Evaluating the infinite sum $ \sum_{k=1}^\infty \frac{H_k}{k^4}$

Question

I was exploring MSE when I got to know about the Euler sums of the general type $\sum_{n=1}^\infty \frac{H_n}{n^k}$.

I managed to prove the cases $k=2,3$. But, I am stuck in $k=4$.

In my approach, I managed to prove that

$$ \sum_{n=1}^\infty \frac{H_n}{n^4} = \sum_{n,k=1}^\infty \frac{1}{n^2k^3} - \sum_{n,k=1}^\infty \frac{1}{n^2k^2(n+k)}$$

For which I used $ H_n = \sum_{k=1}^\infty \frac{k}{n(n+k)}$.

The first sum is trivial, but I failed to evaluate the second one.

I saw quite a few answers on MSE, and I found this one quite satisfactory, but cumbersome at the same time.

I am seeking for a simpler answer. Any help?

Answer 1

It seems that your idea is effectively good.

$$\sum_{n=1}^\infty \sum_{k=1}^\infty \frac{1}{n^2k^2(n+k)}=\sum_{n=1}^\infty \frac{1}{n^2}\sum_{k=1}^\infty \frac{1}{k^2(n+k)}$$ $$\sum_{k=1}^\infty \frac{1}{k^2(n+k)}=\sum_{k=1}^\infty\Bigg[-\frac{1}{n^2 k}+\frac{1}{n^2 (n+k)}+\frac{1}{n k^2}\Bigg] =-\frac 1{n^2} \left(\psi (n+1)-\frac{\pi ^2 }{6}n+\gamma \right)$$

Now (using a CAS for the only difficult part), $$\sum_{k=1}^\infty \frac {\psi (n+1) } {n^4}=-\frac{\pi ^2 }{6}\zeta (3)+3 \zeta (5)-\gamma\frac{ \pi ^4}{90}$$