# Compute $\sum_{n=1}^\infty\frac{H_nH_{2n}}{(2n+1)^3}$

How to prove that

$$\small{\sum_{n=1}^\infty\frac{H_nH_{2n}}{(2n+1)^3}=\frac1{12}\ln^52+\frac{13}{128}\zeta(5)-\frac12\ln^32\zeta(2)+\frac74\ln^22\zeta(3)-\frac{17}{8}\ln2\zeta(4)+2\ln2\operatorname{Li}_4\left(\frac12\right)}$$ whre $$H_n$$ is the harmonic number, $$\zeta$$ is the Riemann zeta function and $$\operatorname{Li}_a(x)$$ is the polylogarithm function.

This problem is proposed by Cornel ( can be found here ) and no solution has been submitted yet. I tried all the tools I used in solving the other tough series but did not work, so I consider this series very hard to crack. Any idea ?

• I am tagging " integration" as logarithmic integrals are very related to harmonic series.

## 1 Answer

Here is a sketch of Cornel's way.

We use that $$\displaystyle \sum_{k=1}^{\infty} \frac{H_k}{(k+1)(k+2n+1)}=\frac{H_{2n}^2+H_{2n}^{(2)}}{4n}$$ and then multiply all by $$1/n^2$$ and consider the summation from $$n=1$$ to $$\infty$$. Later in the process we use another critical step, $$\displaystyle \sum_{n=1}^{\infty} \frac{1}{(2k+2n+1) 2n}=\frac{1}{(2k+1)^2}+\frac{H_{2k}}{2k+1}-\frac{H_k}{2(2k+1)}-\frac{\log(2)}{2k+1}$$.

Essentially, these are almost exactly the steps presented in section 6.59, pages $$530$$-$$532$$, in the book (Almost) Impossible Integrals, Sums, and Series.

So, after simple calculations and rearrangements we arrive at

$$\frac{1}{4} \sum _{n=1}^{\infty } \frac{\left(H_{2 n}\right){}^2}{n^3}+\frac{1}{4} \sum _{n=1}^{\infty } \frac{H_{2 n}^{(2)}}{n^3}$$

$$=\frac{1}{8}\sum _{n=1}^{\infty }\frac{H_n}{n^4}-\frac{1}{4}\sum _{n=1}^{\infty }\frac{H_n H_{2 n}}{n^3}+\frac{\pi^2}{24}\sum _{n=1}^{\infty }\frac{H_{2 n}}{n^2}-4\sum _{n=1}^{\infty }\frac{H_{2 n+1}^2}{(2 n+1)^3}+\frac{\pi^2}{6}\sum _{n=1}^{\infty }\frac{H_{2 n+1}}{(2 n+1)^2}\\+4\sum _{n=1}^{\infty }\frac{H_{2 n+1}}{(2 n+1)^4}+2\sum _{n=1}^{\infty }\frac{H_n H_{2 n}}{(2 n+1)^3}+4 \log (2)\sum _{n=1}^{\infty }\frac{ H_{2 n+1}}{(2 n+1)^3}-\frac{\pi^2}{48}\sum _{n=1}^{\infty }\frac{1}{n^3}-\frac{\pi^2}{6}\sum _{n=1}^{\infty }\frac{1}{(2 n+1)^3}-4 \log (2)\sum _{n=1}^{\infty }\frac{1}{(2 n+1)^4}.$$

Since all the series are known except the desired one, the extraction is immediately achieved.

For example, a solution to the challenging series $$\displaystyle \sum _{n=1}^{\infty }\frac{H_n H_{2 n}}{n^3}$$ is presented in https://math.stackexchange.com/q/3345138.

• (+1) for the impressive work. Indeed all these series are known after we use $2\sum_{n=1}^\infty a_{2n}=\sum_{n=1}^\infty a_{n}+\sum_{n=1}^\infty (-1)^n a_{n}$. – Ali Shather Sep 18 '19 at 12:06