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


*

*$H_n^2-H_n^{(2)}$ is a nicer weight than $H_n^2$, due to the Taylor series of $\log^3(1-x)$;

*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;

*Every series of the $\sum_{n\geq 1}\frac{H_n^{(k)}}{n(n+1)}$ kind can be simply computed by $\text{SBP}$.



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
$$\begin{eqnarray*} \sum_{n\geq 1}\frac{H_n^2}{n(n+1)}=\zeta(3)+2\sum_{n\geq 1}\frac{H_n}{(n+1)^2}&=&\zeta(3)+2\int_{0}^{1}\frac{\log(1-x)\log(x)}{1-x}\,dx \\&\stackrel{\text{IBP}}{=}&\zeta(3)+\int_{0}^{1}\frac{\log^2(1-x)}{x}\,dx\\&=&\zeta(3)+\int_{0}^{1}\frac{\log^2(x)\,dx}{1-x}\\&=&\zeta(3)+\sum_{n\geq 0}\int_{0}^{1}x^n\log^2(x)\,dx\\&=&\zeta(3)+2\,\zeta(3).\end{eqnarray*}$$
A: using the identity  $$ \sum_{n=1}^{\infty}x^n \left(H_n^2-H_n^{(2)}\right)=\frac{\ln^2(1-x)}{1-x}$$
divide both sides by $x$ then integrate w.r.t $x$ from $x=0$ to $y$ , we get
$$ \sum_{n=1}^{\infty}\frac{y^n}{n}\left(H_n^2-H_n^{(2)}\right)=\int_0^y\frac{\ln^2(1-x)}{1-x}\ dx-\frac13 {\ln^3(1-y)}$$
integrate both sides w.r.t $y$ from $y=0$ to $1$ , we get
\begin{align*}
\sum_{n=1}^{\infty}\frac{H_n^2-H_n^{(2)}}{n(n+1)}&=\int_0^1\int_0^y\frac{\ln^2(1-x)}{x}\ dxdy-\frac13\int_0^1{\ln^3(1-y)}\ dy\\
&=\int_0^1\frac{\ln^2(1-x)}{x}\left(\int_x^1dy\right)dx-\frac13\int_0^1{\ln^3(y)}\ dy\\
&=\int_0^1\frac{\ln^2(1-x)}{x}\left(1-x\right)dx-\frac13(-6)\\
&=\int_0^1\frac{\ln^2(1-x)}{x}dx-\int_0^1\ln^2(1-x)\ dx+2\\
&=\int_0^1\frac{\ln^2x}{1-x}dx-(2)+2\\
&=2\zeta(3)\tag{1}
\end{align*} 
\begin{align*}
\sum_{n=1}^{N}\frac{H_n^{(2)}}{n(n+1)}&=\sum_{n=1}^{N}\frac{H_n^{(2)}}{n}-\sum_{n=0}^{N}\frac{H_n^{(2)}}{n+1}\\
&=\sum_{n=1}^{N}\frac{H_n^{(2)}}{n}-\sum_{n=1}^{N+1}\frac{H_{n-1}^{(2)}}{n}\\
&=\sum_{n=1}^{N}\frac{H_n^{(2)}}{n}-\sum_{n=1}^{N+1}\frac{H_{n}^{(2)}}{n}+\sum_{n=1}^{N+1}\frac1{n^3}\\
&=\sum_{n=1}^{N}\frac{H_n^{(2)}}{n}-\sum_{n=1}^{N}\frac{H_{n}^{(2)}}{n}-\frac{H_{N+1}^{(2)}}{N+1}+\sum_{n=1}^{N+1}\frac1{n^3}\\
&=-\frac{H_{N+1}^{(2)}}{N+1}+\sum_{n=1}^{N+1}\frac1{n^3}
\end{align*}
letting $N$ approach $\infty$ yields
$$\sum_{n=1}^{\infty}\frac{H_n^{(2)}}{n(n+1)}=0+\sum_{n=1}^\infty\frac1{n^3}=\zeta(3)\tag{2}$$
plugging $(2)$ in $(1)$ we have
$$\sum_{n=1}^{\infty}\frac{H_n^2}{n(n+1)}=3\zeta(3)$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
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 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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With $\ds{N \in \mathbb{N}_{\ \geq\ 1}\,}$:
\begin{align}
&\bbox[10px,#ffd]{\sum_{n = 1}^{N}
{H_{n}^{2} \over n\pars{n + 1}}} =
\sum_{n = 1}^{N}{H_{n}^{2} \over n} -
\sum_{n = 1}^{N}{H_{n}^{2} \over n + 1}
\\[5mm] = &\
\sum_{n = 1}^{N}{H_{n}^{2} \over n} -
\sum_{n = 2}^{N + 1}{H_{n - 1}^{2} \over n}
\\[5mm] = &\
\sum_{n = 1}^{N}{H_{n}^{2} \over n} -
\sum_{n = 1}^{N}{H_{n}^{2} - 2H_{n}/n + 1/n^{2} \over n} -
{H_{N}^{2} \over N + 1}
\\[5mm] \stackrel{\mrm{as}\ N\ \to\ \infty}{\Huge \to}\,\,\, &\
2\sum_{n = 1}^{\infty}H_{n}\
\overbrace{\bracks{-\int_{0}^{\infty}\ln\pars{x}x^{n - 1}\,\dd x}}
^{\qquad\ds{1 \over n^{2}}}\ -\
\overbrace{\sum_{n = 1}^{\infty}{1 \over n^{3}}}^{\ds{\zeta\pars{3}}} \\[5mm] = &\
-2\int_{0}^{1}\ln\pars{x}\
\overbrace{\sum_{n = 1}^{\infty}H_{n}\
x^{n}}^{\ds{-\,{\ln\pars{1 - x} \over 1 - x}}}\,{\dd x \over x} -
\zeta\pars{3}
\\[5mm] = &\
2\int_{0}^{1}{\ln\pars{x}\ln\pars{1 - x} \over 1 - x}\,\dd x
\\[2mm] + &\
2\int_{0}^{1}{\ln\pars{x}\ln\pars{1 - x} \over x}\,\dd x
 -
\zeta\pars{3}
\\[5mm] = &\
4\int_{0}^{1}{\ln\pars{x}\ln\pars{1 - x} \over x}\,\dd x
 -
\zeta\pars{3}
\\[5mm] = &\ 
-4\int_{0}^{1}\mrm{Li}_{2}'\pars{x}\ln\pars{x}\,\dd x
 -
\zeta\pars{3}
\\[5mm] = &\
4\int_{0}^{1}\ \overbrace{\mrm{Li}_{2}\pars{x} \over x}
^{\ds{\mrm{Li}_{3}'\pars{x}}}\ \,\dd x
 -
\zeta\pars{3} =
\bbx{\large 3\zeta\pars{3}} \\ &
\end{align}
A: Using the generating function
$$\sum_{n=1}^\infty\frac{H_{n}^2}{n}x^{n}=\operatorname{Li}_3(x)-\ln(1-x)\operatorname{Li}_2(x)-\frac13\ln^3(1-x)$$
Integrate both sides from $x=0$ to $1$ we get
$$\sum_{n=1}^\infty\frac{H_{n}^2}{n(n+1)}=\int_0^1\operatorname{Li}_3(x)dx-\int_0^1\ln(1-x)\operatorname{Li}_2(x)dx-\frac13\int_0^1\ln^3(1-x)dx$$
$$=(\zeta(3)-\zeta(2)+1)-(-2\zeta(3)-\zeta(2)+3)-\frac13(-6)=\boxed{3\zeta(3)}$$
A: We have
$$\sum_{n=1}^\infty\frac{H_n}{n+1}x^{n+1}=\frac12\ln^2(1-x)$$
multiply both sides by $-\frac{\ln(1-x)}{x^2}$ then $\int_0^1$ and use that $-\int_0^1 x^{n-1}\ln(1-x)dx=\frac{H_n}{n}$ we get
$$\sum_{n=1}^\infty\frac{H_n^2}{n(n+1)}=-\frac12\int_0^1\frac{\ln^3(1-x)}{x^2}dx=-\frac12\int_0^1\frac{\ln^3(x)}{(1-x)^2}dx$$
$$=-\frac12\sum_{n=1}^\infty n\int_0^1 x^{n-1}\ln^3(x)dx=3\sum_{n=1}^\infty\frac{1}{n^3}=3\zeta(3)$$
