Find $$\sum_{k=1}^n\frac{\left(H_k^{(p)}\right)^2}{k^p}\,,$$ where $H_k^{(p)}=1+\frac1{2^p}+\cdots+\frac1{k^p}$ is the $k$th generalized harmonic number of order $p$.
Cornel proved in his book, (almost) impossible integral, sums and series, the following identity :
$$\sum_{k=1}^n\frac{\left(H_k^{(p)}\right)^2}{k^p}=\frac13\left(\left(H_n^{(p)}\right)^3-H_n^{(3p)}\right)+\sum_{k=1}^n\frac{H_k^{(p)}}{k^{2p}}$$
using series manipulations and he also suggested that this identity can be proved using Abel's summation and I was successful in proving it that way. other approaches are appreciated.
I am posting this problem as its' importance appears when $n$ approaches $\infty$.