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By this post we celebrate Romania’s National Day! It's a new harmonic series of weight $5$ involving harmonic numbers of the type $\overline H$ proposed by Cornel Ioan Valean.

$$\sum_{n=1}^{\infty} \frac{H_n \overline{H}_n}{n^3}$$ $$=\frac{1}{6}\log^3(2)\zeta (2)-\frac{7}{8}\log ^2(2)\zeta (3)+4\log(2)\zeta (4)-\frac{193 }{64}\zeta (5)-\frac{1}{60} \log ^5(2)$$ $$+\frac{3 }{8}\zeta (2) \zeta (3)+2\operatorname{Li}_5\left(\frac{1}{2}\right),$$ where $\overline{H}_n=1-\frac{1}{2}+\cdots+\frac{(-1)^{n-1}}{n}.$


As you'll see later, Cornel derived the series result by combining results from his book, (Almost) Impossible Integrals, Sums, and Series, and his new recent papers.

Question: Do you know if the value of the present series is known in the mathematical literature (published papers, books)? I mention that I'm not interested in solutions.

Some historical facts: According to the book, Mathematics by experiment, by the mathematicians Jonathan Borwein and David Bailey, the series version $\displaystyle \sum_{n=1}^{\infty} \frac{(\overline{H}_n)^2}{(n+1)^3}$ was calculated first in 1994 and you may find the closed-form on page 60.

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  • $\begingroup$ I know you don't want any alternative solutions but could we make an exception with mine? (I think its quite a nice one). If not, where do you suggest I post it? Thanks. $\endgroup$ Dec 22, 2020 at 3:54

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A first solution by Cornel Ioan Valean (described in large steps)

From the paper The calculation of a harmonic series with a weight 5 structure, involving the product of harmonic numbers, $H_n H^{(2)}_{2n}$ by Cornel Ioan Valean, we have that

$$I=\int_0^1 \frac{\log^2(1+x)\operatorname{Li}_2(-x) }{x} \textrm{d}x=\frac{2}{15}\log^5(2)-\frac{2}{3}\log^3(2)\zeta(2)+\frac{7}{4}\log^2(2)\zeta(3)-\frac{1}{8}\zeta(2)\zeta(3)\\-\frac{125}{32}\zeta(5)+4 \log (2)\operatorname{Li}_4\left(\frac{1}{2}\right)+4 \operatorname{Li}_5\left(\frac{1}{2}\right).$$ Integrating by parts, we get $I=-5/16\log(2)\zeta(4)+\displaystyle\frac{1}{2}\underbrace{\int_0^1\frac{(\operatorname{Li}_2(-x))^2}{1+x}\textrm{d}x}_{\displaystyle J }$.

Now, the key step (the magical one) is to observe that the Cauchy product of $\displaystyle \frac{(\operatorname{Li}_2(-x))^2}{1+x}$ can be expressed using a powerful sum in the book, (Almost) Impossible Integrals, Sums, and Series, that is the result in $(4.19)$,

$$ \sum_{k=1}^{n-1} \frac{H_k^{(2)}}{(n-k)^2}=\left(H_n^{(2)}\right)^2-5 H_n^{(4)}+4\sum _{k=1}^n \frac{H_k}{k^3},$$

and then we have $$J=\int_0^1\frac{(\operatorname{Li}_2(-x))^2}{1+x}\textrm{d}x=\sum_{n=1}^{\infty} \int_0^1(-1)^n x^n \left(\left(H_n^{(2)}\right)^2-5 H_n^{(4)}+4\sum _{k=1}^n \frac{H_k}{k^3}\right)\textrm{d}x$$ $$= \sum_{n=1}^{\infty} (-1)^n \frac{\left(H_n^{(2)}\right)^2}{n+1}-5\sum_{n=1}^{\infty} (-1)^n \frac{H_n^{(4)}}{n+1} +4\sum_{n=1}^{\infty} (-1)^n \frac{1}{n+1}\sum _{k=1}^n \frac{H_k}{k^3}.$$

Upon reindexing all series and changing the summation order in the last series (which gives the main series to calculate), everything reduces to using the values of the series $\displaystyle \sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_n^{(2)}}{n^3}$, $\displaystyle \sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_n^{(4)}}{n}$ and $\displaystyle \sum_{n=1}^{\infty}(-1)^{n-1}\frac{(H_n^{(2)})^2}{n}$, which are all given in the book, (Almost) Impossible Integrals, Sums, and Series, on pages $311$ and $529$.

Note that a simple generalization of the series $\displaystyle \sum_{n=1}^{\infty} (-1)^n \frac{H_n^{(4)}}{n}$ may be found in the paper A simple strategy of calculating two alternating harmonic series generalizations by Cornel Ioan Valean.

Full details will be given soon in a new paper.


A bonus of the previous idea: The younger brother of the previous series $\displaystyle \sum_{n=1}^{\infty} \frac{H_n \overline{H}_n}{n^2}$ can be calculated in a similar style. For example, if we use that

$$\begin{equation*} \sum_{k=1}^{n-1} \frac{H_k^{(2)}}{n-k}=\sum_{k=1}^{n} \frac{H_{k}}{k^{2}}+H_n H_n^{(2)}-2H_{n}^{(3)}, \end{equation*}$$

which you may find in the same book on page $287$, then multiply both sides by $(-1)^n/(n+1)$, sum from $n=1$ to $\infty$, next reverse the Cauchy product in the left-hand side, turn everything into a simple integral and calculate it, together with the fact that for the right-hand side you follow the same arrangements strategy as in the previous series and then make use of the generating functions presented in Sect. 4.10, page $284$ of the mentioned book, to extract the needed alternating series, you're done. Exactly, that simple!

Full details will be given soon 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.

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

We start from the beginning with splitting the series based on parity, and then, using Botez-Catalan identity, we have $$\sum_{n=1}^{\infty} \frac{H_n \overline{H}_n}{n^3}=1+\frac{1}{8}\sum_{n=1}^{\infty} \frac{H_{2n} \overline{H}_{2n}}{n^3}+\sum_{n=1}^{\infty} \frac{H_{2n+1} \overline{H}_{2n+1}}{(2n+1)^3}$$ $$=1+\sum_{n=1}^{\infty}\frac{H_{2 n}^2}{(2n)^3}+\sum_{n=1}^{\infty}\frac{H_{2 n+1}^2}{(2 n+1)^3}-\sum_{n=1}^{\infty}\frac{H_n}{(2 n+1)^4}-\sum_{n=1}^{\infty}\frac{H_n H_{2 n}}{(2 n)^3}-\sum_{n=1}^{\infty}\frac{H_n H_{2 n}}{(2 n+1)^3}$$ $$=\frac{1}{6}\log^3(2)\zeta (2)-\frac{7}{8}\log ^2(2)\zeta (3)+4\log(2)\zeta (4)-\frac{193 }{64}\zeta (5)-\frac{1}{60} \log ^5(2)$$ $$+\frac{3 }{8}\zeta (2) \zeta (3)+2\operatorname{Li}_5\left(\frac{1}{2}\right),$$

and it's easy to see the first two series are reducible to series already calculated in the book, , (Almost) Impossible Integrals, Sums, and Series, then the third series is calculated in a generalized form in the paper A new powerful strategy of calculating a class of alternating Euler sums by Cornel Ioan Valean and the last 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.

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