Say I have a number $a$ and I want to find a number $b>a$ where $b \sim U[0,1]$. The value of b is a geometric random variable with well known expectation $\frac{1}{1-a}$
However, consider the number of draws to be a random variable $X$. We want $E[X]$. Then $P(X>x) = a^x$, which implies that $E(X)=\sum_0^\infty P(X>x) = \sum_0^\infty a^x$
However, this does not resolve to $\frac{1}{1-a}$. Am I not allowed to use the inverse CDF to calculate expectation with discrete RVs?