Show that $\sum\limits_{n=1}^{\infty}\frac{(H_{n})^2}{n(n+1)}=3\zeta(3)$, where, for every positive $n$, $H_n=\sum\limits_{k=1}^n\frac1k$

Question

The problem I was considering about is the evaluation of the following series:

\begin{align*} \sum_{n=1}^{\infty}\frac{(H_{n})^2}{n(n+1)} \end{align*}

The attempt I could make was to change $(H_{n})^2$ to $H_{n}(H_{n+1}-\frac{1}{n+1})$ to see if the denominator and the numerator could match each other. Then the given series becomes

\begin{align*} \sum_{n=1}^{\infty}\left(\frac{H_{n}H_{n+1}}{n(n+1)}-\frac{H_n}{n(n+1)^2}\right) \end{align*}

The second term in the above series calculates to $\zeta{(3)}-\zeta{(2)}$. I have no idea how to evaluate the first term. Wolfram alpha gives 3$\zeta(3)$ for the value of the desired series. So the first term must be somehow evaluated to $2\zeta(3)+\zeta(2)$.

Other ways to calculate the desired series are also appreciated, particularly if integrals were used.

Answer 1

In terms of Stirling numbers of the first kind we have $$ \log^2(1-x)=\sum_{n\geq 1}\frac{2H_{n-1}}{n}\,x^n,\qquad -\log^3(1-x) = \sum_{n\geq 1}\frac{3\left(H_{n-1}^2-H_{n-1}^{(2)}\right)}{n}\,x^n$$ hence $$ \sum_{n\geq 1}\frac{3\left(H_{n-1}^2-H_{n-1}^{(2)}\right)}{n(n+1)}=-\int_{0}^{1}\log(1-x)^3\,dx = 6\tag{A}$$

$$\begin{eqnarray*} \sum_{n\geq 1}\frac{H_n^2}{n(n+1)}&=&\sum_{n\geq 1}\frac{H_{n-1}^2+\frac{2H_{n-1}}{n}+\frac{1}{n^2}}{n(n+1)}\\&=&\int_{0}^{1}\text{Li}_3(x)\,dx+\int_{0}^{1}\frac{(1-x)\log^2(1-x)}{x}\,dx+2+\sum_{n\geq 1}\frac{H_{n-1}^{(2)}}{n(n+1)}\\&\stackrel{\text{SBP}}{=}&\left(1-\zeta(2)+\zeta(3)\right)+2(\zeta(3)-1)+2+\sum_{n\geq 1}\left(1-\frac{1}{n+1}\right)\frac{1}{n^2}\\&=&1-\zeta(2)+3\,\zeta(3)+\zeta(2)-1\stackrel{\color{green}{\checkmark}}{=}\color{blue}{3\,\zeta(3)}.\tag{B} \end{eqnarray*}$$

$\text{SBP}$ stands for summation by parts.
This might not be the most efficient approach, but its logic is pretty simple to explain:

1. $H_n^2-H_n^{(2)}$ is a nicer weight than $H_n^2$, due to the Taylor series of $\log^3(1-x)$;
2. Both the integrals $\int_{0}^{1}\frac{x^m \log^{h}(x)}{1-x}\,dx$ and $\int_{0}^{1}\text{Li}_k(x)\,dx$ are elementary;
3. Every series of the $\sum_{n\geq 1}\frac{H_n^{(k)}}{n(n+1)}$ kind can be simply computed by $\text{SBP}$.

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By $\text{SBP}$ directly, $$\begin{eqnarray*} \sum_{n=1}^{N}\frac{H_n^2}{n(n+1)}&=&\left(1-\frac{1}{N+1}\right)H_N^2-\sum_{n=1}^{N-1}\left(1-\frac{1}{n+1}\right)\frac{H_n+H_{n+1}}{n+1}\\&=&H_N^2-\frac{H_N^2}{N+1}-\sum_{n=1}^{N-1}\frac{2H_n}{n+1}+\sum_{n=1}^{N-1}\frac{2H_n}{(n+1)^2}-H_N^{(2)}+H_N^{(3)}\\ &=&-\frac{H_N^2}{N+1}+\sum_{n=1}^{N-1}\frac{2H_n}{(n+1)^2}+H_N^{(3)}\end{eqnarray*}$$ hence $$\begin{eqnarray*} \sum_{n\geq 1}\frac{H_n^2}{n(n+1)}=\zeta(3)+2\sum_{n\geq 1}\frac{H_n}{(n+1)^2}&=&\zeta(3)+2\int_{0}^{1}\frac{\log(1-x)\log(x)}{1-x}\,dx \\&\stackrel{\text{IBP}}{=}&\zeta(3)+\int_{0}^{1}\frac{\log^2(1-x)}{x}\,dx\\&=&\zeta(3)+\int_{0}^{1}\frac{\log^2(x)\,dx}{1-x}\\&=&\zeta(3)+\sum_{n\geq 0}\int_{0}^{1}x^n\log^2(x)\,dx\\&=&\zeta(3)+2\,\zeta(3).\end{eqnarray*}$$