# 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?

Please help.

• 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

## 2 Answers

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.

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.

• @RaymondManzoni Thanks very much Raymond, I will add this to my post if you don't mind. – Start wearing purple Jun 3 '13 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 : $$Cl_n\left(2\,\pi \frac pq\right)=\frac 1{q^n(n-1)!}\begin{cases} \sum_{k=1}^{q-1}\sin\left(2\pi k\frac pq\right)\,\psi_{n-1}\left(\frac kq\right), & n\ \text{even} \\ -\sum_{k=1}^q \cos\left(2\pi k\frac pq\right)\,\psi_{n-1}\left(\frac kq\right), & n\ \text{odd} \end{cases}$$ – 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

You can refer to my calculation in this forum.