How to show that $\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}=\frac{\pi^2}{6}$

Beside telescoping, all you need is to reverse the order of the double sum: $$\sum_{n\ge 1}\frac{\sum_{k=1}^n k^{-1}}{n(n+1)}=\sum_{k\ge 1}k^{-1}\sum_{n\ge k}\frac{1}{n(n+1)}=\sum_{k\ge 1}k^{-2}=\frac{\pi^2}{6}.$$Edit to explain the reversal: we're summing all terms of the form $\frac{k^{-1}}{n(n+1)}$ with $n,\,k$ integers in the set $\{(n,\,k)|n,\,k\ge 1,\,k\le n\}$. But I could equivalently describe this set as $\{(n,\,k)|k\ge 1,\,n\ge k\}$.

We have \begin{align} \sum_{n=1}^\infty\frac{H_n}{n^2+n} &=\sum_{n=1}^\infty H_n\left(\frac1n-\frac1{n+1}\right) \\&=\sum_{n=1}^\infty\sum_{m=1}^n\frac1m\left(\frac1n-\frac1{n+1}\right) \\&=\sum_{m=1}^\infty\frac1m\sum_{n=m}^\infty\left(\frac1n-\frac1{n+1}\right) \\&=\sum_{m=1}^\infty\frac1m\cdot\frac1m \\&=\frac{\pi^2}6. \end{align}

applying summation by parts \begin{align} \sum_{n=1}^N\frac{H_n}{n(n+1)}&=\sum_{n=1}^N\frac{H_n}{n}-\sum_{n=1}^N\frac{H_n}{n+1}\\ &=\sum_{n=1}^N\frac{H_n}{n}-\sum_{n=1}^{N+1}\frac{H_{n-1}}{n}\\ &=\sum_{n=1}^N\frac{H_n}{n}-\sum_{n=1}^{N+1}\frac{H_{n}}{n}+\sum_{n=1}^{N+1}\frac{1}{n^2}\\ &=\sum_{n=1}^N\frac{H_n}{n}-\sum_{n=1}^{N}\frac{H_{n}}{n}-\frac{H_{N+1}}{N+1}+\sum_{n=1}^{N+1}\frac{1}{n^2}\\ &=-\frac{H_{N+1}}{N+1}+\sum_{n=1}^{N+1}\frac{1}{n^2} \end{align} letting $N$ approach $\infty$, we get, $$\sum_{n=1}^\infty\frac{H_n}{n(n+1)}=0+\sum_{n=1}^{\infty}\frac{1}{n^2}=\zeta(2)$$