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

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.
• There is a similar question where denominator is $n(n+2)$ rather than $n(n+1)$: How to find the sum of this infinite series: $\sum_{n=1}^{ \infty } \frac1n \cdot \frac{H_{n+2}}{n+2}$
• A change which looks - at the first glance - only minor leads to $\sum\limits_{n=1}^{\infty} \frac{H_n}{n (n+1)} = \frac{\pi^2}{6}$. See: Striking applications of summation by parts