Find Expected value

Question

Suppose, $X \sim \operatorname{Poisson}(\lambda)$, Then what is $\operatorname{E}\left(\dfrac{1}{X+1}\right)$ ?

My attempt:

Let $Y = \dfrac{1}{X+1}.$ Then

$$ F_Y(t) = \{ Y \leq t\} = \left\{ X \geq \frac{1-t} t \right\} = 1- F_Y \left( \frac{1-t} t \right). $$

is this way okay? Any hint/alternative way to solve this ...

Answer 1

Hint: Substitute and compensate.

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

Can you take it from here or do you need more help?