Computing $\sum_{n\geq 0}n\frac{1}{4^n}$ Can I compute the sum
$$
\sum_{n\geq 0}n\frac{1}{4^n}
$$
by use of some trick?
First I thought of a geometrical series?

The idea is to take the derivative of the geometric series. For $|x|<1$,
$$\frac1{(1-x)^2}=\sum_{n=1}^\infty nx^{n-1}$$
So,
$$\frac x{(1-x)^2}=\sum_{n=1}^\infty nx^n.$$
