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

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?

• In terms of Gieseking's constant $\rm{Cl}_2\left(\frac\pi3\right)$ then, $$\sum_{n=1}^\infty \frac{H_n}{\binom{2n}{n}} = -\frac{2\sqrt{3} \pi}{27}\big(\log (3)-2\big)+\frac{8\sqrt3}{27}\rm{Cl}_2\Big(\frac\pi3\Big)$$ Jun 10, 2019 at 18:12
• Note also that, $$\frac34\sum_{n=1}^\infty \frac{1}{n^3\,\binom {2n}n} =-\zeta(3)+\frac\pi2\,\rm{Cl}_2\Big(\frac\pi3\Big)$$ as mentioned in this post Jun 10, 2019 at 18:29

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

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.

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.

• @RaymondManzoni Thanks very much Raymond, I will add this to my post if you don't mind. Jun 3, 2013 at 17:05
• Glad it helped @O.L.! (I removed my initial comment and can't upvote again as I should!) I found in $2001$ following general correspondence and sent it to E. Weisstein (for the credit : K.S. Kölbig from CERN got it much earlier and probably others, it is straigthforward... once obtained ! :-)) For $0\le p<q$ we have : $$Cl_n\left(2\,\pi \frac pq\right)=\frac 1{q^n(n-1)!}\begin{cases} \sum_{k=1}^{q-1}\sin\left(2\pi k\frac pq\right)\,\psi_{n-1}\left(\frac kq\right), & n\ \text{even} \\ -\sum_{k=1}^q \cos\left(2\pi k\frac pq\right)\,\psi_{n-1}\left(\frac kq\right), & n\ \text{odd} \end{cases}$$ Jun 3, 2013 at 20:21
• @RaymondManzoni Looking at your formula, I suddenly understood that I've already seen this kind of relations. It reminded me some formulas from this paper; $\zeta'(-1,x)$ is a close relative of Barnes and Clausen functions. Jun 3, 2013 at 20:45
• Yes @O.L. this is related and an elaboration of the Gauss formula for $\psi(p/q)$ at the end of Knuth AOCP I. Note that a Kölbig paper appears at the end of your article. The second (more general) Kölbig paper is at CERN. Jun 3, 2013 at 20:55
• @RaymondManzoni: Can you look at this list involving $\rm{Cl}_2\big(\tfrac{\pi}3\big)$? Jun 12, 2019 at 11:15

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 $$S = \frac{2}{27}\left\{ \sqrt{3}\pi (2 - \log 3) + \psi_{1}\left( \frac{1}{3} \right) - \psi_{1}\left( \frac{2}{3} \right) \right\}.$$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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 ..