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
$$H_{kn} = \sum_{j=1}^{kn} \frac{1}{j}$$
(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.