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I was able to get the following result

$$3\sum\limits_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^6}+\sum\limits_{n=1}^\infty\frac{H_nH_n^{(3)}}{n^5}=11\zeta(3)\zeta(6)+\frac52\zeta(4)\zeta(5)-\frac{13}{6}\zeta^3(3)-2\zeta(2)\zeta(7)-5\zeta(9)$$ where $H_n^{(p)}=1+\frac1{2^p}+\cdots+\frac1{n^p}$ is the $n$th generalized harmonic number of order $p$.

based on a nice identity and some manageable Euler sums. Is this result known in the literature? Can we evaluate the terms separately?

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  • $\begingroup$ I haven't seen this result... $\endgroup$ – David G. Stork Aug 2 at 21:32
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In answer to your question, can the sums be evaluated separately? Yes they can. The results for each of these two Euler sums can be found in the 2016 paper Euler sums and integrals of polylogarithm functions by Ce Xu et al.

The results are: $$\sum_{n = 1}^\infty \frac{H_n H^{(2)}_n}{n^6} = \frac{17}{6} \zeta (3) \zeta (6) + \frac{173}{72} \zeta (9) + \frac{1}{4} \zeta (4) \zeta (5) - 3 \zeta (2) \zeta (7) - \frac{2}{3} \zeta^3 (3) \quad \text{(See Eq. 3.18)}$$ and $$\sum_{n = 1}^\infty \frac{H_n H^{(3)}_n}{n^5} = \frac{679}{24} \zeta (9) - 11 \zeta (2) \zeta (7) - \frac{1}{2} \zeta (3) \zeta (6) - \frac{29}{4} \zeta (4) \zeta (5) - \frac{1}{6} \zeta^3 (3).$$

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By the Cauchy product we have,

$$\operatorname{Li}_3^2(x)=\sum_{n=1}^\infty\left(\frac{12H_n}{n^5}+\frac{H_n^{(2)}}{n^4}+\frac{2H_n^{(3)}}{n^3}-\frac{20}{n^6}\right)x^n\tag{1}$$

Divide both sides of $(1)$ by $x$ then integrate from $x=0$ to $1$ to get

\begin{align} S&=\sum_{n=1}^\infty\left(\frac{12H_n}{n^6}+\frac{6H_n^{(2)}}{n^5}+\frac{2H_n^{(3)}}{n^4}-\frac{20}{n^7}\right)=\int_0^1\frac{\operatorname{Li}_3^2(x)}{x}\ dx\\ &=\sum_{n=1}^\infty\frac{1}{n^3}\int_0^1x^{n-1}\operatorname{Li}_3(x)\ dx\quad \text{apply integration by parts}\\ &=\sum_{n=1}^\infty\frac{1}{n^3}\left(\frac{\zeta(3)}{n}-\frac{\zeta(2)}{n^2}+\frac{H_n}{n^3}\right)\\ &\boxed{S=\zeta(3)\zeta(4)-\zeta(2)\zeta(5)+\sum_{n=1}^\infty\frac{H_n}{n^6}} \end{align}

Now multiply both sides of $(1)$ by $\large\frac{\operatorname{Li}_2(x)}{x}$ then integrate from $x=0$ to $1$ to get

\begin{align} I&=\int_0^1\frac{\operatorname{Li}_3^2(x)\operatorname{Li}_2(x)}{x}\ dx=\frac13\operatorname{Li}_3^3(1)=\frac13\zeta^3(3)\\ &=\sum_{n=1}^\infty\left(\frac{12H_n}{n^5}+\frac{6H_n^{(2)}}{n^4}+\frac{2H_n^{(3)}}{n^3}-\frac{20}{n^6}\right)\int_0^1 x^{n-1}\operatorname{Li}_2(x)\ dx\quad \text{apply integration by parts}\\ &=\sum_{n=1}^\infty\left(\frac{12H_n}{n^5}+\frac{6H_n^{(2)}}{n^4}+\frac{2H_n^{(3)}}{n^3}-\frac{20}{n^6}\right)\left(\frac{\zeta(2)}{n}-\frac{H_n}{n^2}\right)\\ &=\zeta(2)S-12\sum_{n=1}^\infty\frac{H_n^2}{n^7}-6\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^6}-2\sum_{n=1}^\infty\frac{H_nH_n^{(3)}}{n^5}+20\sum_{n=1}^\infty\frac{H_n}{n^8} \end{align} Rearranging the terms and plugging the boxed result of $S$ , we get

$$3\sum\limits_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^6}+\sum\limits_{n=1}^\infty\frac{H_nH_n^{(3)}}{n^5}\\=\frac78\zeta(3)\zeta(6)-\frac54\zeta(4)\zeta(5)-\frac16\zeta^3(3)+\frac12\zeta(2)\sum_{n=1}^\infty\frac{H_n}{n^6}-6\sum_{n=1}^\infty\frac{H_n^2}{n^7}+10\sum_{n=1}^\infty\frac{H_n}{n^8}\tag{2}$$

We have

$$\sum_{n=1}^\infty\frac{H_n}{n^6}=4\zeta(7)-\zeta(2)\zeta(5)-\zeta(3)\zeta(4)\tag{3}$$

$$\sum_{k=1}^\infty\frac{H_k}{k^8}=5\zeta(9)-\zeta(2)\zeta(7)-\zeta(3)\zeta(6)-\zeta(4)\zeta(5)\tag{4}$$

$$\sum_{n=1}^\infty\frac{H_n^2}{n^7}=-\zeta(2)\zeta(7)-\frac72\zeta(3)\zeta(6)+\frac13\zeta^3(3)-\frac{5}{2}\zeta(4)\zeta(5)+\frac{55}{6}\zeta(9)\tag{5}$$

The results of $(3)$ and $(4)$ can be obtained from Euler's identity and the result of $(5)$ can be found here.

By substituting the results of $(3)$, $(4)$ and $(5)$ in $(2)$, we get our closed form.

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Special thanks to Cornel for showing us how to expand these polylogarithms using Cauchy product. More such polylogarithmic identities can be found in his book , Almost Impossible Integrals, Sums, and Series.

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