$\sum_{n = 1}^{D - 1} \frac{n}{D - n}$ written as a function of $D$?

Question

$$\sum_{n = 1}^{D - 1} \frac{n}{D - n}$$

Is it possible to reduce this summation to a function of $D$?

Answer 1

$$\sum_{n = 1}^{D - 1} \frac{n}{D - n}\\ =\sum_{n=1}^{D-1}\left( -1 + \frac{D}{D-n}\right)\\ =-D+1 + D\sum_{n=1}^{D-1}\frac1{D-n} \\ =-D+1 + D\sum_{m=1}^{D-1}\frac1{m} \\ = -D+1 + D H_{D-1}$$ where $H_n$ is the $n$th harmonic number.