A special harmonic series discovered by Jonathan Borwein and David Bailey

A second solution by Cornel Ioan Valean (in large steps)

We want to split the series based on parity and the use of Botez-Catalan identity, and then we write

$$\sum_{n=1}^{\infty} \frac{(\overline{H}_n)^2}{n^3}=1+\sum_{n=1}^{\infty} \frac{(\overline{H}_{2n})^2}{(2n)^3}+\sum_{n=1}^{\infty} \frac{(\overline{H}_{2n+1})^2}{(2n+1)^3}$$ $$=1+\frac{1}{8}\sum _{n=1}^{\infty } \frac{H_n^2}{n^3}+\sum _{n=1}^{\infty } \frac{\left(H_{2 n+1}\right){}^2}{(2 n+1)^3}+\sum _{n=1}^{\infty } \frac{H_{2 n}^2}{ (2n)^3}+\sum _{n=1}^{\infty } \frac{H_n^2}{(2 n+1)^3}-2\sum _{n=1}^{\infty } \frac{H_n}{(2 n+1)^4}$$ $$-2\sum _{n=1}^{\infty } \frac{H_n H_{2 n}}{(2n)^3}-2\sum _{n=1}^{\infty } \frac{H_n H_{2 n}}{(2 n+1)^3}$$ $$=\frac{19}{8}\log (2) \zeta (4) +\frac{7}{4} \log ^2(2)\zeta (3)+\frac{1}{3} \log ^3(2)\zeta (2) +\frac{3 }{4}\zeta (2) \zeta (3)-\frac{167 }{32}\zeta (5)$$ $$-\frac{1}{30} \log ^5(2)+4 \operatorname{Li}_5\left(\frac{1}{2}\right),$$ where the first three series may be easily extracted from the book, (Almost) Impossible Integrals, Sums, and Series, for the fourth series see the previous solution, then with respect to the fifth series you may see the generalized series in the paper A new powerful strategy of calculating a class of alternating Euler sums by Cornel Ioan Valean, and the last two series are calculated in the paper On the calculation of two essential harmonic series with a weight 5 structure, involving harmonic numbers of the type $H_{2n}$ by Cornel Ioan Valean.

End of story.

A solution by Cornel Ioan Valean (in large steps)

Let's recall first Botez-Catalan identity, $\displaystyle 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots-\frac{1}{2n}=H_{2n}-H_n$, and then use it to calculate the difference $H_m-\overline{H}_m$, where we consider the cases $m=2n$ and $m=2n+1$. So. for $m=2n$, we have $H_{2n}-\overline{H}_{2n}=H_n$ and for $m=2n+1$, we get $H_{2n+1}-\overline{H}_{2n+1}=H_n.$

Also, we have that $\displaystyle \frac{(H_n-\overline{H}_n)^2}{n^3}=\frac{H_n^2}{n^3}-2\frac{H_n\overline{H}_n}{n^3}+\frac{(\overline{H}_n)^2}{n^3}$. Therefore,

$$\sum_{n=1}^{\infty}\frac{(\overline{H}_n)^2}{n^3}=\sum_{n=1}^{\infty} \frac{(H_n-\overline{H}_n)^2}{n^3}-\sum_{n=1}^{\infty}\frac{H_n^2}{n^3}+2\sum_{n=1}^{\infty}\frac{H_n\overline{H}_n}{n^3}$$ $$\text{\{we split the first series in the right-hand side according to $n$ even and odd\}}$$ $$=\sum_{n=1}^{\infty} \frac{H_n^2-H_n^{(2)}}{(2n+1)^3}+\sum_{n=1}^{\infty} \frac{H_n^{(2)}}{(2n+1)^3}-\frac{7}{8}\sum_{n=1}^{\infty}\frac{H_n^2}{n^3}+2\sum_{n=1}^{\infty}\frac{H_n\overline{H}_n}{n^3}$$ $$=\frac{19}{8}\log (2) \zeta (4) +\frac{7}{4} \log ^2(2)\zeta (3)+\frac{1}{3} \log ^3(2)\zeta (2) +\frac{3 }{4}\zeta (2) \zeta (3)-\frac{167 }{32}\zeta (5)$$ $$-\frac{1}{30} \log ^5(2)+4 \operatorname{Li}_5\left(\frac{1}{2}\right),$$ and at this point all the series are known. The first series from penultimate line can be calculated using Cornel's Master Theorem of Series (Ali Shather nicely explains Cornel's method in this post https://math.stackexchange.com/q/3377671), the second series is immediately obtained by using the Abels's summation (a similar approach to this one https://math.stackexchange.com/q/3259984, and the precise value is also given by Ali Shather in this post Compute $\sum_{n=1}^\infty\frac{H_{n}^2}{(2n+1)^3}$), the third series appears in (Almost) Impossible Integrals, Sums, and Series, and finally, the last series is calculated in this post https://math.stackexchange.com/q/3458445 where you may find Cornel's solution.

End of story.

All details will appear in a new paper.

Update: the evaluation of the series appears in the preprint, Two advanced harmonic series of weight 5 involving skew-harmonic numbers.