Do we have closed form for
$$\sum_{k=1}^\infty (-1)^k\frac{H_k^{(4)}}{k^2}\ ?$$
where $H_k$ is the harmonic number.
I encountered this sum while I was working on calculating $\displaystyle \sum_{k=1}^\infty\frac{H_k^{(2)}}{(2k+1)^4}$ and I found using Abel's summation that
$$\sum_{k=1}^\infty\frac{H_k^{(2)}}{(2k+1)^4}+2\sum_{k=1}^\infty (-1)^k\frac{H_k^{(4)}}{k^2}=\frac{31}{16}\zeta^2(3)-\frac{13}{3}\zeta(6)$$
If the reader is curious about how I got this relation, I just followed the same approach here.
I am familiar with $\displaystyle \sum_{k=1}^\infty (-1)^k\frac{H_k^{(4)}}{k}$ but not the $\displaystyle \frac1{k^2}$ version.
Here is the integral representation of each
Knowing that
$$\sum_{k=1}^\infty H_k^{(2)}x^k=\frac{\operatorname{Li}_2(x)}{1-x}$$
Replace $x$ with $x^2$ and multiply both sides by $-\frac16\ln^3x$ then $\int_0^1$ we have
$$\sum_{k=1}^\infty\frac{H_n^{(2)}}{(2n+1)^4}=-\frac16\int_0^1\frac{\operatorname{Li}_2(x^2)\ln^3x}{1-x^2}\ dx$$
For the other sum , use
$$\sum_{k=1}^\infty H_k^{(4)}x^k=\frac{\operatorname{Li}_4(x)}{1-x}$$
Replace $x$ with $-x$ and multiply both sides by $-\frac{\ln x}{x}$ then $\int_0^1$ we have
$$\sum_{k=1}^\infty(-1)^k \frac{H_k^{(4)}}{k^2}=-\int_0^1\frac{\operatorname{Li}_4(-x)\ln x}{x(1+x)}\ dx$$
In terms of known constants such as $\pi$, $\ln 2$, or function values of the Riemann zeta function $\zeta (x)$ or the polylogarithm $\operatorname{Li}_s (x)$, the answer seems to be no. The reason for this is it can be shown that $$\sum_{n = 1}^\infty \frac{(-1)^n H^{(4)}_n}{n^2} = 4 \sum_{n = 1}^\infty \frac{(-1)^n H_n}{n^5} + 2 \sum_{n = 1}^\infty \frac{(-1)^n H^{(2)}_n}{n^4} - \frac{9}{16} \zeta^2 (3) + \frac{359}{64} \zeta (6),$$ and it is well-known that the alternating Euler sum $\sum_{n = 1}^\infty (-1)^n H_n/n^5$ has no closed form in terms of the known constants I mentioned above.