closed form solution for $ \displaystyle \sum_{x=1}^{k-1} {\frac{1}{x(k-x)}} $

Question

Is there any closed form solution for this summation?

$$ \sum_{x=1}^{k-1} {\frac{1}{x(k-x)}} $$

$k$ is a finite integer constant.

Answer 1

We have that $\frac{1}{x(k-x)} = \frac{1}{k} \left( \frac{1}{x} + \frac{1}{k-x} \right)$. Therefore, $$\sum_{x=1}^{k-1} \frac{1}{x(k-x)} = \frac{1}{k} \left( \sum_{x=1}^{k-1} \frac{1}{x} + \sum_{x=1}^{k-1} \frac{1}{k-x} \right).$$ By reversing the order of summation in the second sum, we get that this is equal to $\frac{2}{k} H_{n-1}$, where $H_{n-1} = \sum_{x=1}^{k-1} \frac{1}{x}$ is the harmonic sum. (There is no simpler closed-form expression for $H_n$, though a great deal is known about it.)