Find the expectation of a Geometric distribution using $\mathbb{E}(X)= \sum_{k=1}^\infty P(X \ge k)$.
Okay I know how to find the expectation using the definition of the geometric distribution $P(X=k)= p \cdot(1-p)^{k-1}$ and I figured that $P(X \ge k)=(1-p)^{k-1}$ but I don't know how to show it.
I know the expectation is $\frac{1}{p}$ but I just get $\mathbb E(X)= \frac{1}{p^2}$ using the method specified in the question.