How to show $\sum_{n=1}^{\infty}\frac{H_{n+1}}{n(n+1)}=2$? I am interested in the proof of
$$\sum_{n=1}^{\infty}\frac{H_{n+1}}{n(n+1)}=2, \quad 
H_{n}=1+\frac{1}{2}+\cdots+\frac{1}{n}$$
This result can be verified by Mathematica or by WolframAlpha
This series can be found as Problem 3.59 (a) in the book Ovidiu Furdui: Limits, Series, and Fractional Part Integrals. Problems in Mathematical Analysis, Springer, 2013, Problem Books in Mathematics; where it is stated in the form
$$\sum_{k=1}^\infty\left(1+\frac12+\frac13+\dots+\frac1{n+1}\right)\frac1{n(n+1)}=2.$$
Some related thoughts:


*

*It is relatively easy to show that the series converges, the $n$-th term is approximately $\frac{\ln n}{n(n+1)}$. So we could use limit comparison test with the series $\frac1{n^\alpha}$ for any $\alpha\in(1,2)$.

*If the numerator is one, the series sums to $1$: How can I prove that $\sum_{n=1}^\infty \frac{1}{n(n+1)} = 1$? Here we expect larger result.

 A: For this solution, we'll make use of the telescoping sum
$$\sum_{n=1}^\infty{1\over n(n+1)}=\sum_{n=1}^\infty\left({1\over n}-{1\over n+1} \right)=1$$
and freely interchange sums, derivatives, and integrals.
First, let 
$$H_{n+1}(x)=x+{x^2\over2}+\cdots+{x^{n+1}\over n+1}$$ 
so that
$$H_{n+1}'(x)=1+x+\cdots+x^n={1-x^{n+1}\over1-x}$$
Now let
$$F(x)=\sum_{n=1}^\infty{H_{n+1}(x)\over n(n+1)}$$
so that
$$\begin{align}
F'(x)&=\sum_{n=1}^\infty{H_{n+1}'(x)\over n(n+1)}\\
&={1\over1-x}\left(\sum_{n=1}^\infty{1\over n(n+1)}-\sum_{n=1}^\infty{x^{n+1}\over n(n+1)} \right)\\
&={1\over1-x}\left(1-\sum_{n=1}^\infty\left({x^{n+1}\over n}-{x^{n+1}\over n+1} \right)\right)\\
&={1\over1-x}\left(1+\sum_{n=2}^\infty{x^n\over n}-x\sum_{n=1}^\infty{x^n\over n} \right)\\
&={1\over1-x}\left(1-x+\sum_{n=1}^\infty{x^n\over n}-x\sum_{n=1}^\infty{x^n\over n} \right)\\
&=1+\sum_{n=1}^\infty{x^n\over n}
\end{align}$$
and thus, since $F(0)=0$, we have
$$\sum_{n=1}^\infty{H_{n+1}\over n(n+1)}=F(1)=\int_0^1F'(x)dx=1+\sum_{n=1}^\infty{1\over n(n+1)}=1+1=2$$
A: Through generating functions:
$$ -\log(1-x)=\sum_{n\geq 1}\frac{x^n}{n},\qquad \frac{-\log(1-x)}{1-x}=\sum_{n\geq 1} H_{n} x^n\tag{1} $$
for any $x\in(0,1)$. By considering that $\int_{0}^{1}-\log(x)\,dx=1$ and that $\int_{0}^{1}x^n(1-x)\,dx=\frac{1}{(n+1)(n+2)}$
$$ -\log(1-x)=\sum_{n\geq 1}H_n x^n(1-x),\qquad 1=\sum_{n\geq 1}\frac{H_{n}}{(n+1)(n+2)}.\tag{2} $$
Since
$$ \sum_{n\geq 1}\frac{1}{(n+1)^2(n+2)}+\frac{1}{(n+1)(n+2)^2}=\sum_{n\geq 1}\frac{1}{(n+1)^2}-\frac{1}{(n+2)^2}=\frac{1}{4}\tag{3}$$
we get
$$ \frac{5}{4}=\sum_{n\geq 1}\frac{H_{n+2}}{(n+1)(n+2)},\qquad \sum_{n\geq 1}\frac{H_{n+1}}{n(n+1)}=\frac{5}{4}+\frac{H_2}{2}=2.\tag{4}$$
A: using the fact that $\displaystyle -\int_0^1x^n\ln(1-x)\ dx=\frac{H_{n+1}}{n+1}$
divide both sides by $n$ then take the sum, we get
$$\sum_{n=1}^\infty\frac{H_{n+1}}{n(n+1)}=-\int_0^1\ln(1-x)\sum_{n=1}^\infty\frac{x^n}{n}\ dx=\int_0^1\ln^2(1-x)\ dx=\int_0^1\ln^2x\ dx=2$$
A: Well.
Since
\begin{align}
\frac{1}{(n+1)n}= \frac{1}{n}-\frac{1}{n+1}
\end{align}
then we see that
\begin{align}
\frac{H_{n+1}}{n(n+1)} =&\ \frac{H_{n+1}}{n} - \frac{H_{n+1}}{n+1}\\
=&\ \frac{H_n}{n}-\frac{H_{n+1}}{n+1}+\frac{1}{n(n+1)}.
\end{align}
Hence we have that
\begin{align}
\sum^\infty_{n=1}\frac{H_{n+1}}{n(n+1)} = \sum^\infty_{n=1}\left(\frac{H_n}{n}-\frac{H_{n+1}}{n+1} \right)+\sum^\infty_{n=1}\left(\frac{1}{n}-\frac{1}{n+1} \right) = H_1 + 1 = 2.
\end{align}
A: $$\sum_{n=1}^N \frac{H_{n+1}}{n(n+1)}
=\sum_{n=1}^N H_{n+1}\left(\frac1n-\frac1{n+1}\right)
=H_2-\frac{H_{N+1}}{N+1}+\sum_{n=2}^{N}\frac{H_{n+1}-H_n}{n}
=H_2-\frac{H_{N+1}}{N+1}+\sum_{n=2}^{N}\frac1{n(n+1)}
\to\frac32+\sum_{n=2}^\infty\frac1{n(n+1)}$$
as $N\to\infty$ etc.
