# An advanced harmonic series of weight $5$ with harmonic numbers $\overline{H}_n$

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.

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

$$I=\int_0^1 \frac{\log^2(1+x)\operatorname{Li}_2(-x) }{x} \textrm{d}x=\frac{2}{15}\log^5(2)-\frac{2}{3}\log^3(2)\zeta(2)+\frac{7}{4}\log^2(2)\zeta(3)-\frac{1}{8}\zeta(2)\zeta(3)\\-\frac{125}{32}\zeta(5)+4 \log (2)\operatorname{Li}_4\left(\frac{1}{2}\right)+4 \operatorname{Li}_5\left(\frac{1}{2}\right).$$ 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.

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.