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, $$\sum_{n=1}^\infty \frac{H_n}{\binom{2n}{n}} = -\frac{2\sqrt{3} \pi}{27}\big(\log (3)-2\big)+\frac{8\sqrt3}{27}\rm{Cl}_2\Big(\frac\pi3\Big)$$ – Tito Piezas III Jun 10 '19 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 – Tito Piezas III Jun 10 '19 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.
• 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 : $$Cl_n\left(2\,\pi \frac pq\right)=\frac 1{q^n(n-1)!}\begin{cases} \sum_{k=1}^{q-1}\sin\left(2\pi k\frac pq\right)\,\psi_{n-1}\left(\frac kq\right), & n\ \text{even} \\ -\sum_{k=1}^q \cos\left(2\pi k\frac pq\right)\,\psi_{n-1}\left(\frac kq\right), & n\ \text{odd} \end{cases}$$ – Raymond Manzoni Jun 3 '13 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. – Start wearing purple Jun 3 '13 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. – Raymond Manzoni Jun 3 '13 at 20:55
• @RaymondManzoni: Can you look at this list involving $\rm{Cl}_2\big(\tfrac{\pi}3\big)$? – Tito Piezas III Jun 12 '19 at 11:15