How to calculate complicated geometrical series? I have a geometrical series (I don't know if its geometrical series or not):
$$
\sum_{n=1}^{\infty }n\rho ^{n}(1-\rho)
$$
how can I simplify it ? ( assume that $ 0 \le \rho \le 1$ )
The last answer in my calculatio should be $\frac{\rho}{1-\rho}$. But I really don't know how ?

$$\sum_{n=1}^{\infty }n\rho ^{n}(1-\rho)=(1-\rho)\rho\frac{d(\sum_{n=1}^{\infty }\rho^n)}{d \rho}$$
Now using Infinite Geometric Series $$\sum_{n=1}^{\infty }\rho^n=\frac \rho{1-\rho}=-1-\frac1{\rho-1}$$ as $|\rho|<1$
