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.

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
$$\sum_{n=1}^{\infty}\frac{H_n^2}{n(n+1)}=3\zeta(3)$$
