A series involving inverses of harmonic numbers

Solution 1:

math.SE, my sincere apologies, but the question was written wrong, $A_n$ is supposed to be $$(2n+2)(E_n)(E_{n+1})$$ not $$(2n+1)(E_n)(E_{n+1})$$ which as you can see from my work makes the summation a lot simpler and easier to find a closed form for.

A derivation on how to solve the problem: $$\begin{align} \sum_{n = 1}^{\infty}\frac{1}{A_n} &= \sum_{n = 1}^{\infty}\frac{4}{(2n+2)(H_n)(H_{n+1})} \\ &= \sum_{n = 1}^{\infty}\frac{4}{(2n+2)}*(n+1)*\left(\frac{1}{H_n} - \frac{1}{H_{n+1}}\right) \\ &= \sum_{n = 1}^{\infty}2\left(\frac{1}{H_n} - \frac{1}{H_{n+1}}\right) \\ &= 2\left(\frac{1}{H_1}\right) \\ &= 2 \end{align}$$