Poisson Distribution with random variable

Question

Let $X$ be a random variable with Poisson distribution with parameter $\lambda$, $\lambda> 0$. Get:

(a) $E \left(\frac{1} {x+1} \right) $

Answer (a) $ \frac{1-e^{-\lambda}}{\lambda}$

I've tried to do it in $n$ ways. Can someone help me?

Answer 1

$$\mathbb{E}\left[\frac{1}{X+1}\right] = \sum_{k=0}^\infty \frac{1}{k+1} e^{-\lambda} \frac{\lambda^k}{k!} = \frac{e^{-\lambda}}{\lambda} \sum_{k=0}^\infty \frac{\lambda^{k+1}}{(k+1)!} = \frac{e^{-\lambda}}{\lambda} (e^{\lambda}-1) = \frac{1-e^{-\lambda}}{\lambda}$$