Urns, balls, and simplifying an infinite series Consider a large number of urns and a large number of people each randomly placing $n \sim Poisson(\lambda)$ balls in urns (people do not place multiple balls in the same urn). Urns receiving multiple balls pick a ball at random. If an individual's balls are chosen by multiple urns, then the individual randomly chooses an urn and we call that a match. Letting z denote the probability that person i is chosen by urn j, conditional on placing a ball in j, I conjecture that the probability that an urn that has chosen a ball forms a match with that ball is given by:
$$\displaystyle\sum_{n\geq 1} \left( \frac{\lambda^ne^{-\lambda}}{n!} \right)\left[\sum_{k=0}^{n-1} {n-1\choose k} z^k(1-z)^{n-1-k}\left(\frac{1}{k+1} \right) \right]$$
Simplifying the expression inside of the brackets yields
$$ \frac{(1-z)^{n-1}}{n} \sum_{k=0}^{n-1}
 {n \choose k+1}\left( \frac{z}{1-z} \right)^k $$
This is close to a familiar looking series for the inner brackets, but I'm not sure how to proceed. Any suggestions or references?
 A: Simplifying the expression inside of the brackets:
$\left[\sum\limits_{k=0}^{n-1} {n-1\choose k} z^k(1-z)^{n-1-k}\left(\frac{1}{k+1} \right) \right]\tag1$
Using the fact that $\frac{1}{k+1}=\int\limits_0^1 t^kdt$ and replace the order the sum and integral we get:
$\int\limits_0^1\sum\limits_{k=0}^{n-1} {n-1\choose k} (zt)^k(1-z)^{n-1-k}dt\tag2$
According to Binomial formula we can write: 
$\int\limits_0^1 (zt+1-z)^{n-1}dt=[\frac{(zt+1-z)^n}{zn}]_{t=0}^1=\frac{1}{zn}-\frac{(1-z)^n}{zn}\tag3$
So the result of the sum inside the brakets is: 
$\frac{1}{zn}-\frac{(1-z)^n}{zn}\tag4$
Regarding the whole expression: 
$\sum\limits_{n=1}^\infty \frac{\lambda^ne^{-\lambda}}{n!}\big(\frac{1}{zn}-\frac{(1-z)^n}{zn}\big)\tag5$
Using the same trick as in (2) we have: 
$\frac{e^{-\lambda}}{z}\int\limits_0^1 \frac{1}{t}\sum\limits_{n=1}^\infty \frac{\lambda^n}{n!}\big(t^n-t^n(1-z)^n\big)\tag6$
Known that $\sum\limits_{n=1}^\infty \frac{x^n}{n!}=\sum\limits_{n=0}^\infty \frac{x^n}{n!}-1;$ we can form the(5) in the following way:
$\frac{e^{-\lambda}}{z}\int\limits_0^1 \big(\frac{e^{\lambda t}}{t}-\frac{e^{\lambda t(1-z)}}{t}\big)dt \tag7$
This integral can be solved by using the integral functions.
