What is this Basel-like sum of inverses?

Question

I ran across this sum and thought it looked very familiar but can't place it and had trouble searching for it. Does anyone recognize it or have an idea where I should look? I would like to know its sum.

$$F(n) = \sum_{i=1}^{n-1} \frac{1}{i(n-i)}$$

Thanks!

Answer 1

$$\sum_{i=1}^{n-1}\frac1{i(n-i)}=\frac1n\sum_{i=1}^{n-1}\left[\frac1i+\frac1{n-i}\right]=\frac1n\sum_{i=1}^{n-1}\frac1i+\sum_{i=1}^{n-1}\frac1{n-i}=\frac2n\sum_{i=1}^{n-1}\frac1i=\frac2nH_{n-1}$$ Where $H_k$ denotes the $k^\text{th}$ harmonic number.