Sum of Geometric infinite series

Question

How do I solve this:

$$\sum_{k=1}^{\infty}k(1-p)^{k-1}$$

I forgot how to do this, or the formula I need to use. Could not find it online for some reason.

Answer 1

We have $$\sum_{k\geq0}x^{k}=\frac{1}{1-x},\left|x\right|<1 $$ hence taking the derivative $$\sum_{k\geq0}kx^{k-1}=\frac{1}{\left(1-x\right)^{2}} $$ then take $x=1-p$.

Answer 2

This is not really a geometric series. However $$\sum_{k=0}^{\infty} (1-p)^{k}=\dfrac{1}{1-(1-p)}=\dfrac{1}{p}$$ provided $|1-p|<1$. This is an analytic function of $p$ in the disc of convergence, so it is allowed to differentiate term by term, which gives you the series you're interested in.

Answer 3

Provided $0<p<2$, then we can write that $$\sum_{k=1}^{\infty}(1-p)^{k}=\frac{1}{1-(1-p)}=\frac1p$$ By taking the derivative with respect to $p$, we get $$\sum_{k=1}^{\infty}k(1-p)^{k-1}=\frac{1}{p^2}$$