Let X be Poisson with Parameter λ
Compute the mean of (1+X)^-1.
thanks
Start by noting that $\displaystyle E\left(\frac{1}{1+X} \right) = \sum_{x = 0}^{+\infty} \frac{1}{1 + x} \cdot \frac{\lambda^x e^{-\lambda}}{x!} = \sum_{x = 0}^{+\infty}\frac{\lambda^x e^{-\lambda}}{(x+1)!} = \frac{1}{\lambda} \sum_{x = 0}^{+\infty}\frac{\lambda^{x+1} e^{-\lambda}}{(x+1)!}$.
$\displaystyle \sum_{k = 0}^{+\infty}\frac{\lambda^{k} e^{-\lambda}}{k!} = 1$ will be useful in calculating the above sum ....
Try to look further than your nose ^^
$\displaystyle \frac{1}{\lambda} \sum_{x = 0}^{+\infty}\frac{\lambda^{x+1} e^{-\lambda}}{(x+1)!}$
Make the change $\displaystyle k=x+1$
and $\displaystyle 1=\sum_{k = 0}^{+\infty}\frac{\lambda^{k} e^{-\lambda}}{k!}=e^{-\lambda}+ \sum_{k=1}^\infty \frac{\lambda^k e^{-\lambda}}{k!}$
$\displaystyle \frac{1}{\lambda} \sum_{x = 0}^{+\infty} \frac{\lambda^{x+1} e^{-\lambda}}{(x+1)!} = \frac{1}{\lambda} \left( \frac{\lambda e^{-\lambda}}{1!} + \frac{\lambda^{2} e^{-\lambda}}{2!} + \, .... \right) = \frac{1}{\lambda} \left( 1 - e^{-\lambda}\right)$ using $\displaystyle \sum_{k = 0}^{+\infty}\frac{\lambda^{k} e^{-\lambda}}{k!} = 1$.