Compute the mean of $(1 + X)^{-1}$ where $X$ is Poisson$(\lambda)$ Question
Let $X$ be Poisson with parameter $\lambda$.  Compute the mean of $(1 + X)^{-1}$. (Introduction to Probability Theory, Hoel, pp. 104)
Answer Key
The answer key shows that the mean is $\lambda^{-1}(1 - e^{-\lambda})$
My Solution
$$E(1 + X)^{-1} = \sum\limits_{j = 1}^\infty (1 + j)^{-1} \dfrac{\lambda^{(1 + j)^{-1}}e^{-\lambda}}{(1 + j)^{-1}!}$$
$$= e^{-\lambda} \sum\limits_{j = 1}^\infty \dfrac{\lambda^{(1 + j)^{-1}}}{((1 + j)^{-1} - 1)!}$$
$$= \lambda^{-1} e^{-\lambda} \sum\limits_{j = 1}^\infty \dfrac{\lambda^{(1 + j)^{-1} - 1}}{((1 + j)^{-1} - 1)!}$$
$$= \lambda^{-1} e^{-\lambda}(e^{\lambda} - 1)$$
$$= \lambda^{-1}(1 - e^{-\lambda})$$
EDIT: I found why I got the answer incorrect.  Thanks for those who answer the question! :)
 A: Your solution is not quite correct:
$$
  \mathsf{E}\left(\frac{1}{1+X}\right) = \sum_{k=0}^\infty \frac{1}{1+k} \Pr\left(X=k\right) = \sum_{k=0}^\infty \frac{1}{1+k} \frac{\lambda^k \mathrm{e}^{-\lambda}}{k!} = \frac{\mathrm{e}^{-\lambda}}{\lambda} \sum_{k=0}^\infty \frac{\lambda^{k+1} }{(k+1)!} = \frac{\mathrm{e}^{-\lambda}}{\lambda} \left( \sum_{k=1}^\infty \frac{\lambda^{k} }{k!}  \right)  = \frac{\mathrm{e}^{-\lambda}}{\lambda} \left( \sum_{k=0}^\infty \frac{\lambda^{k} }{k!} -1 \right) = \frac{\mathrm{e}^{-\lambda}}{\lambda} \left(\mathrm{e}^\lambda -1 \right)
$$
A: We do the calculation. The expectation is 
$$\sum_{n=0}^\infty \frac{1}{1+n}e^{-\lambda}\frac{\lambda^n}{n!}.$$
This is 
$$e^{-\lambda} \sum_{n=0}^\infty \frac{\lambda^n}{(n+1)!}.$$
Divide, multiply by $\lambda$. We get 
$$\frac{e^{-\lambda}}{\lambda}\sum_{n=0}^\infty \frac{\lambda^{n+1}}{(n+1)!}.$$
The inner sum above is equal to  $e^{\lambda}-1$, so the whole thing simplifies to
$$\frac{1}{\lambda}(1-e^{-\lambda}).$$ 
A: Your formula for expectation has a problem, you should not plug in $1+X$ inverse in the probability as you are summing over the sample space, you should be plugging in $X$ only.
