Double Euler sum $ \sum_{k\geq 1} \frac{H_k^{(2)} H_k}{k^3} $ I proved the following result 
$$\displaystyle \sum_{k\geq 1}  \frac{H_k^{(2)} H_k}{k^3} =- \frac{97}{12} \zeta(6)+\frac{7}{4}\zeta(4)\zeta(2) + \frac{5}{2}\zeta(3)^2+\frac{2}{3}\zeta(2)^3$$
After consideration of powers of polylogarithms.
You can refer to the following thread .
My question is : are there any papers in the literature which dealt with that result?
Are my evaluations worth publishing ?
 A: The following new solution is proposed by Cornel Ioan Valean. Based on a few ideas presented in the book, (Almost) Impossible Integrals, Sums, and Series, like the Cauchy product of $(\operatorname{Li}_2(x))^2$, that is $\displaystyle  (\operatorname{Li}_2(x))^2=4\sum_{n=1}^{\infty}x^n\frac{H_n}{n^3}+2\sum_{n=1}^{\infty}x^n\frac{H_n^{(2)}}{n^2}-6\sum_{n=1}^{\infty}\frac{x^n}{n^4}$,  where if we multiply both sides by $\displaystyle \frac{\log(1-x)}{x}$ and then integrate from $x=0$ to $x=1$, using that $\displaystyle \int_{0}^{1}x^{n-1}\log(1-x)\textrm{d}x=-\frac{H_{n}}{n}$, we get 
\begin{equation*}
\int_0^1  \frac{\log(1-x)}{x}(\operatorname{Li}_2(x))^2 \textrm{d}x=-\frac{1}{3}(\operatorname{Li}_2(x))^3\biggr|_{x=0}^{x=1}=-\frac{35}{24}\zeta(6)
\end{equation*}
\begin{equation*}
=6\sum_{n=1}^{\infty} \frac{H_n}{n^5}-4\sum_{n=1}^{\infty} \frac{H_n^2}{n^4}-2\sum_{n=1}^{\infty} \frac{H_nH_n^{(2)}}{n^3}
\end{equation*}
\begin{equation*}
=5\zeta^2(3)-\frac{17}{3}\zeta(6)-2\sum_{n=1}^{\infty} \frac{H_nH_n^{(2)}}{n^3},
\end{equation*}
where the first sum comes from the classical generalization, $
\displaystyle 2\sum_{k=1}^\infty \frac{H_k}{k^n}=(n+2)\zeta(n+1)-\sum_{k=1}^{n-2} \zeta(n-k) \zeta(k+1), \ n\in \mathbb{N},\  n\ge2$, and the second sum, $\displaystyle \sum_{n=1}^{\infty} \frac{H_n^2}{n^4}=\frac{97}{24}\zeta(6)-2\zeta^2(3)$, is calculated in the mentioned book or in this article.
To conclude, we have 
\begin{equation*}
\sum_{n=1}^{\infty}\frac{H_n H_n^{(2)}}{n^3}=\frac{1}{2}\left(5\zeta^2(3)-\frac{101}{24}\zeta(6)\right).
\end{equation*}
Note the present solution circumvents the necessity of using the value of the series $\displaystyle \sum_{n=1}^{\infty} \left(\frac{H_n}{n}\right)^3$.
