TITLE: How to show $\sum_{n=1}^{\infty}\frac{H_{n+1}}{n(n+1)}=2$?
QUESTION [8 upvotes]: 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.

REPLY [2 votes]: 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
$$\sum_{n=1}^\infty\frac{H_{n+1}}{n(n+1)}=-\int_0^1\ln(1-x)\sum_{n=1}^\infty\frac{x^n}{n}\ dx=\int_0^1\ln^2(1-x)\ dx=\int_0^1\ln^2x\ dx=2$$