Attempt to find $E\left(\dfrac{1}{1+X}\right)$ for Poisson distribution with parameter $\lambda$:
$$E\left(\frac{1}{1+X}\right) = \sum\limits_{x=0}^{\infty} \frac{e^{-\lambda}\lambda^{x}}{x!(x+1)}$$
$$= \frac{1}{\lambda}\sum\limits_{x=0}^{\infty} \frac{e^{-\lambda}\lambda^{x+1}}{(x+1)!}$$
$= \dfrac{1}{\lambda}$ as the sigma just sums to $1.$
But this doesn't appear to be the correct answer as per my textbook, is there something I'm overlooking? Thanks!
You have got one key part wrong at the end.
Indeed, when you write $\frac 1{\lambda} \sum_{x=0}^{\infty} \frac{e^{-\lambda} \lambda^{x+1}}{(x+1)!}$, the sum is not $1$, since the sum might go over $x = 0$ to $\infty$ but here we have $\mathbf{x+1}$ as the variable, not $x$, in the Poisson summation. So, we are actually skipping the term in the sum which corresponds to $0$. We must account for this.
So infact, the summation is actually $1 - \frac{e^{-\lambda}\lambda^0}{0!} = 1 - e^{-\lambda}$, thence giving the answer $\frac{1 - e^{-\lambda}}{\lambda}$, which should be correct.
