[Math] Expectation of $\frac{1}{x+1}$ of Poisson distribution

Question

As the title states, I'm trying to find the expecteed value of $\frac{1}{x+1}$ where $X \sim \mathrm{Poisson}(\lambda)$

My attempt:
\begin{align}
&\sum \frac{1}{x+1} \cdot \frac{e^{-λ}\cdot λ^x}{x!} \\
\implies& \sum \frac{x+1}{(x+1)^2} \cdot \frac{e^{-λ}\cdot λ^x}{x!} \\
\implies& e^{-λ} \cdot \sum \frac{\lambda^{x+1}}{(x+1)!} \cdot \lambda
\end{align}

I don't know how to further progress from here … Help is appreciated

Answer 1

You should recognize that

$$\sum_{n=0}^{\infty} \frac{\lambda^{n+1}}{(n+1)!} = e^{\lambda}-1$$

So your result is

$$E\left ( \frac{1}{N+1}\right) = \frac{1-e^{-\lambda}}{\lambda}$$