {L{i_2}\left( { - \frac{{1 + i\sqrt 3 }}{2}} \right)} \right)}$$ :) :)

Note. 1) The result numerically is the same as the above solution of My !! 2) With the above method, of the function generator, the series $$\displaystyle{\sum\limits_{n = 1}^\infty {\frac{{{H_n}}}{{\left( {\begin{array}{*{20}{c}} {2n}\\ n \end{array}} \right)}} \cdot {a^n}} }$$ for various values of $\displaystyle{a}$ !! have fun ..