Finding a closed-form for the sum $\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}H_{2n}}{n^{4}}$ 
Let $\mathcal{S}$ denote the sum of the following alternating series:
$$\mathcal{S}:=\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}H_{2n}}{n^{4}}\approx-1.392562725547,$$
where $H_{n}$ denotes the $n$-th harmonic number.
Question: Is it possible to obtain a closed-form expression for $\mathcal{S}$ in terms of elementary functions and well-known special functions such as polylogarithms?

My approach was to use the following integral representation for the $n$-th harmonic number to convert the sum $\mathcal{S}$ into an integral via the technique of summing under the integral sign:
$$H_{n}=\int_{0}^{1}\mathrm{d}t\,\frac{1-t^{n}}{1-t};~~~\small{n\in\mathbb{Z}_{\ge0}}.$$
We find
$$\begin{align}
\mathcal{S}
&=\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}H_{2n}}{n^{4}}\\
&=\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}}{n^{4}}\int_{0}^{1}\mathrm{d}t\,\frac{1-t^{2n}}{1-t}\\
&=\int_{0}^{1}\mathrm{d}t\,\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}}{n^{4}}\cdot\frac{1-t^{2n}}{1-t}\\
&=\int_{0}^{1}\mathrm{d}t\,\frac{1}{1-t}\left[\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}}{n^{4}}-\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}t^{2n}}{n^{4}}\right]\\
&=\int_{0}^{1}\mathrm{d}t\,\frac{\operatorname{Li}_{4}{\left(-1\right)}-\operatorname{Li}_{4}{\left(-t^{2}\right)}}{1-t}\\
&=-\int_{0}^{1}\mathrm{d}t\,\frac{2\ln{\left(1-t\right)}\operatorname{Li}_{3}{\left(-t^{2}\right)}}{t};~~~\small{I.B.P.}\\
&=2\operatorname{Li}_{2}{\left(1\right)}\operatorname{Li}_{3}{\left(-1\right)}-\int_{0}^{1}\mathrm{d}t\,\frac{4\operatorname{Li}_{2}{\left(t\right)}\operatorname{Li}_{2}{\left(-t^{2}\right)}}{t};~~~\small{I.B.P.}\\
&=-\frac32\zeta{\left(2\right)}\,\zeta{\left(3\right)}-4\int_{0}^{1}\mathrm{d}t\,\frac{\operatorname{Li}_{2}{\left(t\right)}\operatorname{Li}_{2}{\left(-t^{2}\right)}}{t}.\\
\end{align}$$
Any ideas how to proceed from here?

 A: Consider
$$\psi \left( -z \right)+\gamma \underset{z\to n}{\mathop{=}}\,\frac{1}{z-n}+{{H}_{n}}+\sum\limits_{k=1}^{\infty }{\left( {{\left( -1 \right)}^{k}}H_{n}^{k+1}-\zeta \left( k+1 \right) \right){{\left( z-n \right)}^{k}}}, n\ge 0$$
Then
$$\left( \psi \left( -2z \right)+\gamma  \right)\frac{\pi f\left( z \right)}{\sin \left( \pi z \right)}\underset{2z\to 2k+1}{\mathop{=}}\,\frac{{{\left( -1 \right)}^{k}}\pi f\left( k+\tfrac{1}{2} \right)}{2\left( z-\tfrac{2k+1}{2} \right)}+O\left( 1 \right)$$
On the other hand we also have
$$\left( \psi \left( -2z \right)+\gamma  \right)\frac{\pi f\left( z \right)}{\sin \left( \pi z \right)}\underset{2z\to 2k}{\mathop{=}}\,\frac{{{\left( -1 \right)}^{k}}f\left( k \right)}{2{{\left( z-k \right)}^{2}}}+\frac{{{\left( -1 \right)}^{k}}\left\{ {{H}_{2k}}f\left( k \right)+\tfrac{1}{2}f'\left( k \right) \right\}}{z-k}+O\left( 1 \right), k\ge 0$$
Similarly
$$\left( \psi \left( -2z \right)+\gamma  \right)\frac{\pi f\left( z \right)}{\sin \left( \pi z \right)}\underset{z\to -k}{\mathop{=}}\,\frac{{{\left( -1 \right)}^{k}}\left( \psi \left( 2k \right)+\gamma  \right)f\left( -k \right)}{z+k}, k>0$$
The only other residues are those due to $f$ which we assume has one pole at the origin of order at least $2$.  The sum of residues over the entire plane is zero.  Hence
$$\begin{align}
  & \sum\limits_{k=1}^{\infty }{{{\left( -1 \right)}^{k}}{{H}_{2k}}f\left( k \right)} \\ 
 & =-\frac{1}{2}\sum\limits_{k=1}^{\infty }{{{\left( -1 \right)}^{k}}f'\left( k \right)}-\frac{\pi }{2}\sum\limits_{k=0}^{\infty }{{{\left( -1 \right)}^{k}}f\left( k+\tfrac{1}{2} \right)}-\sum\limits_{k=1}^{\infty }{{{\left( -1 \right)}^{k}}\left( \psi \left( 2k \right)+\gamma  \right)f\left( -k \right)}\\&-\underset{z=0}{\mathop{res}}\,\left( \psi \left( -2z \right)+\gamma  \right)\frac{\pi f\left( z \right)}{\sin \left( \pi z \right)} \\ 
\end{align}$$
Letting $f\left( z \right)=\frac{1}{{{z}^{4}}}$ we find
$$\begin{align}
  & \sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k}}{{H}_{2k}}}{{{k}^{4}}}} \\ 
 & =-\frac{\pi }{192}\left\{ {{\psi }^{\left( 3 \right)}}\left( \tfrac{1}{4} \right)-{{\psi }^{\left( 3 \right)}}\left( \tfrac{3}{4} \right) \right\}+\frac{2}{3}{{\pi }^{2}}\zeta \left( 3 \right)+\frac{113}{8}\zeta \left( 5 \right)-\sum\limits_{k=1}^{\infty }{{{\left( -1 \right)}^{k}}\frac{{{H}_{2k-1}}}{{{k}^{4}}}} \\ 
\end{align}$$
Now
$$\sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k}}{{H}_{2k}}}{{{k}^{4}}}}=A-\sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k}}{{H}_{2k-1}}}{{{k}^{4}}}}\Rightarrow \sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k}}\left( {{H}_{2k}}+{{H}_{2k-1}} \right)}{{{k}^{4}}}}=A$$
But we can write this as
$$\sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k}}\left( \frac{1}{2k}+2{{H}_{2k-1}} \right)}{{{k}^{4}}}}=A$$
hence
$$\sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k}}{{H}_{2k-1}}}{{{k}^{4}}}}=\frac{1}{2}A+\frac{15}{64}\zeta \left( 5 \right)$$
We have then
$$\sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k}}{{H}_{2k}}}{{{k}^{4}}}}=\frac{\pi }{384}\left\{ {{\psi }^{\left( 3 \right)}}\left( \tfrac{3}{4} \right)-{{\psi }^{\left( 3 \right)}}\left( \tfrac{1}{4} \right) \right\}+\frac{{{\pi }^{2}}}{3}\zeta \left( 3 \right)+\frac{437}{64}\zeta \left( 5 \right)$$
