Alternating harmonic sum $\sum_{k\geq 1}\frac{(-1)^k}{k^3}H_k$

Let us first recall that harmonic numbers have generating function \begin{align} \sum_{k=1}^{\infty}H_kx^k=-\frac{\ln(1-x)}{1-x}, \end{align} and therefore \begin{align} S=\sum_{k=1}^{\infty}\frac{(-1)^k}{k^3}H_k&=\frac{1}{2}\sum_{k=1}^{\infty}(-1)^kH_k\int_0^{\infty}e^{-kx}x^2dx=\\ &=-\frac{1}{2}\int_0^{\infty}\frac{\ln(1+e^{-x})}{1+e^{-x}}x^2dx. \end{align} Mathematica knows how to evaluate the last integral in terms of zeta values and polylogarithms. Its answer is $$S=-\frac{11\pi^4}{360}+\frac{\ln^42-\pi^2\ln^22}{12}+2\mathrm{Li}_4\left(\frac12\right)+\frac{7\ln 2}{4}\zeta(3).$$ It is unlikely that it can be simplified further: Wolfram Alpha proposes alternative expressions for $\mathrm{Li}_{2}\left(\frac12\right)$ and $\mathrm{Li}_{3}\left(\frac12\right)$ in terms of elementary functions and zeta values, but does not suggest anything simpler for $\mathrm{Li}_{4}\left(\frac12\right)$.

Related problems: (I), (II), (III). Your sum is a special case of the following general case which I derived an integral representation for it

$$ A(p,q) =\sum_{k=1}^{\infty} \frac{(-1)^{k+1}H^{(p)}_k}{k^q} = \frac{\left( -1 \right) ^{q}}{\Gamma(q)}\int _{0}^{1}\!{\frac { \left( \ln\left( u \right)\right)^{q-1}{\mathrm{Li}_{p}(-u)} }{ u\left( 1+u \right) }}{du}. $$

where $ \mathrm{Li}_{p}(z) $ is the polylogarithm function. So, letting $p=1$ and $q=3$ in the above formula gives an integral representation for your sum

$$ A(1,3) =\sum_{k=1}^{\infty} \frac{(-1)^{k+1}H^{(1)}_k}{k^3} = \frac{\left( -1 \right) ^{3}}{\Gamma(3)}\int _{0}^{1}\!{\frac { \left( \ln\left( u\right) \right)^{3-1}{\mathrm{Li}_{1}(-u)} }{ u (1+u) }}{du}.$$

$$ \implies A(1,3) =\frac{1}{2}\int _{0}^{1}\!{\frac { \left( \ln \left( u \right) \right) ^{2} \ln \left( 1+u \right) }{u\left(1+u\right)}}{du} \sim 0.8592471579. $$

See here for related techniques.

Note:

1) $$ \mathrm{Li}_{1}(-u)=-\ln(1+u). $$

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{k\ \geq\ 1}{\pars{-1}^{k} \over k^{3}}H_{k} & = \sum_{k = 1}^{\infty}\pars{-1}^{k}H_{k}\ \overbrace{\bracks{{1 \over 2}\int_{0}^{1}\ln^{2}\pars{x}x^{k - 1}\,\dd x}} ^{\ds{1 \over k^{3}}} \\[5mm] & = {1 \over 2}\int_{0}^{1}\ln^{2}\pars{x} \bracks{\sum_{k = 1}^{\infty}H_{k}\pars{-x}^{k}}\,{\dd x \over x} \\[5mm] & = {1 \over 2}\int_{0}^{1}\ln^{2}\pars{x} \bracks{-\,{\ln\pars{1 + x} \over 1 + x}}\,{\dd x \over x} = -\,{1 \over 2}\int_{0}^{1} {\ln^{2}\pars{x}\ln\pars{1 + x} \over \pars{1 + x}x}\,\dd x \\[5mm] & = {1 \over 2}\int_{0}^{1}{\ln^{2}\pars{x}\ln\pars{1 + x} \over 1 + x}\,\dd x - {1 \over 2}\int_{0}^{1}{\ln^{2}\pars{x}\ln\pars{1 + x} \over x}\,\dd x \\[1cm] & = {1 \over 6}\int_{0}^{1}{3\ln^{2}\pars{x}\ln\pars{1 + x} - 3\ln\pars{x}\ln^{2}\pars{1 + x} \over 1 + x}\,\dd x \\[5mm] & + {1 \over 2}\int_{0}^{1}{\ln\pars{x}\ln^{2}\pars{1 + x} \over 1 + x}\,\dd x + {1 \over 2}\int_{0}^{-1}\mrm{Li}_{2}'\pars{x}\ln^{2}\pars{-x}\,\dd x \\[1cm] & = {1 \over 6}\int_{0}^{1}{\ln^{3}\pars{x} \over 1 + x}\,\dd x - {1 \over 6}\int_{0}^{1}{\ln^{3}\pars{1 + x} \over 1 + x}\,\dd x \\[5mm] &- {1 \over 6}\int_{0}^{1}\ln^{3}\pars{x \over 1 + x}\,{\dd x \over 1 + x} - {1 \over 6}\int_{0}^{1}{\ln^{3}\pars{1 + x} \over x}\,\dd x \\[5mm] & - \int_{0}^{-1}\mrm{Li}_{3}'\pars{x}\ln\pars{-x}\,\dd x \\[1cm] & = -\,{1 \over 6}\int_{0}^{-1}{\ln^{3}\pars{-x} \over 1 - x}\,\dd x - {1 \over 24}\,\ln^{4}\pars{2} - {1 \over 6}\int_{0}^{1/2}{\ln^{3}\pars{x} \over 1 - x}\,\dd x \\[5mm] & +{1 \over 6}\int_{1}^{2}{\ln^{3}\pars{x} \over 1 - x}\,\dd x\ +\ \underbrace{\quad\int_{0}^{-1}\mrm{Li}_{4}'\pars{x}\,\dd x\quad} _{\ds{= \,\mrm{Li}_{4}\pars{-1} = -\,{7 \over 720}\,\pi^{4}}}\label{1}\tag{1} \end{align}

The remaining integrals are evaluated by successive integration by parts. Namely,

\begin{align} \int{\ln^{3}\pars{\pm x} \over 1 - x}\,\dd x & = -\ln\pars{1 - x}\ln^{3}\pars{\pm x} - 3\int\mrm{Li}_{2}'\pars{x}\ln^{2}\pars{\pm x}\,\dd x \\[5mm] & = -\ln\pars{1 - x}\ln^{3}\pars{\pm x} - 3\,\mrm{Li}_{2}\pars{x}\ln^{2}\pars{\pm x} + 6\int\mrm{Li}_{3}'\pars{x}\ln\pars{\pm x}\,\dd x \\[1cm] & = -\ln\pars{1 - x}\ln^{3}\pars{\pm x} - 3\,\mrm{Li}_{2}\pars{x}\ln^{2}\pars{\pm x} + 6\,\mrm{Li}_{3}\pars{x}\ln\pars{\pm x} \\[5mm] & - 6\int\mrm{Li}_{4}'\pars{x}\,\dd x \\[1cm] & = -\ln\pars{1 - x}\ln^{3}\pars{\pm x} - 3\,\mrm{Li}_{2}\pars{x}\ln^{2}\pars{\pm x} + 6\,\mrm{Li}_{3}\pars{x}\ln\pars{\pm x} \\[5mm] & - 6\,\mrm{Li}_{4}\pars{x}\label{2}\tag{2} \end{align}

\begin{align} S&=\sum_{n=1}^\infty\frac{(-1)^n H_n}{n^3}=\frac12\int_0^1\frac{\ln^2x}{x}\sum_{n=1}^\infty H_n(-x)^n\ dx=-\frac12\underbrace{\int_0^1\frac{\ln^2x\ln(1+x)}{x(1+x)}\ dx}_{x=(1-y)/y}\\ &=\frac12\underbrace{\int_{1/2}^1\frac{\ln^2((1-x)/x)\ln(x)}{1-x}\ dx}_{x=1-y}=\frac12\int_0^{1/2}\frac{\ln^2(x/(1-x))\ln(1-x)}{x}\ dx\\ &=\frac12\left(\int_0^{1/2}\frac{\ln^2x\ln(1-x)}{x}\ dx+\int_0^{1/2}\frac{\ln^3(1-x)}{x}\ dx\right)-\int_0^{1/2}\frac{\ln x\ln^2(1-x)}{x}\ dx\\ &=\frac12\left(I_1+I_2\right)-I_3 \end{align} Applying IBP for the first integral by setting $dv=\ln^2x/x$ and $u=\ln(1-x)$ and letting $x=1-y$ for the second integral, we have:

\begin{align} I_1+I_2&=\frac13\ln^42+\frac13\int_0^{1/2}\frac{\ln^3x}{1-x}\ dx+\int_{1/2}^1\frac{\ln^3x}{1-x}\ dx\\ &=\frac13\ln^42+\int_0^1\frac{\ln^3x}{1-x}\ dx-\frac23\int_0^{1/2}\frac{\ln^3x}{1-x}\ dx\\ &=\frac13\ln^42-6\zeta(4)-\frac23\sum_{n=1}^\infty\int_0^{1/2}x^{n-1}\ln^3x\ dx\\ &=\frac13\ln^42-6\zeta(4)+\frac23\sum_{n=1}^\infty\left(\frac{\ln^32}{n2^n}+\frac{3\ln^22}{n^22^n}+\frac{6\ln2}{n^32^n}+\frac{6}{n^42^n}\right)\\ &=4\operatorname{Li_4}\left(\frac12\right)+4\ln2\operatorname{Li_3}\left(\frac12\right)+2\ln^22\operatorname{Li_2}\left(\frac12\right)+\ln^42-6\zeta(4) \end{align} Applying IBP for the third integral by setting $dv=\ln x/x$ and $u=\ln^2(1-x)$, \begin{align} I_3=\int_0^{1/2}\frac{\ln x\ln^2(1-x)}{x}\ dx&=\frac12\ln^42+\underbrace{\int_0^{1/2}\frac{\ln^2x\ln(1-x)}{1-x}\ dx}_{x=1-y}\\ &=\frac12\ln^42+\int_{1/2}^1\frac{\ln x\ln^2(1-x)}{x}\ dx \end{align} By adding the third integral to both sides, we get: \begin{align} I_3&=\frac14\ln^42+\frac12\int_0^1\frac{\ln x\ln^2(1-x)}{x}\ dx\\ &=\frac14\ln^42+\sum_{n=1}^\infty\left(\frac{H_n}{n}-\frac{1}{n^2}\right)\int_0^1 x^{n-1}\ln x\ dx\\ &=\frac14\ln^42+\sum_{n=1}^\infty\left(\frac{H_n}{n}-\frac{1}{n^2}\right)\left(-\frac{1}{n^2}\right)\\ &=\frac14\ln^42+\zeta(4)-\sum_{n=1}^\infty\frac{H_n}{n^3}\\ &=\frac14\ln^42-\frac14\zeta(4) \end{align} Grouping $I_1, I_2$ and $I_3$: \begin{align} S&=2\operatorname{Li_4}\left(\frac12\right)+2\ln2\operatorname{Li_3}\left(\frac12\right)+\ln^22\operatorname{Li_2}\left(\frac12\right)-\frac{11}4\zeta(4)+\frac14\ln^42\\ &=2\operatorname{Li_4}\left(\frac12\right)-\frac{11}4\zeta(4)+\frac74\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac{1}{12}\ln^42 \end{align} note that we used $$\operatorname{Li_3}\left( \frac12\right)=\frac78\zeta(3)-\frac12\ln2\zeta(2)+\frac16\ln^32$$ $$\operatorname{Li_2}\left( \frac12\right) =\frac12\zeta(2)-\frac12\ln^22$$

Bonus:

By Cauchy product we have

$$\operatorname{Li}_2^2(x)=4\sum_{n=1}^\infty x^n\frac{H_n}{n^3}+2\sum_{n=1}^\infty x^n\frac{H_n^{(2)}}{n^2}-6\operatorname{Li}_4(x)$$

set $x=-1$ and rearrange the term to have

$$\sum_{n=1}^\infty(-1)^n\frac{H_n^{(2)}}{n^2}=\frac12\operatorname{Li}_2^2(-1)+3\operatorname{Li}_4(-1)-2\sum_{n=1}^\infty(-1)^n\frac{H_n}{n^3}$$

substiuting the value of $\sum_{n=1}^\infty(-1)^n\frac{H_n}{n^3}$ along with the special values of $\operatorname{Li}_2(-1)=-\frac12\zeta(2)$ and $\operatorname{Li}_4(-1)=-\frac78\zeta(4)$ we get

$$\sum_{n=1}^{\infty}(-1)^n\frac{H_n^{(2)}}{n^2}=-4\operatorname{Li}_4\left(\frac12\right)+\frac{51}{16}\zeta(4)-\frac72\ln2\zeta(3)+\ln^22\zeta(2)-\frac16\ln^42$$

Here we compute the quantity \begin{equation} A(1,q) := \sum\limits_{k=1}^\infty \frac{(-1)^{k+1} H_k}{k^q} = \frac{(-1)^{q+1}}{(q-1)!} \int\limits_0^1 \frac{[\log(u)]^{q-1} \cdot \log(1+u)}{u(u+1)} d u \end{equation} We have: \begin{eqnarray} A(1,q) &=& \left.\frac{(-1)^{q+1}}{(q-1)!} \frac{\partial^{q-1}}{\partial \theta_1^{q-1}} \frac{\partial^{1}}{\partial \theta_2^{1}} \int\limits_0^1 u^{\theta_1-1} \cdot (u+1)^{\theta_2-1} du \right|_{\theta_1=0,\theta_2=0}\\ &=& \left.\frac{(-1)^{q+1}}{(q-1)!} \frac{\partial^{q-1}}{\partial \theta_1^{q-1}} \frac{\partial^{1}}{\partial \theta_2^{1}} \frac{2^{\theta_2-1}}{\theta_1} F_{2,1}[1, 1-\theta_2,1+\theta_1; 1/2] \right|_{\theta_1=0,\theta_2=0} \\ &=&\left.\frac{(-1)^{q+1}}{(q-1)!} \frac{\partial^{q-1}}{\partial \theta_1^{q-1}} \frac{\log(2) F_{2,1}[1,1,1+\theta_1;1/2] - F_{2,1}^{(0,1,0,0)}[1,1,1+\theta_1;1/2]}{2 \theta_1} \right|_{\theta_1=0} \\ &=& \left.\frac{1}{2} \sum\limits_{l=0}^{q-1} \frac{(-1)^l}{l!} \left[\log(2) \cdot F_{2,1}^{(0,0,l,0)}[1,1,1+\theta_1;1/2] - F_{2,1}^{(0,1,l,0)}[1,1,1+\theta_1;1/2]\right] \frac{1}{\theta_1^{q-l}} \right|_{\theta_1=0} \end{eqnarray} In the second line we integrated by parts and used the definition of the hypergeometric function. In the third line we differentiated with respect to $\theta_2$ and took the limit $\theta_2 \rightarrow 0$ and we defined \begin{equation} F_{2,1}^{(0,p,q,0)}[a,b,c;x] := \frac{\partial^p}{\partial b^p} \frac{\partial^q}{\partial c^q} F_{2,1}[a,b,c;x] \end{equation} and finally in the last line we applied the chain rule. Now clearly the last expression on the right hand side is singular at zero. Therefore we need to reduce it to the common denominator and then apply the d'Hospital rule. Here the common denominator is $\theta_1^q$ and we have to differentiate $q$-times both the numerator and the denominator. The final result is quite simple. It reads: \begin{equation} A(1,q) = \frac{(-1)^{q+1}}{2^q q!} \left[ q \log(2) \left( \Psi^{(q-1)}(\frac{1}{2}) - \Psi^{(q-1)}(1) \right) - 2^{q-1} F_{2,1}^{(0,1,q,0)}[1,1,1,;1/2] \right] \end{equation} The last thing that remains to be checked if the higher derivatives of the hipergeometric function all reduce to polygamma function values at unity and at one half. I will check this later.

Now, using the integral representation of the hipergeometric function we fairly easy find the derivative in question. It reads: \begin{equation} \frac{1}{2} \frac{(-1)^{q-1}}{q!} F_{2,1}^{(0,1,q,0)}[1,1,1;1/2] = \left(1+\log(2)\right) Li_q(-1) - \frac{(-1)^{q-1}}{q!} \int\limits_0^1 \frac{\log(\xi)^q \cdot \log(1+\xi)}{(1+\xi)^2} d\xi \end{equation} Inserting this into the equation for $A(1,q)$ after using Interesting connection between polylogarithms and polygamma functions. we get: \begin{equation} A(1,q) = -Li_q(-1) + \frac{(-1)^{q-1}}{q!} \int\limits_0^1 \frac{\log(\xi)^q \cdot \log(1+\xi)}{(1+\xi)^2} d\xi \end{equation} and finally using Antiderivative of a function involving logarithms and a fraction. we get the final result for $q=3$. We have: \begin{eqnarray} &&A(1,q) = -Li_q(-1) + \frac{(-1)^{q-1}}{q!} \cdot \\ && \!\!\!\!\!\!\!\left( -6 \text{Li}_4\left(\frac{1}{2}\right)+6 \text{Li}_4(2)-\frac{9 \zeta (3)}{2}-\frac{21}{2} \zeta (3) \log (2)+\frac{\pi ^4}{20}-\frac{\log ^4(2)}{4}+i \pi \log ^3(2)-\frac{1}{2} \pi ^2 \log ^2(2)\right) \\ &&=0.8592471579285906155... \end{eqnarray}