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.
 A: (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 $\sum_{n=1}^\infty \frac{1}{n(n+1)}  + 1$.

A: Hint:
$$
\begin{align}
\sum_{n=1}^\infty\frac{H_{n+1}}{n(n+1)}
&=\sum_{n=1}^\infty\sum_{k=1}^{n+1}\frac1k\left(\frac1n-\frac1{n+1}\right)\\
&=\sum_{n=1}^\infty\frac1{n+1}\left(\frac1n-\frac1{n+1}\right)
+\sum_{n=1}^\infty\sum_{k=1}^n\frac1k\left(\frac1n-\frac1{n+1}\right)\\
&=\sum_{n=1}^\infty\left(\frac1n-\frac1{n+1}-\frac1{(n+1)^2}\right)
+\sum_{k=1}^\infty\sum_{n=k}^\infty\frac1k\left(\frac1n-\frac1{n+1}\right)
\end{align}
$$
Now its simply summing telescoping series (three times).
A: Noting that
$$\frac{1}{n(n + 1)} = \frac{1}{n} - \frac{1}{n + 1},$$
the sum may be rewritten as
$$\sum_{n = 1}^\infty \frac{H_{n+1}}{n(n + 1)} = \sum_{n = 1}^\infty \left (\frac{H_{n + 1}}{n} - \frac{H_{n + 1}}{n + 1} \right ).\tag1 \label1$$
Now as the harmonic numbers satisfy the recurrence relation
$$H_{n + 1} = H_n + \frac{1}{n + 1},$$
we have
\begin{align}
\sum_{n = 1}^\infty \frac{H_{n+1}}{n(n + 1)} &= \sum_{n = 1}^\infty \left [\frac{H_n}{n(n+1)} + \frac{1}{n(n +1)} - \frac{1}{(n + 1)^2} \right ]\\
&= \sum_{n = 1}^\infty \frac{H_n}{n(n + 1)} + \sum_{n = 1}^\infty \frac{1}{n(n + 1)} - \underbrace{\sum_{n = 1}^\infty \frac{1}{(n + 1)^2}}_{n \, \mapsto n - 1}\\
&= \sum_{n = 1}^\infty \frac{H_n}{n(n + 1)} + \sum_{n = 1}^\infty \frac{1}{n(n+1)} - \sum_{n = 1}^\infty \frac{1}{n^2} + 1\tag2\\
&= \zeta (2) + 1 - \zeta (2) + 1\\
&= 2
\end{align}
In (2) an evaluation for the first sum can be found here while the second sum telescopes and has a sum equal to one.
A: using the fact that $\displaystyle\int_0^1 x^{n}\ln(1-x)\ dx=-\frac{H_{n+1}}{n+1}$
divide both sides by $-n$ then take the sum, 
\begin{align}
S&=\sum_{n=1}^\infty\frac{H_{n+1}}{n(n+1)}=-\int_0^1\ln(1-x)\left(\sum_{n=1}^\infty\frac{x^n}{n}\right)\ dx\\
&=\int_0^1\ln^2(1-x)\ dx=\int_0^1\ln^2x\ dx=2
\end{align}
