This is supposed to be easy, but somehow I am kinda stuck on this.
We can write
$$\sum_{k = 1}^{\infty}kp^{k-1} = \sum_{k=0}^{\infty}kp^{k-1}$$
But one little trick to prove that the mean of a geometric random variable equals $\frac{1}{p}$ is to claim
$$\sum_{k=1}^{\infty}kp^{k-1} = \sum_{k=0}^{\infty}(k+1)p^k$$
However, I think the second equation is not true since it implies that
$$\sum_{k=0}^{\infty}(k+1)p^k = \sum_{k=0}^{\infty}kp^{k-1}$$ which is apparently false.
Does anyone know how to solve this?
As hinted in the comments, the formula $$\sum_{k=0}^{\infty}(k+1)p^k = \sum_{k=0}^{\infty}kp^{k-1}$$ can be proved by rewriting $$\sum_{k=0}^{\infty}kp^{k-1}$$ as $$\sum_{k=1}^{\infty}kp^{k-1},$$ since the first term (for $k = 0$) is just $$0\cdot{p^{0-1}} = 0.$$