# Closed form for $\sum^\infty_{n=1}\frac{H_n}{2^n\,(2n+1)^2}$

(This is a slight variation of another question, already answered)

Can we find a closed form of the following series? $$S=\sum^\infty_{n=1}\frac{H_n}{2^n\,(2n+1)^2}\tag1$$ Using some non-rigorous numerical methods, I found a conjectured form: \begin{align} S &\stackrel?=\sqrt{2}\left[\frac{\ln^32}{10}-\frac{\ln^3\!\left(1+\sqrt2\right)}{10}+\frac{27}{40}\,\ln\left(1+\sqrt{2}\right)\cdot\ln^22 \right. \\ & \hspace{5mm} \left. -\pi^2\left(\frac{5\ln2}{24}+\frac{\ln\left(1+\sqrt2\right)}{20}\right)+\operatorname{Li}_2\!\left(\frac1{\sqrt2}\right)\cdot\ln2+\frac65\,\operatorname{Li}_3\!\left(\frac1{\sqrt2}\right) \right. \\ & \hspace{10mm} \left. -\frac3{10}\,\operatorname{Li}_3\!\left(\frac{2-\sqrt{2}}4\right)+\frac3{10}\,\operatorname{Li}_3\!\left(\frac{2+\sqrt{2}}4\right)-\frac{21}{160}\zeta(3)\right]\tag2 \end{align} It it possible to prove this result or further simplify it?

• Using an integral representation $H_n=\int_0^1\frac{1-x^n}{1-x}dx$ the series $S$ can be converted to an integral containing dilogarithms, that can be evaluated and simplified using Mathematica to this form that numerically matches my conjectured result $(2)$ and contains similar terms. – Vladimir Reshetnikov Aug 31 '15 at 20:16
• Shouldn't you then add that form to the body of the question? – Vincenzo Oliva Aug 31 '15 at 20:50
• Related questions: (1), (2). – Vladimir Reshetnikov Aug 31 '15 at 21:19
• A way for a proof could be as follows: 1.) Use $\int_{0}^{1}dx \log(x)x^{2 n}=\frac{1}{(2n+1)^2}$ 2.) Use the generating function $\sum_1^{\infty}x^n H_n=\frac{-\log(1-x)}{1-x}$ 3.) Check that the resulting integrals have antiderivatives in form of Polylogs (differentiate mathematica results or integration by parts) 4.) Simplify the answer using special values of Polylogs, functional identies etc. Maybe someone can build a real answer out of that.... – tired Sep 1 '15 at 10:05
• Difficult problem. But you can bound the sum from above and below using elementary methods. – user98186 Jan 2 '16 at 16:46

I will here derive an expression for the more general problem of evaluating $\sum_{n=1}^\infty \frac{H_n z^{2n+1}}{(2n+1)^2}$ for $|z| < 1$ in terms of polylogarithms. This derivation uses the same idea as outlined by tired in the comments. The derivation contains the expression for several integrals which people might find interesting, however the derivation contains so many terms that I had to rely on mathematical software to get everything right so it's not very illuminating and it does not fully answer the question.

Mathematica code used to evaluate and simplify the integrals can be found here.

We start using the identity $\int_0^1x^{2n}\log(x){\rm d}x = -\frac{1}{(2n+1)^2}$ the sum can be written as an integral $$\sum_{n=1}^\infty \frac{H_n z^{2n+1}}{(2n+1)^2} = -z\int_0^1\sum_{n=1}^\infty (zx)^{2n}H_n\log(x)\,{\rm d}x = \int_0^1\frac{\log(x)\log(1-(zx)^2)}{1-(zx)^2}z{\rm d}x$$

Splitting the $\log$ and a partial fraction decomposition gives us $$\sum_{n=1}^\infty \frac{H_n z^{2n+1}}{(2n+1)^2} = \frac{I(z) - I(-z) + J(z) - J(-z)}{2}$$

where $I(z)$ is given by (see the end of this post for steps to derive this)

$$I(z)=\int_0^1\frac{\log(x)\log(1+zx)}{1+zx}z{\rm d}x = \int_0^z\frac{\log(x)\log(1+x)}{1+x}{\rm d}x - \log(z)\int_0^z\frac{\log(1+x)}{1+x}{\rm d}x \\= \text{Li}_3\left(\frac{1}{1+z}\right)+\text{Li}_2\left(\frac{1}{1+z}\right) \log(1+z)-\zeta(3)+\frac{\log^3(1+z)}{3}-\frac{\log(z)\log^2(1+z)}{2}$$ and using various polylog identities (see the end of this post) we get $$I(-z) = \text{Li}_3(1-z)-\text{Li}_2(1-z) \log (1-z)-\zeta(3)-\frac{\log(z)\log^2(1-z)}{2}$$

The last integral we need is $$J(z) = \int_0^1\frac{\log(x)\log(1-zx)}{1+zx}z{\rm d}x = \int_0^z\frac{\log(x)\log(1-x)}{1+x}{\rm d}x - \log(z)\int_0^z\frac{\log(1-x)}{1+x}{\rm d}x \\= -\text{Li}_3\left(\frac{1-z}{2}\right)+\text{Li}_3\left(\frac{z-1}{2z}\right)-\text{Li}_3\left(\frac{z-1}{z}\right)-\text{Li}_3(-z)+\left[\text{Li}_2\left(\frac{1-z}{2}\right)-\text{Li}_2\left(\frac{z-1}{2z}\right)+\text{Li}_2\left(\frac{z-1}{z}\right)\right]\log\left(\frac{1-z}{2z}\right)+\text{Li}_2(-z)\log(2z)\\+\frac{7\zeta(3)}{8}-\frac{1}{6}\pi^2\log(2)+\frac{1}{2}\log(2)\log(z)\log(2z)+\frac{1}{12}\pi^2\log(z)$$ and again using polylog identities we find $$J(-z) = -\text{Li}_3(z)-\text{Li}_3\left(\frac{z}{z+1}\right)+\text{Li}_3\left(\frac{2z}{z+1}\right)-\text{Li}_3\left(\frac{z+1}{2}\right)+\text{Li}_2(z) \log (2z)+\left[-\text{Li}_2\left(\frac{z}{z+1}\right)+\text{Li}_2\left(\frac{2z}{z+1}\right)+\text{Li}_2\left(\frac{z+1}{2}\right)\right] \log\left(\frac{z+1}{2 z}\right)-\frac{1}{2} \log (2) \log^2(z+1)-\frac{1}{2} \log ^2(2) \log (z)+\log (2) \log (2 z) \log(z+1)+\frac{1}{12} \pi ^2 \log (z)+\frac{7 \zeta(3)}{8}-\frac{\log ^3(2)}{3}$$

The reason I have choosen to give separate expressions for $\pm z$ is to make sure all terms are real when $z\in (0,1)$ so that one can apply these directly without any knowledge about the analytical continuation of the polylogarithmic functions. The expressions for $I(z),J(z)$ above have been compared to numerical results and agree perfectly.

Adding up we get (this equation is also double checked numerically)

$$\matrix{2\sum_{n=1}^\infty \frac{H_nz^{2n+1}}{(2n+1)^2} &=& -\text{Li}_3\left(\frac{1-z}{2}\right)-\text{Li}_3(1-z)-\text{Li}_3(-z)+\text{Li}_3(z)-\text{Li}_3\left(\frac{z}{z-1}\right)\\&&+\text{Li}_3\left(\frac{2z}{z-1}\right)+\text{Li}_3\left(\frac{1}{1+z}\right)+\text{Li}_3\left(\frac{z}{1+z}\right)\\&&-\text{Li}_3\left(\frac{2z}{1+z}\right)+\text{Li}_3\left(\frac{1+z}{2}\right)+\text{Li}_2\left(\frac{1}{1+z}\right) \log (1+z)\\&&+\text{Li}_2(1-z) \log(1-z)+\text{Li}_2(-z) \log(2z)-\text{Li}_2(z)\log (2z)\\&&+\left[\text{Li}_2\left(\frac{1-z}{2}\right)-\text{Li}_2\left(\frac{z}{z-1}\right)+\text{Li}_2\left(\frac{2 z}{z-1}\right)\right] \log \left(\frac{1-z}{2z}\right)\\&&+\left[\text{Li}_2\left(\frac{z}{1+z}\right)-\text{Li}_2\left(\frac{2z}{1+z}\right)-\text{Li}_2\left(\frac{z+1}{2}\right)\right] \log\left(\frac{1+z}{2z}\right)\\&&-\log \left(\frac{1+z}{1-z}\right) \left[\frac{1}{2} \log\left(\frac{z}{2}\right)\log\left(1-z^2\right)+\log (2) \log (2z)\right]+\frac{1}{3}\log^3(1+z)}$$

and right here I lost my appetite for working more on this!

$$\bf \text{Useful polylog identities:}$$ For real $w>1$ we have

$$\text{Li}_2(w)+\text{Li}_2\left(\frac{1}{w}\right)=-\frac{\log^2(w)}{2}-i\pi \log (w)+\frac{\pi ^2}{3}$$

$$\text{Li}_3(w)-\text{Li}_3\left(\frac{1}{w}\right) = -\frac{\log^3(w)}{6}-\frac{1}{2} i \pi\log ^2(w)+\frac{1}{3} \pi^2 \log(w)$$

$$\bf \text{Derivation of I(z)}$$

Take $t = \log (1 + x)$ then the integral becomes $$\int_0^z\frac{\log(x)\log(1+x)}{1+x}{\rm d}x = \int_0^{\log (1 + z)} x^2 + x\log (1 + e^{-x}) {\rm d} x$$ Now expanding the $\log$ in a Taylor series and integrating term-by-term gives us $$\frac {\log^3 (1 + z)} {3} - \sum_ {n=1}^\infty\left[\frac {1} {n^3} - \frac {1} {n^3 (1 + z)^n} - \frac {\log (1 + z)} {n^2 (1 + z)^n}\right]$$ The sum can then be evaluated using the definition of the polylogarithm and the $\zeta$ function.

• I understood everything, except for the method used to evaluate $\displaystyle\int\frac{\log(x)\log(1-x)}{1+x}~dx.$ – Lucian Jan 4 '16 at 8:13
• @Lucian Yeah that is though one. I tried substituting $w = \log (1 - w)$ and then expanding both the $\log$ and $1/(1+x)$ in series. This leads to simple integrals, but a nightmare double sum with too many term. I just gave up and evaluated the integral directly using software. – Winther Jan 4 '16 at 9:44