I have a random variable $N$ that follows the Poisson distribution pattern where $\lambda > 0$, and I have to find the mean for $1/(N+1)$ where $N$ can take any of all the natural numbers $\{0, 1, 2, ..., +\infty\}$
The problem is that I am stuck finding the mean for $1/(N+1)$, where I got that the mean is the: $\sum_{i=0}^n \frac{1}{(i+1)}\cdot P(Z=\frac{1}{(i+1)})$
The problem resides in not knowing what distribution it follows thus how can I find the formula for $P(Z=z)$ where Z is random variable from the set $\{1/1, 1/2, 1/3, 1/4, ..., 0\}$
SOLUTION:
Thanks to drhab and infinite excuses for any unwanted caused strain, I realized that the formula for mean in my course work is more generalized so:
$E(Z) = \sum_{i=0}^n \frac{1}{(i+1)}\cdot P(N=i) =\sum_{i=0}^n \frac{1}{(i+1)} \cdot e^{-\lambda}\frac{\lambda^{i}}{i!} =\sum_{i=0}^n e^{-\lambda}\frac{\lambda^{i}}{(i+1)!} =\frac{1}{\lambda}\sum_{i=0}^n e^{-\lambda}\frac{\lambda^{i+1}}{(i+1)!} =\frac{1}{\lambda}\cdot(1-e^{-\lambda}\frac{\lambda^{0}}{(0)!}) =\frac{1}{\lambda}\cdot(1-e^{-\lambda}) $
Cheers
Hint:
You can find $\mathsf EZ$ much more easily.
Remember that: $$\mathsf Ef(N)=\sum_{n=0}^{\infty}f(n)\mathsf P(N=n)$$
Apply this on $f$ prescribed by $n\mapsto\frac1{n+1}$.