How i can to calculate the harmonic partial series: $\sum_{k=1}^n\frac{\left(H_k\right)^2-H^{(2)}_k}{k}$? I have trying to calculate $$S_n=\sum_{k=1}^n\frac{\left(H_k\right)^2-H^{(2)}_k}{k} $$
I'm trying to apply Abel's Summation, i.e.: $$ \sum_{k=m}^n a_kb_k=A_nb_n-A_{m-1}b_m-\sum_{k=m}^{n-1}A_k\left(b_{k+1}-b_k\right)\tag{1}$$
where $\displaystyle A_k=\sum_{i=1}^k a_i$. Taking $m=1$, $\displaystyle a_k=\frac1k\to A_k=H_k$ and $\displaystyle b_k=\left(H_k\right)^2-H^{(2)}_k$ in $(1)$, i got that:
\begin{align*}
S_n&=H_n\left\{ \left(H_k\right)^2-H^{(2)}_k\right\}-\sum_{k=1}^{n-1}H_k\left\{ \left(H_{k+1}\right)^2-H^{(2)}_{k+1}-\left(H_{k}\right)^2+H^{(2)}_{k}\right\} \\
&=H_n\left\{ \left(H_k\right)^2-H^{(2)}_k\right\}-\sum_{k=1}^{n-1}H_k\left\{ \left(H_{k+1}+H_k\right)\left(H_{k+1}-H_k\right)-H^{(2)}_{k+1}+H^{(2)}_{k}\right\}\tag{2}\\
&=H_n\left\{ \left(H_k\right)^2-H^{(2)}_k\right\}-\sum_{k=1}^{n-1}H_k\left\{ \left(2H_{k}+\frac1{k+1}\right)\left(\frac1{k+1}\right)-\frac1{(k+1)^2}\right\}\\
&=H_n\left\{ \left(H_k\right)^2-H^{(2)}_k\right\}-2\sum_{k=1}^{n-1}\frac{\left(H_k\right)^2}{k+1}\color{red}{-\sum_{k=1}^{n-1}\frac{H_k}{(k+1)^2}}\color{red}{+\sum_{k=1}^{n-1}\frac{H_k}{(k+1)^2}}\\
&=H_n\left\{ \left(H_k\right)^2-H^{(2)}_k\right\}-2\sum_{k=1}^{n }\frac{\left(H_{k-1}\right)^2}{k}
\end{align*}
Notice that, in $(2)$ i used the facts $\displaystyle H_{k+1}=H_k+\frac{1}{k+1}$ and $\displaystyle H_{k+1}^{(2)}=H_k^{(2)}+\frac{1}{(k+1)^2}$. My difficulty here is to evaluate the last sum above. Thank you in advance for any tip or solution and i accept any approach for the original problem.

First, I wondered if the sign between the two harmonic numbers in the denominator is correct. Supposing this is the case, then we have

$$\sum_{k=1}^n\frac{H_k^2-H_k^{(2)}}{k}=\frac{1}{3}(H_n^3-3 H_nH_n^{(2)}-4 H_n^{(3)})+2\sum_{k=1}^n\frac{H_k}{k^2},\tag1$$

where for the last sum there is no known simpler form (e.g., to express it in terms of harmonic numbers).
During the calculations, I used that
$$\sum_{k=1}^n\frac{H_k^2+H_k^{(2)}}{k}=\frac{1}{3}(H_n^3+3 H_nH_n^{(2)}+2 H_n^{(3)}),$$
which appears in (Almost) Impossible Integrals, Sums, and Series (for a proof, see pages $61$-$62$).
Also, by Abel's summation we immediately we have that
$$\sum_{k=1}^n\frac{H_k}{k^2}+\sum_{k=1}^n\frac{H_k^{(2)}}{k}=H_n H_n^{(2)}+H_n^{(3)},$$
which is useful to get the result in $(1)$.
