Help on $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{H_n^3}{n}$ From this post:
The sum: $$\sum_{n=1}^{\infty}(-1)^{n+1}\frac{H_n^3}{n+1}=-\frac{9}{8}\zeta(3)\ln(2)+\frac{\pi^4}{288}-\frac{\ln^4(2)}{4}+\frac{\pi^2}{8}\ln^2(2)$$ 
by moving the $n+1$ back to $n$
What is the sum of: $$\sum_{n=1}^{\infty}(-1)^{n+1}\frac{H_n^3}{n}?$$ 
I guess it could be easy, but I cannot see it.
Can anyone please with this sum? I needs it to solve one of my current problem I am working on.

Here might be a start. Observe that$$\begin{align*}\sum\limits_{n\geq1}(-1)^{n+1}\frac {H_n^3}{n+1} & =\frac {H_1^3}2-\frac {H_2^3}3+\frac {H_3^3}4-\cdots\\ & =\sum\limits_{n\geq2}(-1)^n\frac {H_{n-1}^3}n\end{align*}$$
Now use the fact that $H_n=H_{n-1}+\tfrac 1n$. Therefore, the original sum becomes$$\begin{align*}\sum\limits_{n\geq1}(-1)^{n+1}\frac {H_n^3}{n+1} & =\sum\limits_{n\geq2}(-1)^n\frac {1}n\left[H_n-\frac 1n\right]^3\\ & =\sum\limits_{n\geq2}(-1)^n\frac 1n\left[H_n^3-\frac {3H_n^2}n+\frac {3H_n}{n^2}-\frac 1{n^3}\right]\\ & =\sum\limits_{n\geq2}(-1)^{n}\frac {H_n^3}{n}-3\sum\limits_{n\geq2}(-1)^n\frac {H_n^2}{n^2}+3\sum\limits_{n\geq2}(-1)^{n}\frac {H_n}{n^3}+\frac {7\pi^4}{720}\\ & =-\sum\limits_{n\geq2}(-1)^{n+1}\frac {H_n^3}n+3\sum\limits_{n\geq2}(-1)^{n+1}\frac {H_n^2}{n^2}+3\sum\limits_{n\geq2}(-1)^{n}\frac {H_n}{n^3}+\frac {7\pi^4}{720}\end{align*}$$
The second sum can be found with the help of this question. Reindex the identity to start the sum from two to get

$$\sum\limits_{n\geq2}(-1)^n\frac {H_n}{n^3}\color{red}{=1+2\operatorname{Li}_4\left(\frac 12\right)+\frac {7\log 2}4\zeta(3)-\frac {11\pi^4}{360}+\frac {\log^42}{12}-\frac {\pi^2\log^22}{12}}$$

In a similar fashion, some digging on MSE gives this question which might be a big help in tackling the second sum:$$\sum\limits_{n\geq2}(-1)^n\frac {H_n^2}{n^2}$$
