The calculation of the series $\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}}{n^4}$ Happy New Year 2020, Romania
In this recent post, Evaluate $\int_0^1 \frac{\arctan x\ln^2 x}{1+x^2}\,dx$, the proposed integral reduces to the calculation of $\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}}{n^3}$ which is known in the literature. Now, what can we say about the more advanced version of it, the one with $n^4$ in the denominator?

$$\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}}{n^4}$$

Can we do it by series manipulations? It looks like a new series in the literature.
A good note: With the previous result in hand which we combine with some of the Cornel's work, we immediately arrive at two delightful series results,

$$i) \ \sum _{n=1}^{\infty }(-1)^{n-1} \frac{ H_{2 n}^{(2)}}{n^3}=\frac{61 }{192}\pi ^2 \zeta (3)+\frac{1973 }{128}\zeta (5)+\frac{\pi ^5}{16}-\frac{1}{128} \pi  \psi ^{(3)}\left(\frac{1}{4}\right);$$
$$ii) \ \sum _{n=1}^{\infty } (-1)^{n-1} \frac{ H_{2 n}^{(3)}}{n^2}=\frac{\pi ^3 G}{8}+\frac{1}{64}\pi ^2 \zeta (3)-\frac{2997 }{256}\zeta (5)-\frac{\pi ^5}{32}+\frac{1}{256} \pi  \psi ^{(3)}\left(\frac{1}{4}\right).$$

How would you go proving these last two results?
Another good note: Remaining on the ground with alternating harmonic series of weight $5$ and harmonic numbers of the type $H_{2n}$, we might be curious to know what the value of the series $\displaystyle \sum _{n=1}^{\infty }(-1)^{n-1} \frac{ H_{2 n}^{(4)}}{n}$ is.

$$\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}^{(4)}}{n}$$
$$=4\zeta(5)-\frac{3}{128}\zeta(2)\zeta(3)-\frac{7}{128}\log(2)\zeta(4)+\frac{\pi^5}{192}-\frac{\pi^3}{16}G-\frac{\pi}{1536}\psi^{(3)}\left(\frac{1}{4}\right).$$

This last series is elegantly calculated in A simple strategy of calculating two alternating harmonic series generalizations where one may also find its generalization with respect to the order of the harmonic number.
 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
$$\begin{align}
  & \sum\limits_{k=1}^{\infty }{{{\left( -1 \right)}^{k}}{{H}_{2k}}f\left( k \right)} \\ 
 & =-\frac{1}{2}\sum\limits_{k=1}^{\infty }{{{\left( -1 \right)}^{k}}f'\left( k \right)}-\frac{\pi }{2}\sum\limits_{k=0}^{\infty }{{{\left( -1 \right)}^{k}}f\left( k+\tfrac{1}{2} \right)}-\sum\limits_{k=1}^{\infty }{{{\left( -1 \right)}^{k}}\left( \psi \left( 2k \right)+\gamma  \right)f\left( -k \right)}\\&-\underset{z=0}{\mathop{res}}\,\left( \psi \left( -2z \right)+\gamma  \right)\frac{\pi f\left( z \right)}{\sin \left( \pi z \right)} \\ 
\end{align}$$
This goes a long way to generalise such results as you can now pick any power by simply choosing f appropriately.  For example 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
$$\sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k}}{{H}_{2k}}}{{{k}^{4}}}}=\frac{\pi }{384}\left\{ {{\psi }^{\left( 3 \right)}}\left( \tfrac{3}{4} \right)-{{\psi }^{\left( 3 \right)}}\left( \tfrac{1}{4} \right) \right\}+\frac{{{\pi }^{2}}}{3}\zeta \left( 3 \right)+\frac{437}{64}\zeta \left( 5 \right)$$
