Evaluate $ \sum_{n=1}^{\infty}\frac{(-1)^{n}H_{n}^{(3)}}{2n+1}$ and $\sum_{n=1}^{\infty}\frac{(-1)^{n}H_{n}}{(2n+1)^{3}}$ I ran into an Euler sum that appears to be rather tough.  Apparently, it does indeed have a closed form, so I assume it is doable.  May be a false assumption, though. :)
$\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n}H_{n}^{(3)}}{2n+1}=\frac{\pi^{2}}{6}G+4\beta(4)-\frac{3\pi}{2}\zeta(3)$
$G\displaystyle$ is the Catalan and $\displaystyle\beta(4)=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)^{4}}$,   the Dirichlet Beta. 
and $\displaystyle H_{n}^{(3)}=\sum_{k=1}^{n}\frac{1}{k^{3}}$
I thought about giving this a go using the digamma kernel as mentioned in Flajolet's famous paper on evaluating Euler sums via contours.  Near the end of his paper, he mentions sums of this type but does not go into detail on them.
He mentions $\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n}H_{n}}{(2n+1)^{3}}=3\beta(4)-\frac{7\pi}{16}\zeta(3)-\frac{\pi^{3}}{16}log(2)$,
claiming it can be done by using the kernel $\displaystyle\pi\csc(\pi z)(\psi(-z)+\gamma)$.
So, do you think this one can be done with this method or any other?. I would like to think so.
 A: Evaluation of $\displaystyle\sum_{n=1}^\infty\frac{(-1)^n H_n^{(3)}}{2n+1}$:
Recall the identity
$$\sum_{n=1}^\infty H_n^{(3)}x^n=\frac{\text{Li}_3(x)}{1-x}$$
where if we replace $x$ by $-x^2$ then $\int_0^1$ we have
$$\sum_{n=1}^\infty\frac{(-1)^n H_n^{(3)}}{2n+1}=\int_0^1\frac{\text{Li}_3(-x^2)}{1+x^2}dx=I$$
$$=\int_0^\infty\frac{\text{Li}_3(-x^2)}{1+x^2}dx-\underbrace{\int_1^\infty\frac{\text{Li}_3(-x^2)}{1+x^2}dx}_{x\to 1/x}$$
$$=\int_0^\infty\frac{\text{Li}_3(-x^2)}{1+x^2}dx-\int_0^1\frac{\text{Li}_3(-1/x^2)}{1+x^2}dx$$
By adding $I$ to both sides we have
$$2I=\int_0^\infty\frac{\text{Li}_3(-x^2)}{1+x^2}dx+\int_0^1\frac{\text{Li}_3(-x^2)-\text{Li}_3(-1/x^2)}{1+x^2}dx$$
Use $\text{Li}_3(-x)-\text{Li}_3\left(-\frac1x\right)=-\zeta(2)\ln(x)-\frac16\ln^3(x)$ and replace $x$ by $x^2$ we get
$$2I=\int_0^\infty\frac{\text{Li}_3(-x^2)}{1+x^2}dx+\int_0^1\frac{-2\zeta(2)\ln x-\frac43\ln^3x}{1+x^2}dx$$
$$=\int_0^\infty\frac{\text{Li}_3(-x^2)}{1+x^2}dx-2\zeta(2)\underbrace{\int_0^1\frac{\ln x}{1+x^2}dx}_{-G}-\frac43\underbrace{\int_0^1\frac{\ln^3x}{1+x^2}dx}_{-6\beta(4)}$$
Divide by $2$
$$I=\frac12\int_0^\infty\frac{\text{Li}_3(-x^2)}{1+x^2}dx+G\zeta(2)+4\beta(4)$$
For the remaining integral, write $\text{Li}_3(x)=\frac12\int_0^1\frac{x\ln^2y}{1-xy}dy$ and replace $x$ by $-x^2$
$$\Longrightarrow \int_0^\infty\frac{\text{Li}_3(-x^2)}{1+x^2}dx=\int_0^1\ln^2y\left(-\frac12\int_0^\infty\frac{x^2}{(1+x^2)(1+x^2y)}dx\right)dy$$
$$=\int_0^1\ln^2y\left(-\frac{\pi}{4}\frac{1}{\sqrt{y}+y}\right)dy\overset{\sqrt{y}=x}{=}-2\pi\int_0^1\frac{\ln^2x}{1+x}dx=-3\pi\zeta(3)$$
Thus
$$\sum_{n=1}^{\infty}\frac{(-1)^{n}H_{n}^{(3)}}{2n+1}=4\beta(4)+G\zeta(2)-\frac{3\pi}{2}\zeta(3)$$

The sum $\displaystyle\sum_{n=1}^\infty\frac{(-1)^nH_n}{(2n+1)^3}$ is already computed here.
