I want to evaluate the following series
$$ \sum^\infty_{n=1} (-1)^{n+1} \frac{H_n}{n^3}. $$
From Wolfram MathWorld equation (19) we have the following related identity: $$ \sum^\infty_{n=1} \frac{H_n}{n^3} = \frac{5}{4}\zeta(4) = \frac{\pi^4}{72}. $$
My attempt:
$$ \begin{align} \sum^\infty_{n=1} (-1)^{n+1} \frac{H_n}{n^3} &= \sum^\infty_{n=1}\frac{(-1)^{n+1}}{n^3}\sum^\infty_{k=1}\left(\frac{1}{k}-\frac{1}{n+k}\right) \\ &= \sum^\infty_{n=1}\frac{(-1)^{n+1}}{n^2}\sum^\infty_{k=1}\frac{1}{k(n+k)} \\ &= \frac{1}{2} \left[ \sum^\infty_{n=1}\sum^\infty_{k=1} \frac{(-1)^{n+1}}{n^2 k(n+k)} + \sum^\infty_{n=1}\sum^\infty_{k=1} \frac{(-1)^{k+1}}{k^2 n(n+k)} \right], \\ \end{align} $$ exchanging the role of $n$ and $k$ in the second sum. Then $$ \begin{align} \sum^\infty_{n=1} (-1)^{n+1} \frac{H_n}{n^3} &= \frac{1}{2} \sum^\infty_{n=1}\sum^\infty_{k=1} \frac{(-1)^{n+1}k + (-1)^{k+1}n}{n^2 k^2(n+k)} \\ &= \frac{1}{2} \sum^\infty_{n=1}\sum^\infty_{k=1} \frac{(-1)^{n+1}(n+k) + ((-1)^{k+1} - (-1)^{n+1}) n}{n^2 k^2(n+k)} \\ &= \frac{1}{2}\left(\sum^\infty_{n=1}\frac{(-1)^{n+1}}{n^2}\right)\left(\sum^\infty_{k=1}\frac{1}{k^2}\right) + \frac{1}{2}\sum^\infty_{n=1} (-1)^{n+1} \frac{H_n}{n^3} - \frac{1}{2}\sum^\infty_{k=1} \sum^\infty_{n=1}\frac{(-1)^{n+1}}{k^2n(n+k)}, \end{align} $$ and $$ \begin{align} \sum^\infty_{n=1} (-1)^{n+1} \frac{H_n}{n^3} &= \frac{\pi^4}{72} - \frac{1}{2}\sum^\infty_{k=1} \sum^\infty_{n=1}\frac{(-1)^{n+1}}{k^2n(n+k)}. \end{align} This is where I got stuck, since the second series doesn't seem any easier to evaluate.
