TITLE: Closed form of the sum $\sum_{n=1}^{\infty}\frac{H_n}{n^x}$
QUESTION [5 upvotes]: Some days ago I derived the identity
$$\sum_{n=1}^{\infty}\frac{H_n}{n^2}=2\zeta(3)$$
where $H_n$ is the $n$th Harmonic number. Other related identities include
$$\sum_{n=1}^{\infty}\frac{H_n}{n^3}=\frac{\pi^4}{72}$$
$$\sum_{n=1}^{\infty}\frac{H_n}{n^4}=\frac{-1}{6}\pi^2\zeta(3)+3\zeta(5)$$
$$\sum_{n=1}^{\infty}\frac{H_n}{n^5}=\frac{1}{540}(\pi^6-270\zeta(3)^2)$$
(I proved all except the last one) Now I wondered if there is a closed form for the generalized form
$$\sum_{n=1}^{\infty}\frac{H_n}{n^x}$$
For some real number $x$. Here is my try: The well known identity $\psi(n+1)=H_n-\gamma$ gives
$$\sum_{n=1}^{\infty}\frac{H_n}{n^x}=\sum_{n=1}^{\infty}\frac{\psi(n+1)}{n^x}-\gamma\zeta(x)$$
But I don't know what to do further. Any help would be appreciated.

REPLY [3 votes]: Euler proved the following result:
Theorem For integer $q\geq 2$, we have $$\sum_{n=1}^\infty \frac{H_n}{n^q}=\left( 1+\frac{q}{2}\right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1) \zeta(q-k)$$
A proof using the residue theorem can be found in the paper: "Euler sums and contour integral representations" by Philippe Flajolet and Bruno Salvy.