How to evaluate the composed euler sum $\sum_{n=1}^\infty \frac{H_{kn}}{n^2}$ Is there a closed form for the following ?

$$\sum_{n=1}^\infty \frac{H_{kn}}{n^2}$$

I suspect it is easy for $k={1,2}$ ; but the complexity might increase for greater values
Can we generalize 

$$\sum_{n=1}^\infty \frac{H_{kn}}{n^q}$$

where the harmonic numbers are defined as
$$H_{kn} = \sum_{j=1}^{kn} \frac{1}{j}$$
 A: (This is more of a big comment than a solution.)
Let $p,q\in\mathbb{N}^{+}$, with $q>1$. We can derive an integral representation for the sum as follows:
$$\begin{align}
S{\left(p,q\right)}
&=\sum_{n=1}^{\infty}\frac{H_{pn}}{n^q}\\
&=\sum_{n=1}^{\infty}\frac{1}{n^q}\int_{0}^{1}\frac{1-t^{pn}}{1-t}\,\mathrm{d}t\\
&=\int_{0}^{1}\left[\sum_{n=1}^{\infty}\frac{1}{n^q}\left(\frac{1-t^{pn}}{1-t}\right)\right]\,\mathrm{d}t\\
&=\int_{0}^{1}\frac{1}{1-t}\left[\sum_{n=1}^{\infty}\frac{\left(1-t^{pn}\right)}{n^q}\right]\,\mathrm{d}t\\
&=\int_{0}^{1}\frac{1}{1-t}\left[\sum_{n=1}^{\infty}\frac{1}{n^q}-\sum_{n=1}^{\infty}\frac{\left(t^{p}\right)^n}{n^q}\right]\,\mathrm{d}t\\
&=\int_{0}^{1}\frac{\zeta{\left(q\right)}-\operatorname{Li}_{q}{\left(t^{p}\right)}}{1-t}\,\mathrm{d}t\\
&=\int_{0}^{1}\frac{\zeta{\left(q\right)}-p^{q-1}\sum_{k=0}^{p-1}\operatorname{Li}_{q}{\left(e^{\frac{2\pi i k}{p}}t\right)}}{1-t}\,\mathrm{d}t.\\
\end{align}$$
The formula is clearly useful for plugging in explicit small values for $p,$ and $q$, but perhaps it could be of some use in solving the general case as well.
