<s>[INST] Calculating the summation $\sum_{n=1}^{\infty}\frac{H_{n+1}}{n(n+1)}$ I need to find explicitly the following summation
$$\sum_{n=1}^{\infty}\frac{H_{n+1}}{n(n+1)}, \quad 
H_{n}=1+\frac{1}{2}+\cdots+\frac{1}{n}$$
From Mathematica, I checked that the answer is $2$. The same result is returned by WolframAlpha.
A thought from afar:
$$
\begin{align}
\frac{H_{n+1}}{n(n+1)}
&=H_{n+1}\left(\frac1n-\frac1{n+1}\right)\tag1\\
&=\color{#C00}{H_{n+1}}\color{#090}{(H_n-H_{n-1})}-H_{n+1}(H_{n+1}-H_n)\tag2\\
&=\color{#C00}{\frac1{n+1}}\color{#090}{\frac1n}+\color{#C00}{H_n}\color{#090}{(H_n-H_{n-1})}-H_{n+1}(H_{n+1}-H_n)\tag3\\
&=\frac1n-\frac1{n+1}+\frac{H_n}{n}-\frac{H_{n+1}}{n+1}\tag4\\
&=\frac{H_n+1}{n}-\frac{H_{n+1}+1}{n+1}\tag5
\end{align}
$$
What was done:
$(1)$: partial fractions
$(2)$: used $\frac1n=H_n-H_{n-1}$ and $\frac1{n+1}=H_{n+1}-H_n$
$(3)$: used $\color{#C00}{H_{n+1}=\frac1{n+1}+H_n}$ and $\color{#090}{\frac1n=H_n-H_{n-1}}$
$(4)$: partial fractions and $\frac1n=H_n-H_{n-1}$ and $\frac1{n+1}=H_{n+1}-H_n$
$(5)$: combined terms
Now this looks like Clement C's hint.
 [/INST] (Big) hint:
$$\frac{H_{n+1}}{n(n+1)}
= \frac{H_{n+1}}{n}-\frac{H_{n+1}}{n+1}
= \frac{1}{n(n+1)} + \frac{H_{n}}{n}-\frac{H_{n+1}}{n+1}$$
so that you can get a telescopic series: for any $N\geq 1$,
$$
\sum_{n=1}^N \frac{H_{n+1}}{n(n+1)} 
= \sum_{n=1}^N \frac{1}{n(n+1)}  + \sum_{n=1}^N \frac{H_{n}}{n}  - \sum_{n=2}^{N +1}\frac{H_{n}}{n}
$$
Spoiler:

 The value you should arrive at is $2$, which is $\sum_{n=1}^\infty \frac{1}{n(n+1)}  + 1$.

</s>