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?

Yes, it's just a geometric series of geometric series:
$$
\sum_{n=1}^{\infty}nx^{-n}=\sum_{k=1}^{\infty}\sum_{n=k}^{\infty}x^{-n}=\sum_{k=1}^{\infty}\frac{x^{-k}}{1-1/x}=\frac{x^{-1}}{(1-1/x)^2}=\frac{x}{(x-1)^2}.
$$
So your sum is $4 / 9$.
