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 ?

Using Arithmetico-geometric sequence,  $$\sum_{n=1}^\infty\left[a+(n-1)d\right]r^{n-1}=\frac a{1-r}+\frac{rd}{(1-r)^2}\text{  for } |r|<1$$
We have $$\sum_{n=1}^{\infty}n\rho^n(1-\rho)=(1-\rho)\rho\left(\sum_{n=1}^{\infty}n \rho^{n-1}\right)$$
Here,  we have $\displaystyle a=d=1,r=\rho$
