# A series involving inverses of harmonic numbers

How would I solve this question:

If $$E_n = \frac{1}{2} + \frac{1}{4} + \frac{1}{6}+ \cdots +\frac{1}{2n}$$ and $$A_n = (2n+1)(E_n)(E_{n+1})$$ Find $$\sum_{n = 1}^{\infty}\frac{1}{A_n}$$

My try: We know that $$E_n = \frac{H_n}{2}$$ where $H_n$ is the harmonic series. So $$\sum_{n = 1}^{\infty}\frac{1}{A_n} = \sum_{n = 1}^{\infty}\frac{4}{(2n+1)(H_n)(H_{n+1})}$$ And I managed to simplify $$\frac{1}{(H_n)(H_{n+1})} = (n+1)\left(\frac{1}{H_n} - \frac{1}{H_{n+1}}\right)$$

But that's how far I've gotten. I get stuck after this. Any help/hints is greatly appreciated.

## 1 Answer

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}