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
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 $(1)$.