Seeking general formula for Euler Sum $\sum\limits _{ n=1 }^{ \infty }{ \frac {{\left({H}_{n}\right)}^{p}}{{n}^{q}}}$ with $q$ even and $p$ odd

I am wondering if there is a formula for

$\displaystyle \sum _{ n=1 }^{ \infty }{ \frac { { \left( { H }_{ n } \right) }^{ p } }{ { n }^{ q } } }$ with 'q' being an EVEN positive integer and 'p' being an ODD positive integer.

I found the following formula

$$\sum_{n=1}^\infty \frac{H_n}{n^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k)$$

and it is cited that Euler proved it. I found a proof, here on MSE as well.

This is what piqued my curiosity. I imagine it gets a lot more complicated when the exponent 'p' is no longer just a $1$, but I have much to learn before I could possibly tackle such a proof.

Upon further research, I have discovered that in 1994-1995 some mathematicians proved that when $p=2$ and $q=2$ the series results in the closed form $\frac{17}{4}\zeta(4)$

• This might interest you. – Pablo Rotondo Jun 16 '16 at 9:05
• That is an excellent tip on a fantastic paper !! Thanks – Bob Kadylo Jun 16 '16 at 9:32