Find expected value of $\frac{1}{1+\xi}$ Find expected value of $\frac{1}{1+\xi}$ if $\xi$ has Poisson distribution.
My try:
$$E\left[ \frac{1}{1+\xi}\right]=\sum_{k=0}^{\infty}{x}_{k}{p}_{k}=\sum_{k=0}^{\infty}\frac{1}{k+1}\frac{\lambda^ke^{-\lambda}}{k!}$$
Is it right?
 A: Observe that
$$
\sum_{k=0}^\infty\frac{1}{k+1}\frac{\lambda^ke^{-\lambda}}{k!}=\frac{1}{\lambda}\sum_{k=0}^\infty\frac{\lambda^{k+1}e^{-\lambda}}{(k+1)!}=\frac{1}{\lambda}\sum_{k={\color{red}1}}^\infty\frac{\lambda^ke^{-\lambda}}{k!}=\frac{1}{\lambda}\left(1-e^{-\lambda}\right).
$$
A: I wanted to point out a probabilistic interpretation and manipulation. 
$\xi$ is the number of events in a Poisson process of rate $\lambda$ in the time interval $[0,1]$. These events split the interval into $\xi+1$ subintervals which are identically distributed. The expected length of each subinterval is $E [ \frac{1}{1+\xi}]$. But this is also equal to the expectation of the length first subinterval: the expectation of the  minimum between $T$, the time of the first event of the  Poisson process,  and the time $1$. Recall that  $T$ is exponential with parameter $\lambda$. Summarizing:  
$$E[ \frac{1}{1+\zeta} ] = E[T\wedge 1].$$ 
To compute the RHS, observe that 
$$ E [ T] = E [ T \wedge 1] + E [ (T-1), T>1] = E [ T\wedge 1]+ E[ T-1 | T>1] P(T>1)$$ 
By the memoryless property of exponential random variables,  $T-1$ conditioned on $T>1$ is also exponential $\lambda$, so we obtain
$$ E[T] = E[T\wedge 1] + E[T]P(T>1) \Rightarrow E[T\wedge 1] = E[T] (1-P(T>t) )=\frac{1}{\lambda} (1-e^{-\lambda}).$$ 
A: By the law of the unconscious statistician,
$$
\mathbb{E}\left[\frac{1}{1+\xi}\right]
 = \sum_{k=0}^\infty \frac{p_k}{1+k}
 = e^{-\lambda} \sum_{k=0}^\infty \frac{\lambda^k}{k!(1+k)}
 = \frac{1}{\lambda e^\lambda}
   \sum_{k=0}^\infty \frac{\lambda^{k+1}}{(1+k)!}
 = \frac{e^\lambda - 1}{\lambda e^\lambda}
$$
A: Right, and:
$$ \sum_{k\geq 0}\frac{\lambda^k e^{-\lambda}}{(k+1)k!} = \frac{e^{-\lambda}}{\lambda}\sum_{k\geq 0}\frac{\lambda^{k+1}}{(k+1)!}=\frac{e^{-\lambda}}{\lambda}\left(e^{\lambda}-1\right)=\color{red}{\frac{1-e^{-\lambda}}{\lambda}}.$$
A: To start where you left off,
$$\sum_{k=0}^{\infty}\frac{1}{k+1}\frac{\lambda^ke^{-\lambda}}{k!}$$
$$=\sum_{k=0}^{\infty}\frac{\lambda^ke^{-\lambda}}{(k+1)!}$$
$$=\frac{e}{\lambda}\sum_{k=0}^{\infty}\frac{\lambda^{k+1}e^{-(\lambda+1)}}{(k+1)!}$$
$$=\frac{e}{\lambda}(\sum_{k=0}^{\infty}\frac{\lambda^{k}e^{-\lambda}}{k!}-1)$$
Can you take it from there?
