Prove $\sum_{k=1}^n\frac{\left(H_k^{(p)}\right)^2}{k^p}=\frac13((H_n^{(p)})^3-H_n^{(3p)})+\sum_{k=1}^n\frac{H_k^{(p)}}{k^{2p}}$ 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 $\infty$.
 A: using Abel's summation $\ \displaystyle\sum_{k=1}^n a_k b_k=A_nb_{n+1}+\sum_{k=1}^{n}A_k\left(b_k-b_{k+1}\right)$ where $\displaystyle A_n=\sum_{i=1}^n a_i$
letting $\ \displaystyle a_k=\frac{1}{k^p}$ and $\ \displaystyle b_k=\left(H_k^{(p)}\right)^2$, we get
\begin{align}
S&=\sum_{k=1}^n\frac{\left(H_k^{(p)}\right)^2}{k^p}=\sum_{i=1}^n\frac{\left(H_{n+1}^{(p)}\right)^2}{i^p}+\sum_{k=1}^n\left(\sum_{i=1}^k\frac1{i^p}\right)\left(\left(H_k^{(p)}\right)^2-\left(H_{k+1}^{(p)}\right)^2\right)\\
&=\left(H_{n+1}^{(p)}\right)^2H_n^{(p)}+\sum_{k=1}^n\left(H_k^{(p)}\right)\left(\left(H_k^{(p)}\right)^2-\left(H_{k+1}^{(p)}\right)^2\right)\\
&=\left(H_{n+1}^{(p)}\right)^2H_n^{(p)}+\sum_{k=1}^{n+1}\left(H_{k-1}^{(p)}\right)\left(\left(H_{k-1}^{(p)}\right)^2-\left(H_{k}^{(p)}\right)^2\right)\\
&=\left(H_{n+1}^{(p)}\right)^2H_n^{(p)}-\sum_{k=1}^{n+1}\left(H_{k-1}^{(p)}\right)\left(\frac{2H_k^{(p)}}{k^p}-\frac1{k^{2p}}\right)\\
&=\left(H_{n+1}^{(p)}\right)^2H_n^{(p)}-\sum_{k=1}^{n}\left(H_{k}^{(p)}-\frac1{k^p}\right)\left(\frac{2H_k^{(p)}}{k^p}-\frac1{k^{2p}}\right)-\left(H_{n}^{(p)}\right)\left(\frac{2H_{n+1}^{(p)}}{(n+1)^p}-\frac1{{(n+1)}^{2p}}\right)\\
&=\underbrace{\left(H_{n+1}^{(p)}\right)^2H_n^{(p)}-\left(H_{n}^{(p)}\right)\left(\frac{2H_{n+1}^{(p)}}{(n+1)^p}-\frac1{{(n+1)}^{2p}}\right)}_{\Large\left(H_n^{(p)}\right)^3}-2S+3\sum_{k=1}^n\frac{H_k^{(p)}}{k^{(2p)}}-H_n^{(3p)}\\
&=-2S+3\sum_{k=1}^n\frac{H_k^{(p)}}{k^{(2p)}}+\left(H_n^{(p)}\right)^3-H_n^{(3p)}
\end{align}
which follows  $$S=\frac13\left(\left(H_n^{(p)}\right)^3-H_n^{(3p)}\right)+\sum_{k=1}^n\frac{H_k^{(p)}}{k^{2p}}$$
