Sum of infinite series $ \sum_{k=0}^{\infty} kp(1-p)^k $

Question

$ \sum_{k=0}^{\infty} kp(1-p)^k , p_\in<0,1>$
I know that the sum is
$ \frac{1-p}{p} $
but i don't know how to arrive to it.
There is a resemblence to geometric series but i don't know how to deal with the $k$. $ \sum_{k=0}^{\infty} kp(1-p)^k = p(1-p)\sum_{k=0}^{\infty} k(1-p)^{k-1}$
Sum of the series is
$\sum_{k=0}^{\infty} k(1-p)^{k-1} = \frac{1}{p^2}$
but once again, i don't know how to deal with the $k$.

Answer 1

Hint: $p$ is a constant; so, factor it out and let's just consider $$ \sum_{k=0}^{\infty}k(1-p)^k. $$ More generally, let's let $x$ be a variable with $\lvert x\rvert<1$. Then consider $$ \sum_{k=0}^{\infty}kx^k. $$ Then you can write $$ \sum_{k=0}^{\infty}kx^k=\sum_{k=1}^{\infty}kx^k=x\sum_{k=1}^{\infty}kx^{k-1}=x\frac{d}{dx}\left[\sum_{k=0}^{\infty}x^k\right]=x\frac{d}{dx}\left[\frac{1}{1-x}\right]=\cdots $$ Once you have a result, you can plug $x=1-p$ back into it, multiply the result by $p$, and you've got it.