Expected Value of (1/X) for Poisson X for x>0

Question

I know how to calculate the E(1/(x+1) for Poisson X, but it gets more complicated when it is just E(1/X). Does the answer reduce to a closed form, or is it an infinite sum?

Answer 1

$\Pr[X = 0] = e^{-\lambda} > 0$, therefore $$\operatorname{E}[1/X] = \sum_{x=0}^\infty \frac{1}{x} \Pr[X = x] = \frac{1}{0} e^{-\lambda} + \sum_{x=1}^\infty \frac{1}{x} \Pr[X = x] = \infty.$$