Defining $X\sim Poisson(\lambda)$, I'd like to find $\operatorname{E}(\frac{1}{1+X})$.
I did the following, but i'm not sure if it's right: \begin{align} \operatorname{E}\left(\frac{1}{1+X}\right) &= \sum_{x=0}^{\infty}\frac{e^{-\lambda}\lambda^{x}}{x!(x+1)} = \sum_{x=0}^{\infty}\frac{e^{-\lambda}\lambda^{x+1}}{(x+1)!\lambda} = e^{-\lambda}\frac{1}{\lambda}\sum_{x=0}^{\infty}\frac{\lambda^{x+1}}{(x+1)}\\ \end{align} Making a substitution in the sum index: $y=x-1$ \begin{align} e^{-\lambda}\frac{1}{\lambda}\sum_{y=-1}^{\infty}\frac{\lambda^{y}}{y!} =e^{-\lambda}\frac{1}{\lambda}\left(1+\sum_{y=0}^{\infty}\frac{\lambda^{y}}{y!} \right)=\frac{1}{\lambda} e^{-\lambda} (e^{\lambda}+1) = \frac{1}{\lambda}+\frac{e^{-\lambda}}{\lambda} \end{align} I would thankful if someone pointed me in the right direction, i could not find any posts about this.
\begin{align} \operatorname{E}\left(\frac{1}{1+X}\right) &= \sum_{x=0}^{\infty}\frac{e^{-\lambda}\lambda^{x}}{x!(x+1)} \\&= \sum_{x=0}^{\infty}\frac{e^{-\lambda}\lambda^{x+1}}{(x+1)!\lambda} \\&= e^{-\lambda}\frac{1}{\lambda}\sum_{x=0}^{\infty}\frac{\lambda^{x+1}}{(x+1)!}\\ \\&= e^{-\lambda}\frac{1}{\lambda}\sum_{r=1}^{\infty}\frac{\lambda^{r}}{r!}, \text{where }r=x+1\\ \\&= e^{-\lambda}\frac{1}{\lambda}\left(\sum_{r=0}^{\infty}\frac{\lambda^{r}}{r!}-1\right)\\ \\&= \frac{1}{\lambda}\left(\sum_{r=0}^{\infty}e^{-\lambda}\frac{\lambda^{r}}{r!}-e^{-\lambda}\right)\\ \\&= \frac{1}{\lambda}\left(1-e^{-\lambda}\right)\\ \end{align}