<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] $\newcommand{\Li}{\operatorname{Li}}$Starting with a generating function for the Harmonic Numbers

$$-\frac{\log(1-x)}{1-x}~=~\sum_{n=1}^\infty H_nx^n\tag1$$

we divide both sides by $x$ and integrate afterwards to get
$$\small\begin{align*}
\int-\frac{\log(1-x)}{x(1-x)}\mathrm dx&=\int\sum_{n=1}^\infty H_nx^{n-1}\mathrm dx\\
\Li_2(x)+\frac12\log^2(1-x)+c&=\sum_{n=1}^\infty \frac{H_n}nx^{n}
\end{align*}$$
Plug in $x=0$ to determine the value of $c$ which turns out to be $0$ aswell. Integrating this equation yields to
$$\small\begin{align*}
\int\Li_2(x)+\frac12\log^2(1-x)\mathrm dx&=\int\sum_{n=1}^\infty \frac{H_n}nx^{n}\mathrm dx\\
x\Li_2(x)-x+(x-1)\log(1-x)+\frac12(x-1)(\log^2(1-x)-2\log(1-x)+2)+c&=\sum_{n=1}^\infty\frac{H_n}{n(n+1)}x^{n}
\end{align*}$$
The new constant again equals $0$ due the same argumentation as above. Now we can plug in $x=1$ to compute the value of your given series which overall is given by $\Li_2(1)$ and this is well-known to equal $\zeta(2)$.

$$\therefore~\sum_{n=1}^\infty\frac{H_n}{n(n+1)}~=~\frac{\pi^2}6\tag2$$

</s>