TITLE: How to show that $\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}=\frac{\pi^2}{6}$
QUESTION [6 upvotes]: 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.

REPLY [3 votes]: 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,
$$\sum_{n=1}^\infty\frac{H_n}{n(n+1)}=0+\sum_{n=1}^{\infty}\frac{1}{n^2}=\zeta(2)$$