<s>[INST] How to show that $\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}=\frac{\pi^2}{6}$ Wolfram Alpha shows that
$$\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}=\zeta(2)=\frac{\pi^2}{6}$$
I want to prove this.
Attempt:
I tried to treat this as a telescoping series:
$$\begin{align}
\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}&=\sum_{n=1}^{\infty}H_n\left(\frac{1}{n}-\frac{1}{n+1}\right)\\
&=H_{1}\left(1-\frac{1}{2}\right)+H_{2}\left(\frac{1}{2}-\frac{1}{3}\right)+H_{3}\left(\frac{1}{3}-\frac{1}{4}\right)\\
&=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{6}+\frac{1}{3}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\frac{1}{9}-\frac{1}{12}
\end{align}$$
I think this method is not quite useful, so I tried  another one:
$$H_n=\int_{0}^{1}\frac{1-t^n}{1-t}dt$$
Then,
$$\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}=\sum_{n=1}^{\infty}\frac{1}{n^2+n}\int_{0}^{1}\frac{1-t^n}{1-t}dt$$
At this point, I do not know how to proceed.
 [/INST] 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 $\{(n,\,k)|k\ge 1,\,n\ge k\}$.
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