Finding an expected value Below is a problem that I did from Chapter 4 of the book "Intoduction to Probability Theory". The book was written by Hoel, Port and Stone. The answer I got is wrong. I would like to know what I
did wrong.
Thanks,
Bob  
Problem:
Let $X$ be a Poisson with parameter $\lambda$. Compute the mean of
$(1+X)^{-1}$.
Answer:
The density function for the Poisson distribution is:
    $$f(x) = \frac{\lambda^x e ^ {-x}}{x!}$$
Let $u$ be the mean that we seek.
\begin{eqnarray*}
u &=& \sum_{x = 0}^{\infty} \frac{\lambda^x e ^ {-x}}{x!((1+x))} = 
 \sum_{x = 0}^{\infty} \frac{\lambda^x e ^ {-x}}{(x+1)!} \\
u &=& \sum_{x = 1}^{\infty} \frac{\lambda^x e ^ {-(x-1)}}{(x)!} \\
u &=& \sum_{x = 1}^{\infty} \frac{\lambda^x e ^ {-x + 1}}{(x)!} \\
u &=& e \sum_{x = 1}^{\infty} \frac{\lambda^x e ^ {-x}}{(x)!} \\
\end{eqnarray*}
Observe that when $\lambda$ is very large that $u$ is very large. Therefore, I conclude
that I am already wrong. The books answer is:
$$ \lambda^{-1}(1-e^{-\lambda}) $$
 A: Your $f(x)$ is not correct.  The term $e^{-x}$ should be $e^{-\lambda}$.  
From your first line to the second, when you offset $x$ by $1$, you should have $\lambda^{x-1}$ in the numerator.  That would lead you to pull a factor $\lambda^{-1}$ out of the sum, fixing that part.  
Then the final sum would be $1$ if it went from $0$ to $\infty$, so it is $1$ minus the $x=0$ term giving $1-e^{-\lambda}$
A: \begin{eqnarray*}
u &=& \sum_{x = 0}^{\infty} \frac{\lambda^x e ^ {-\color{red}{\lambda}}}{x!((1+x))}\\ &=& 
 \sum_{x = 0}^{\infty} \frac{\lambda^x e ^ {-\lambda}}{(x+1)!} \\
 &=& \sum_{x = 1}^{\infty} \frac{\lambda^{x-1} e ^ {-\lambda}}{x!} \\
 &=& \frac 1\lambda\sum_{x = 1}^{\infty} \frac{\lambda^x e ^ {-\lambda}}{x!} 
\\ &=& \frac 1\lambda\left(\sum_{x = 0}^{\infty} \frac{\lambda^x e ^ {-\lambda}}{x!}-e^{-\lambda}\right) \\ &\vdots&
\end{eqnarray*}
A: The Poisson distribution is
$$f(x) = \frac{\lambda^x e ^ {-{\color{red}{\lambda}}}}{x!}, x \in \mathbb{N}. $$
Therefore, the expected value $u$ of $(1+X)^{-1}$ is:
$$u  = \sum_{x = 0}^{\infty} \frac{\lambda^x e ^ {-\lambda}}{x!(1+x)} = e ^ {-\lambda}\sum_{x = 0}^{\infty} \frac{\lambda^x}{(1+x)!}.$$
Now, do a substitution $t = x+1$:
$$u  = e ^ {-\lambda}\sum_{t = 1}^{\infty} \frac{\lambda^{t-1} }{t!} = \frac{e^{-\lambda}}{\lambda}\sum_{t = 1}^{\infty} \frac{\lambda^{t}}{t!} = \frac{e^{-\lambda}}{\lambda}\left(\sum_{t = 0}^{\infty} \frac{\lambda^{t}}{t!} - 1\right).$$
It is well-known that:
$$\sum_{t = 0}^{\infty} \frac{\lambda^{t} }{t!} = e^{\lambda}.$$
You can check this fact here.
Finally:
$$u = \frac{e^{-\lambda}}{\lambda}(e^{\lambda}-1) = \frac{1}{\lambda}(1-e^{-\lambda}) = \lambda^{-1}(1-e^{-\lambda}).$$
