# Infinite Series $\sum_{n=1}^\infty\frac{H_n}{n^32^n}$

I'm trying to find a closed form for the following sum $$\sum_{n=1}^\infty\frac{H_n}{n^3\,2^n},$$ where $H_n=\displaystyle\sum_{k=1}^n\frac{1}{k}$ is a harmonic number.

Could you help me with it?

In the same spirit as Robert Israel's answer and continuing Raymond Manzoni's answer (both of them deserve the credit because of inspiring my answer) we have $$\sum_{n=1}^\infty \frac{H_nx^n}{n^2}=\zeta(3)+\frac{1}{2}\ln x\ln^2(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x).$$ Dividing equation above by $x$ and then integrating yields \begin{align} \sum_{n=1}^\infty \frac{H_nx^n}{n^3}=&\zeta(3)\ln x+\frac12\color{red}{\int\frac{\ln x\ln^2(1-x)}{x}\ dx}+\color{blue}{\int\frac{\ln(1-x)\operatorname{Li}_2(1-x)}x\ dx}\\&+\operatorname{Li}_4(x)-\color{green}{\int\frac{\operatorname{Li}_3(1-x)}x\ dx}.\tag1 \end{align} Using IBP to evaluate the green integral by setting $u=\operatorname{Li}_3(1-x)$ and $dv=\frac1x\ dx$, we obtain \begin{align} \color{green}{\int\frac{\operatorname{Li}_3(1-x)}x\ dx}&=\operatorname{Li}_3(1-x)\ln x+\int\frac{\ln x\operatorname{Li}_2(1-x)}{1-x}\ dx\qquad x\mapsto1-x\\ &=\operatorname{Li}_3(1-x)\ln x-\color{blue}{\int\frac{\ln (1-x)\operatorname{Li}_2(x)}{x}\ dx}.\tag2 \end{align} Using Euler's reflection formula for dilogarithm $$\operatorname{Li}_2(x)+\operatorname{Li}_2(1-x)=\frac{\pi^2}6-\ln x\ln(1-x),$$ then combining the blue integral in $(1)$ and $(2)$ yields $$\frac{\pi^2}6\int\frac{\ln (1-x)}{x}\ dx-\color{red}{\int\frac{\ln x\ln^2(1-x)}{x}\ dx}=-\frac{\pi^2}6\operatorname{Li}_2(x)-\color{red}{\int\frac{\ln x\ln^2(1-x)}{x}\ dx}.$$ Setting $x\mapsto1-x$ and using the identity $H_{n+1}-H_n=\frac1{n+1}$, the red integral becomes \begin{align} \color{red}{\int\frac{\ln x\ln^2(1-x)}{x}\ dx}&=-\int\frac{\ln (1-x)\ln^2 x}{1-x}\ dx\\ &=\int\sum_{n=1}^\infty H_n x^n\ln^2x\ dx\\ &=\sum_{n=1}^\infty H_n \int x^n\ln^2x\ dx\\ &=\sum_{n=1}^\infty H_n \frac{\partial^2}{\partial n^2}\left[\int x^n\ dx\right]\\ &=\sum_{n=1}^\infty H_n \frac{\partial^2}{\partial n^2}\left[\frac {x^{n+1}}{n+1}\right]\\ &=\sum_{n=1}^\infty H_n \left[\frac{x^{n+1}\ln^2x}{n+1}-2\frac{x^{n+1}\ln x}{(n+1)^2}+2\frac{x^{n+1}}{(n+1)^3}\right]\\ &=\ln^2x\sum_{n=1}^\infty\frac{H_n x^{n+1}}{n+1}-2\ln x\sum_{n=1}^\infty\frac{H_n x^{n+1}}{(n+1)^2}+2\sum_{n=1}^\infty\frac{H_n x^{n+1}}{(n+1)^3}\\ &=\frac12\ln^2x\ln^2(1-x)-2\ln x\left[\sum_{n=1}^\infty\frac{H_{n+1} x^{n+1}}{(n+1)^2}-\sum_{n=1}^\infty\frac{x^{n+1}}{(n+1)^3}\right]\\&+2\left[\sum_{n=1}^\infty\frac{H_{n+1} x^{n+1}}{(n+1)^3}-\sum_{n=1}^\infty\frac{x^{n+1}}{(n+1)^4}\right]\\ &=\frac12\ln^2x\ln^2(1-x)-2\ln x\left[\sum_{n=1}^\infty\frac{H_{n} x^{n}}{n^2}-\sum_{n=1}^\infty\frac{x^{n}}{n^3}\right]\\&+2\left[\sum_{n=1}^\infty\frac{H_{n} x^{n}}{n^3}-\sum_{n=1}^\infty\frac{x^{n}}{n^4}\right]\\ &=\frac12\ln^2x\ln^2(1-x)-2\ln x\left[\sum_{n=1}^\infty\frac{H_{n} x^{n}}{n^2}-\operatorname{Li}_3(x)\right]\\&+2\left[\sum_{n=1}^\infty\frac{H_{n} x^{n}}{n^3}-\operatorname{Li}_4(x)\right]. \end{align} Putting all together, we have \begin{align} \sum_{n=1}^\infty \frac{H_nx^n}{n^3}=&\frac12\zeta(3)\ln x-\frac18\ln^2x\ln^2(1-x)+\frac12\ln x\left[\sum_{n=1}^\infty\frac{H_{n} x^{n}}{n^2}-\operatorname{Li}_3(x)\right]\\&+\operatorname{Li}_4(x)-\frac{\pi^2}{12}\operatorname{Li}_2(x)-\frac12\operatorname{Li}_3(1-x)\ln x+C.\tag3 \end{align} Setting $x=1$ to obtain the constant of integration, \begin{align} \sum_{n=1}^\infty \frac{H_n}{n^3}&=\operatorname{Li}_4(1)-\frac{\pi^2}{12}\operatorname{Li}_2(1)+C\\ \frac{\pi^4}{72}&=\frac{\pi^4}{90}-\frac{\pi^4}{72}+C\\ C&=\frac{\pi^4}{60}. \end{align} Thus \begin{align} \sum_{n=1}^\infty \frac{H_nx^n}{n^3}=&\frac12\zeta(3)\ln x-\frac18\ln^2x\ln^2(1-x)+\frac12\ln x\left[\sum_{n=1}^\infty\frac{H_{n} x^{n}}{n^2}-\operatorname{Li}_3(x)\right]\\&+\operatorname{Li}_4(x)-\frac{\pi^2}{12}\operatorname{Li}_2(x)-\frac12\operatorname{Li}_3(1-x)\ln x+\frac{\pi^4}{60}.\tag4 \end{align} Finally, setting $x=\frac12$, we obtain \begin{align} \sum_{n=1}^\infty \frac{H_n}{2^nn^3}=\color{purple}{\frac{\pi^4}{720}+\frac{\ln^42}{24}-\frac{\ln2}8\zeta(3)+\operatorname{Li}_4\left(\frac12\right)}, \end{align} which matches Cleo's answer.

References :

$[1]\$ Harmonic number

$[2]\$ Polylogarithm

• Even though you seem to have made a minor error somewhere (which I'm unable to locate), your answer is impressive and deserves an upvote. +1 Aug 26, 2014 at 21:06
• Thanks for the compliment and the upvote @RandomVariable, it means a lot coming from you. The error comes from the constant of integration, but it has been fixed. I think it's correct now. :) Aug 26, 2014 at 21:51
• Very instructive. A pleasure to follow your elaboration. 1+ Aug 26, 2014 at 21:59
• It seems you no longer need verification, but as I promised in chat the other day, I've numerically verified your answer out to the first thousand digits. (I'll delete this comment once you see it.)
– user98602
Aug 27, 2014 at 22:03
• @Tunk-Fey I have the feeling, that you are Cleo. You post the solution with that account first, and after that the details. The top of the iceberg is that you comment funny questions under Cleo's solution, for example "Who you really are?" or something like that. Sep 4, 2014 at 22:33

$$\sum_{n=1}^\infty\frac{H_n}{n^3\,2^n}=\frac{\pi^4}{720}+\frac{\ln^42}{24}-\frac{\ln2}8\zeta(3)+\operatorname{Li}_4\left(\frac12\right).$$

• Could I ask a favor ? What steps did you use to arrive to this beautiful result ? Thanks. Aug 26, 2014 at 7:12
• Instead of asking how do you arrive to this answer, I am interesting in knowing more about you. Who are you? Aug 26, 2014 at 14:49
• -1: It's clearly not useful.
– user146010
Aug 26, 2014 at 21:50
• @Tunk-Fey Clearly a wizard. Aug 27, 2014 at 1:01
• Yet again, we see $\ln2$ acting as a regularized value for $\zeta(1)$. Aug 28, 2014 at 20:14

Note: Please note the top voted answer by @Tunk-Fey is regrettably not correct. Contrary to his claim his final expression (4) when evaluated at $x=\frac{1}{2}$ does not match @Cleo's answer but differs by $\frac{\pi^4}{120}$ from the correct identity: \begin{align*} \sum_{n=1}^\infty \frac{H_n}{n^32^n}&=-\frac{1}{8}\ln 2\zeta(3)+\frac{1}{24}\ln^4(2)+\frac{\pi^4}{720}+ \operatorname{Li}_4\left(\frac{1}{2}\right)\\ &\stackrel{.}{=}0.55824 \end{align*} A rather detailed analysis of the deviation from the correct result is provided in this answer.

Nevertheless it was a pleasure to review his answer which contains nice and instructive aspects. Here I provide a solution in a similar spirit which hopefully overcomes the problems of his answer.

Raymond Manzoni's has nicely demonstrated that for $|x|<1$ \begin{align*} \sum_{n=1}^\infty \frac{H_nx^n}{n^2}&=\zeta(3)+\frac{1}{2}\ln x\ln^2(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)\\ &\qquad+\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x) \end{align*}

This result is our starting point.

\begin{align*} \sum_{n=1}^\infty \frac{H_nx^n}{n^3}&=\int\sum_{n=1}^\infty \frac{H_nx^{n-1}}{n^2}dx\\ &=\zeta(3)\ln(x)+\frac{1}{2}\int\frac{1}{x}\ln x\ln^2(1-x)dx+\int\frac{\ln(1-x)}{x}\operatorname{Li}_2(1-x)dx\\ &\qquad+\int\frac{1}{x}\operatorname{Li}_3(x)dx-\int\frac{1}{x}\operatorname{Li}_3(1-x)dx+C\tag{1}\\ \end{align*}

At first we consider $\int\frac{1}{x}\operatorname{Li}_3(1-x)dx$. Integration by parts with $u=\frac{1}{x}$ and $dv=\operatorname{Li}_3(1-x)dx$ gives

\begin{align*} \int\frac{1}{x}\operatorname{Li}_3(1-x)dx&=\ln x\operatorname{Li}_3(1-x)+\int\frac{\ln x}{1-x}\operatorname{Li}_2(1-x)dx\\ &=\ln x\operatorname{Li}_3(1-x)+\frac{1}{2}\operatorname{Li}_2^2(1-x)+C \end{align*} Once again integration by parts on the RHS with $u=\frac{\ln x}{1-x}$ and $dv=\operatorname{Li}_2(1-x)dx$ gives \begin{align*} \int\frac{\ln x}{1-x}\operatorname{Li}_2(1-x)dx&=\operatorname{Li}_2^2(1-x) -\int\frac{\ln x}{1-x}\operatorname{Li}_2(1-x)dx\\ \Longrightarrow\int\frac{\ln x}{1-x}\operatorname{Li}_2(1-x)dx&=\frac{1}{2}\operatorname{Li}_2^2(1-x)+C \end{align*}

It follows \begin{align*} \int\frac{1}{x}\operatorname{Li}_3(1-x)dx&=\operatorname{Li}_3(1-x)\ln x+\frac{1}{2}\operatorname{Li}_2^2(1-x)+C \end{align*}

and we obtain substituting this result in (1) and noting that \begin{align*} \int\frac{1}{x}\operatorname{Li}_3(x)dx=\operatorname{Li}_4(x)+C \end{align*}

\begin{align*} \sum_{n=1}^\infty \frac{H_nx^n}{n^3}&=\zeta(3)\ln x+\frac{1}{2}\int\frac{1}{x}\ln x\ln^2(1-x)dx+\int\frac{\ln(1-x)}{x}\operatorname{Li}_2(1-x)dx\\ &\qquad+\operatorname{Li}_4(x)-\left(\operatorname{Li}_3(1-x)\ln x+\frac{1}{2}\operatorname{Li}_2^2(1-x)\right)+C\tag{2}\\ \end{align*}

The next step is to calculate $\int\frac{1}{x}\ln x\ln^2(1-x)dx$. We use Euler's reflection formula \begin{align*} \operatorname{Li}_2(x)+\operatorname{Li}_2(1-x)=\frac{\pi^2}{6}-\ln x\ln(1-x) \end{align*} to split the integral into parts which can either be directly calculated or which can be transformed to the remaining integral. We obtain using the reflection formula

\begin{align*} \int&\frac{1}{x}\ln x\ln^2(1-x)dx\\ &=\int\frac{\ln(1-x)}{x}\left(\frac{\pi^2}{6}-\operatorname{Li}_2(x)-\operatorname{Li}_2(1-x)\right)\\ &=-\frac{\pi^2}{6}\operatorname{Li}_2(x)-\int\frac{\ln(1-x)}{x}\operatorname{Li}_2(x)dx -\int\frac{\ln(1-x)}{x}\operatorname{Li}_2(1-x)dx\\ &=-\frac{\pi^2}{6}\operatorname{Li}_2(x)+\frac{1}{2}\operatorname{Li}_2^2(x)dx -\int\frac{\ln(1-x)}{x}\operatorname{Li}_2(1-x)dx \end{align*}

Putting this result into (2) we get

The most complex and cumbersome part is the remaining integral in (3). With the help of Wolfram Alpha a rather lengthy result is provided. After some simplifications we obtain \begin{align*} \int&\frac{\ln(1-x)}{x}\operatorname{Li}_2{(1-x)}dx\\ &=-\frac{1}{2}\ln^2(1-x)\ln^2x+\ln(1-x)\ln^3x-\frac{1}{4}\ln^4x\\ &\qquad-\operatorname{Li}_2(1-x)\left(\ln^2(1-x)-\ln(1-x)\ln x\right)+\operatorname{Li}_2(x)\ln^2 x\\ &\qquad-\operatorname{Li}_2\left(1-\frac{1}{x}\right)\left(\ln^2(1-x)-2\ln(1-x)\ln x+\ln^2 x\right)+\frac{1}{2}\operatorname{Li}_2^2(1-x)\\ &\qquad+2\left(\operatorname{Li}_3\left(1-\frac{1}{x}\right)\left(\ln(1-x)-\ln x\right)+\operatorname{Li}_3(1-x)\ln(1-x) -\operatorname{Li}_3(x)\ln x\right)\\ &\qquad-2\left(\operatorname{Li}_4(1-x)+\operatorname{Li}_4\left(1-\frac{1}{x}\right)-\operatorname{Li}_4(x)\right)+C\\ \end{align*}

Finally substituting this expression into (3) and doing some more simplifications we obtain

From (4) we can now determine the integration constant $C$. In order to do so we calculate $C$ by taking the limit as $x\rightarrow 1$. Most of the terms vanish and noting that according to this answer \begin{align*} \sum_{n=1}^\infty \frac{H_n}{n^3}=\frac{\pi^4}{72} \end{align*} we obtain respecting that $\operatorname{Li}_2(1)=\frac{\pi^2}{6}$ and $\operatorname{Li}_4(1)=\frac{\pi^4}{90}$

\begin{align*} \frac{\pi^4}{72}&=\frac{1}{2}\operatorname{Li}_2(1)\left(-\frac{\pi^2}{6}\right)+\frac{1}{4}\operatorname{Li}^2_2(1)+2\operatorname{Li}_4(1)+C\\ &=-\frac{\pi^4}{72}+\frac{\pi^4}{144}+\frac{2\pi^4}{90}+C\\ \text{it follows}\qquad C&=-\frac{\pi^4}{720} \end{align*}

Setting $x=\frac{1}{2}$ in (4) we finally obtain with $C=-\frac{\pi^4}{720}$ and noting that \begin{align*} \operatorname{Li}_2\left(\frac{1}{2}\right)&=\frac{\pi^{2}}{12}-\frac{1}{2}\ln^2(2)\\ \operatorname{Li}_3\left(\frac{1}{2}\right)&=\frac{7}{8}\zeta(3)+\frac{1}{6}\ln^3(2)-\frac{\pi^{2}}{12}\ln 2\\ \operatorname{Li}_4(-1)&=-\frac{7\pi^4}{720} \end{align*}

\begin{align*} \sum_{n=1}^\infty \frac{H_n}{n^32^n}&=-\zeta(3)\ln(2)+\frac{1}{8}\ln^4(2) +\frac{1}{2}\operatorname{Li}_2\left(\frac{1}{2}\right)\left(\ln^2(2)-\frac{\pi^2}{6}\right)\\ &\qquad+\operatorname{Li}_3\left(\frac{1}{2}\right)\ln 2-\operatorname{Li}_4(-1)+\operatorname{Li}_4\left(\frac{1}{2}\right)-\frac{\pi^4}{720}\\ &=-\frac{1}{8}\ln 2\zeta(3)+\frac{1}{24}\ln^4(2)+\frac{\pi^4}{720}+ \operatorname{Li}_4\left(\frac{1}{2}\right)\\ &\stackrel{.}{=}0.55824 \end{align*} and the claim follows.

Note: Two aspects remain open. The important one is a derivation of \begin{align*} \int&\frac{\ln(1-x)}{x}\operatorname{Li}_2{(1-x)}dx \end{align*} without support from WA. It would also be nice to find some further simplifications of the final expression (4).

• Sorry, I cannot see any difference between Tunk-Fey's final result, Cleo's result and yours except the order of terms. They all numerically evaluate to $0.55823730083320863825151737933247...$ that agrees with approximate numerical summation in Mathematica. Jul 11, 2016 at 5:50
• @VladimirReshetnikov: Cleo's answer is correct. But Tunk-Feys final expression (4) does not give this answer, when evaluated at $x=\frac{1}{2}$. Contrary to his claim his final expression evaluates to $1.36998$. See my referred answer at the beginning of my answer and the last comments to Tunk-Feys answer. Jul 11, 2016 at 6:51
• Oh, I see. Indeed, the expression $(4)$ gives a wrong result when evaluated at $x=1/2$. The final $\color{purple}{\text{purple}}$ expression is correct, but it does not agree with $(4)$. Jul 11, 2016 at 16:27
• @VladimirReshetnikov: Correct! It's not obvious at the first glance. Jul 11, 2016 at 16:29
• I was lost in Tunk-Fey Eqtn. $\left(2\right)$ where he sets $x \mapsto 1 - x$ in an $\underline{indefinite}$ integral. I was waiting he recovers somehow later this step. Maybe, it's related to what you pointed out. Jul 11, 2016 at 19:42

Start with the series $$\sum_{n=1}^\infty H_n z^n = - \dfrac{\ln(1-z)}{1-z} = f_0(z)$$

Then (according to Maple 18) $$\sum_{n=1}^\infty \dfrac{H_n}{n} z^n = \int_0^z \dfrac{f_0(t)}{t}\; dt = \operatorname{Li}_{2}(1-z) + \dfrac{\ln(1-z)^2}{2} = f_1(z)$$

$$\displaystyle \sum_{n=1}^\infty \dfrac{H_n}{n^2} z^n = \int_0^z \dfrac{f_1(t)}{t} dt$$

$$= \zeta \left( 3 \right) +\dfrac{1}{2}\, \ln^2 (1-z) \ln \left( z \right) +\ln (1-z) \operatorname{Li}_{2} (z) -\operatorname{Li}_{3}(1-z) + \operatorname{Li}_{3}(z)$$

But for the next integration it fails to find a closed form. $$\sum_{n=1}^\infty \dfrac{H_n}{n^3} z^n = \int_0^z f_2(t)\; dt$$

• If you're going to do it in terms of a definite integral involving polylogarithms, you could take note that: $$H_{n}=\int_{0}^1\frac{1}{1-x}(1-x^n) dx$$ Then divide through by $n^32^n$ and sum over $\mathbb{N}$ so that: $$\sum_{n=1}^\infty \frac{H_{n}}{n^32^n}=\int_{0}^1\frac{(Polylog[3,1/2]-Polylog[3,x/2])}{x-1}dx$$ Aug 26, 2014 at 0:20
• It is also interesting to notice that $$\operatorname{Li}_3(1/2)=\frac{1}{24} (-2 \pi^2 \log 2 + 4 \log^3 2 + 21 \zeta(3))$$ appears also in my previous formula. Aug 26, 2014 at 0:32

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{n = 1}^{\infty}{H_{n} \over n^{3}\,2^{n}} & = \sum_{n = 1}^{\infty}{H_{n} \over 2^{n}} \bracks{{1 \over 2}\int_{0}^{1}\ln^{2}\pars{x}\,x^{n - 1}\,\dd x} = {1 \over 2}\int_{0}^{1}\ln^{2}\pars{x}\sum_{n = 1}^{\infty} \bracks{H_{n}\pars{x \over 2}^{n}}{\dd x \over x} \\[5mm] &= {1 \over 2}\int_{0}^{1}\ln^{2}\pars{x} \bracks{-\,{\ln\pars{1 - x/2} \over 1 - x/2}}\,{\dd x \over x} = -\,{1 \over 2}\int_{0}^{1/2} {\ln^{2}\pars{2x}\ln\pars{1 - x} \over \pars{1 - x}x}\,\dd x \\[5mm] & = -\,{1 \over 2}\int_{0}^{1/2}{\ln^{2}\pars{2x}\ln\pars{1 - x} \over x}\,\dd x - {1 \over 2}\int_{0}^{1/2}{\ln^{2}\pars{2x}\ln\pars{1 - x} \over 1 - x}\,\dd x \\[5mm] & = {1 \over 2}\int_{0}^{1/2}\mrm{Li}_{2}'\pars{x}\ln^{2}\pars{2x}\,\dd x - {1 \over 2}\int_{1/2}^{1}{\ln^{2}\pars{2\bracks{1 - x}}\ln\pars{x} \over x} \,\dd x \\[1cm] & = -\int_{0}^{1/2}\mrm{Li}_{3}'\pars{x}\ln\pars{2x}\,\dd x \\[5mm] & - {1 \over 2}\,\ln^{2}\pars{2}\int_{1/2}^{1}{\ln\pars{x} \over x}\,\dd x - \ln\pars{2}\int_{1/2}^{1}{\ln\pars{1 - x}\ln\pars{x} \over x}\,\dd x - {1 \over 2}\int_{1/2}^{1}{\ln^{2}\pars{1 - x}\ln\pars{x} \over x}\,\dd x \\[1cm] & = \int_{0}^{1/2}\mrm{Li}_{4}'\pars{x}\dd x + {1 \over 4}\,\ln^{4}\pars{2} + \ln\pars{2}\int_{1/2}^{1}\mrm{Li}_{2}'\pars{x}\ln\pars{x}\,\dd x - {1 \over 2}\int_{1/2}^{1}{\ln^{2}\pars{1 - x}\ln\pars{x} \over x}\,\dd x \\[1cm] & = \mrm{Li}_{4}\pars{1 \over 2} + {1 \over 4}\,\ln^{4}\pars{2} + \ln\pars{2}\bracks{% \mrm{Li}_{2}\pars{1 \over 2}\ln\pars{2} -\int_{1/2}^{1}\mrm{Li}_{3}'\pars{x}\,\dd x} \\[5mm] & - {1 \over 2}\int_{1/2}^{1}{\ln^{2}\pars{1 - x}\ln\pars{x} \over x}\,\dd x \\[1cm] & = \mrm{Li}_{4}\pars{1 \over 2} + {1 \over 4}\,\ln^{4}\pars{2} + \ln\pars{2}\bracks{% \mrm{Li}_{2}\pars{1 \over 2}\ln\pars{2} - \mrm{Li}_{3}\pars{1} + \mrm{Li}_{3}\pars{1 \over 2}} \\[5mm] & - {1 \over 2}\int_{1/2}^{1}{\ln^{2}\pars{1 - x}\ln\pars{x} \over x}\,\dd x \end{align}

Since values of $\ds{\,\mrm{Li}_{2}\pars{1/2}}$ and $\ds{\,\mrm{Li}_{3}\pars{1/2}}$ are well known and $\ds{\,\mrm{Li}_{3}\pars{1} = \zeta\pars{3}}$:

\begin{align} \sum_{n = 1}^{\infty}{H_{n} \over n^{3}\,2^{n}} & = -\,{1 \over 12}\,\ln^{4}\pars{2} - {1 \over 8}\,\ln\pars{2}\zeta\pars{3} + \,\mrm{Li}_{4}\pars{1 \over 2} - {1 \over 2}\ \underbrace{\int_{1/2}^{1}{\ln^{2}\pars{1 - x}\ln\pars{x} \over x}\,\dd x} _{\ds{\equiv\ \mc{I}}} \label{1}\tag{1} \end{align}

$\ds{\large\mc{I}:\ ?}$. \begin{align} \mc{I} & \equiv \int_{1/2}^{1}{\ln^{2}\pars{1 - x}\ln\pars{x} \over x}\,\dd x \\[5mm] & = {1 \over 3}\int_{1/2}^{1}\!{\ln^{3}\pars{1 - x} \over x}\dd x - {1 \over 3}\int_{1/2}^{1}\!{\ln^{3}\pars{x} \over x}\dd x - {1 \over 3}\int_{1/2}^{1}\!\ln^{3}\pars{1 - x \over x}{\dd x \over x} + \int_{1/2}^{1}\!{\ln\pars{1 - x}\ln^{2}\pars{x} \over x}\,\dd x \\[5mm] & = {1 \over 3}\int_{0}^{1/2}{\ln^{3}\pars{x} \over 1 - x}\dd x + {1 \over 12}\,\ln^{4}\pars{2} + {1 \over 3}\int_{0}^{-1}{\ln^{3}\pars{-x} \over 1 - x}\,\dd x - \int_{1/2}^{1}\mrm{Li}_{2}'\pars{x}\ln^{2}\pars{x}\,\dd x \\[1cm] & = {1 \over 3}\bracks{-\ln^{4}\pars{2} - 3\int_{0}^{1/2}\mrm{Li}_{2}'\pars{x}\ln^{2}\pars{x}\dd x} + {1 \over 12}\,\ln^{4}\pars{2} - \int_{0}^{-1}\mrm{Li}_{2}'\pars{x}\ln^{2}\pars{-x}\,\dd x \\[5mm] & -\int_{1/2}^{1}\mrm{Li}_{2}'\pars{x}\ln^{2}\pars{x}\,\dd x \\[1cm] & = -\,{1 \over 4}\,\ln^{4}\pars{2} -\int_{0}^{1}\mrm{Li}_{2}'\pars{x}\ln^{2}\pars{x}\,\dd x - \int_{0}^{-1}\mrm{Li}_{2}'\pars{x}\ln^{2}\pars{-x}\,\dd x \end{align}

The remaining integrals can be straightforward evaluated by successive integration by parts and by using the $\ds{\,\mrm{Li}_{s}}$ recursive property. Namely,

\begin{align} &\int\mrm{Li}_{2}'\pars{x}\ln^{2}\pars{\pm x}\,\dd x = \mrm{Li}_{2}\pars{x}\ln^{2}\pars{\pm x} - 2\int\mrm{Li}_{3}'\pars{x}\ln\pars{\pm x}\,\dd x \\[5mm] & = \mrm{Li}_{2}\pars{x}\ln^{2}\pars{\pm x} - 2\,\mrm{Li}_{3}\pars{x}\ln\pars{\pm x} + 2\int\mrm{Li}_{4}'\pars{x}\,\dd x \\[5mm] & =\ \bbox[15px,#ffe,border:1px dotted navy]{\ds{% \mrm{Li}_{2}\pars{x}\ln^{2}\pars{\pm x} - 2\,\mrm{Li}_{3}\pars{x}\ln\pars{\pm x} + 2\,\mrm{Li}_{4}\pars{x}}} \end{align} such that $$\mc{I} \equiv \int_{1/2}^{1}{\ln^{2}\pars{1 - x}\ln\pars{x} \over x}\,\dd x =\ \bbox[15px,#ffe,border:1px dotted navy]{\ds{% -\,{1 \over 4}\,\ln^{4}\pars{2} - {\pi^{4} \over 360}}}\label{2}\tag{2}$$

With \eqref{1} and \eqref{2}: \begin{align} \sum_{n = 1}^{\infty}{H_{n} \over n^{3}\,2^{n}} & = -\,{1 \over 12}\,\ln^{4}\pars{2} - {1 \over 8}\,\ln\pars{2}\zeta\pars{3} + \,\mrm{Li}_{4}\pars{1 \over 2} - {1 \over 2} \bracks{-\,{1 \over 4}\,\ln^{4}\pars{2} - {\pi^{4} \over 360}} \\[5mm] & =\ \bbox[25px,#ffe,border:1px dotted navy]{\ds{% {1 \over 720}\,\pi^{4} + {1 \over 24}\,\ln^{4}\pars{2} - {1 \over 8}\,\ln\pars{2}\zeta\pars{3} + \,\mrm{Li}_{4}\pars{1 \over 2}}}\ \approx 0.5582 \end{align}

• Lengthy, but crystal clear. Aug 30, 2017 at 18:35

You can have instead the equivalent integral representation

$$I = \int_{0}^{1}\frac{\ln^2(u)\ln(1-u/2)}{u(u-2)}du \sim .5582373010.$$

Alternative approach:

first we start with proving the following equality that appeared as Problem 11921 in The American Mathematical Monthly 2016 proposed by Cornel Ioan Valean: $$\begin{equation*} S=\ln^22\sum_{n=1}^{\infty}\frac{H_n}{(n+1) 2^{n+1}}+\ln2\sum_{n=1}^{\infty}\frac{H_n}{(n+1)^2 2^n}+\sum_{n=1}^{\infty}\frac{H_n}{(n+1)^3 2^n}=\frac14\ln^42+\frac14\zeta(4) \end{equation*}$$ Proof: lets start with the following integral $$I=\displaystyle \int_{1/2}^{1} \frac{\ln(1-x)\ln^2x}{1-x}\,dx$$

By using

$$\frac{\ln(1-x)}{1-x}=-\displaystyle \sum_{n=1}^{\infty}H_n x^n$$

we can write

$$I=-\sum_{n=1}^{\infty}H_n\int_{1/2}^{1}x^n \ln^2x\,dx$$

$$=-\sum_{n=1}^{\infty}H_n\left( -\frac{\ln^22}{(n+1)2^{n+1}}-\frac{\ln2}{(n+1)2^{n+1}}-\frac{1}{(n+1)^32^n}+\frac{2}{(n+1)^3}\right)$$

$$=S-2\sum_{n=1}^{\infty}\frac{H_n}{(n+1)^3} \tag{1}$$

On the other hand, upon integrating by parts we obtain $$\begin{equation*} I=\frac12\ln^42+\int_{1/2}^1 \frac{\ln^2x\ln(1-x)}{x}\,dx\overset{x\mapsto 1-x}{=}\frac12\ln^42+\int_0^{1/2}\frac{\ln^2x\ln(1-x)}{1-x}\,dx \end{equation*}$$ Adding the integral $$I=\int_{1/2}^{1}\frac{\ln^2x\ln(1-x)}{1-x}\,dx\$$ to both sides

$$2I=\frac12\ln^42+\int_0^1 \frac{\ln^2x\ln(1-x)}{1-x}\,dx=\frac12\ln^42-\sum_{n=1}^{\infty}H_n\int_{0}^{1}x^n\ln^2x\,dx$$ $$=\frac12\ln^42-2\sum_{n=1}^{\infty}\frac{H_n}{(n+1)^3}\Longrightarrow I=\frac14\ln^42-\sum_{n=1}^{\infty}\frac{H_n}{(n+1)^3} \tag{2}$$

combining $$(1)$$ and $$(2)$$ yields

$$S=\frac14\ln^42+\sum_{n=1}^{\infty}\frac{H_n}{(n+1)^3}=\frac14\ln^42-\zeta(4)+\sum_{n=1}^{\infty}\frac{H_n}{n^3}$$

subbing $$\sum_{n=1}^{\infty}\frac{H_n}{n^3}=\frac54\zeta(4)$$ completes the proof.

Using the proved equality: \begin{align*} \frac14\ln^42+\frac14\zeta(4)&=\ln^22\sum_{n=1}^{\infty}\frac{H_n}{(n+1) 2^{n+1}}+\ln2\sum_{n=1}^{\infty}\frac{H_n}{(n+1)^2 2^n}+\sum_{n=1}^{\infty}\frac{H_n}{(n+1)^3 2^n}\\ &=\ln^22\sum_{n=1}^{\infty}\frac{H_{n-1}}{n 2^n}+2\ln2\sum_{n=1}^{\infty}\frac {H_{n-1}}{n^2 2^n}+2\sum_{n=1}^{\infty}\frac{H_{n-1}}{n^3 2^n}\\ &=\ln^22\sum_{n=1}^{\infty}\frac{H_n}{n 2^n}+2\ln2\sum_{n=1}^{\infty}\frac {H_n}{n^2 2^n} +2\sum_{n=1}^{\infty}\frac{H_n}{n^3 2^n}-\ln^22\sum_{n=1}^{\infty}\frac{1}{2^n n^2}\\ &\quad -2\ln2\sum_{n=1}^{\infty}\frac{1}{ n^32^n}-2\sum_{n=1}^{\infty}\frac{1}{n^42^n} \end{align*} rearrange the terms to get

$$\sum_{n=1}^{\infty}\frac{H_n}{n^3 2^n}=-\ln2\sum_{n=1}^{\infty}\frac{H_n}{n^2 2^n}-\frac12\ln^22\sum_{n=1}^{\infty}\frac{H_n}{n 2^n}+\operatorname{Li_4}\left( \frac12\right)+\ln2\operatorname{Li_3}\left( \frac12\right)\\+\frac12\ln^22\operatorname{Li_2}\left( \frac12\right)+\frac18\zeta(4)+\frac18\ln^42$$

plugging the values of the first and second sum proved here and here respectively, along with the values of $$\displaystyle\operatorname{Li_3}\left(\frac12\right)$$ and $$\displaystyle\operatorname{Li_2}\left(\frac12\right)$$ we obtain \begin{align} \sum_{n=1}^\infty \frac{H_n}{2^nn^3}=\color{blue}{\operatorname{Li}_4\left(\frac12\right)+\frac18\zeta(4)-\frac18\ln2\zeta(3)+\frac1{24}\ln^42}, \end{align}

By first finding the following integral by using the algebraic identity $$a^2b=\frac{1}{6}\left(a+b\right)^3-\frac{1}{6}\left(a-b\right)^3-\frac{1}{3}b^3$$ one can easily prove avoiding Euler sums that: $$\int _0^1\frac{\ln ^2\left(1-x\right)\ln \left(1+x\right)}{1+x}\:dx=-\frac{1}{4}\zeta \left(4\right)+2\ln \left(2\right)\zeta \left(3\right)-\ln ^2\left(2\right)\zeta \left(2\right)+\frac{1}{4}\ln ^4\left(2\right)$$ Now: $$\int _0^1\frac{\ln ^2\left(1-x\right)\ln \left(1+x\right)}{1+x}\:dx=\frac{1}{2}\ln \left(2\right)\int _0^1\frac{\ln ^2\left(x\right)}{1-\frac{x}{2}}\:dx+\frac{1}{2}\int _0^1\frac{\ln ^2\left(x\right)\ln \left(1-\frac{x}{2}\right)}{1-\frac{x}{2}}\:dx$$ $$=2\ln \left(2\right)\sum _{k=1}^{\infty }\frac{1}{k^3\:2^k}-2\sum _{k=1}^{\infty }\frac{H_k}{k^3\:2^k}+2\sum _{k=1}^{\infty }\frac{1}{k^4\:2^k}$$ $$=2\ln \left(2\right)\operatorname{Li}_3\left(\frac{1}{2}\right)-2\sum _{k=1}^{\infty }\frac{H_k}{k^3\:2^k}+2\operatorname{Li}_4\left(\frac{1}{2}\right)$$ $$=\frac{7}{4}\ln \left(2\right)\zeta \left(3\right)-\ln ^2\left(2\right)\zeta \left(2\right)+\frac{1}{3}\ln ^4\left(2\right)-2\sum _{k=1}^{\infty }\frac{H_k}{k^3\:2^k}+2\operatorname{Li}_4\left(\frac{1}{2}\right)$$ By making use of the result we find: $$\sum _{k=1}^{\infty }\frac{H_k}{k^3\:2^k}=\frac{1}{8}\zeta \left(4\right)+\operatorname{Li}_4\left(\frac{1}{2}\right)-\frac{1}{8}\ln \left(2\right)\zeta \left(3\right)+\frac{1}{24}\ln ^4\left(2\right)$$