I'm interesting in knowing whether there is a closed-form expression for the following series:
$\displaystyle\sum_{n\geq1}\frac{1}{n!}\lambda^n e^{-\lambda} \left[{n \choose k}z^k(1-z)^{n-k} \right] $
I've made some progress simplifying…
$ = \left( \frac{z}{1-z} \right)^ke^{-\lambda}\displaystyle\sum_{n\geq1} \frac{1}{k!(n-k)!}\left[ \lambda(1-z) \right]^n$
…but I'm not sure where to go from here.
Best Answer
I am not sure that you have already had the prove. I start from:
$\left( \frac{z}{1-z} \right)^k \frac{e^{-\lambda}}{k!}\displaystyle\sum_{n\geq1} \frac{1}{(n-k)!}\left[ \lambda(1-z) \right]^n=e^{-\lambda} \frac{{z}^k}{k!}\displaystyle\sum_{n\geq1} \frac{\left[ \lambda(1-z) \right]^{n-k}}{(n-k)!}\tag1$
Reindexing the series, introducing $m=n-k$
$e^{-\lambda} \frac{{z}^k}{k!}\sum\limits_{m=n-k}^\infty \frac{\left[ \lambda(1-z) \right]^{m}}{m!}\tag2$
As n start from 1 to $\infty$, m will start from
$1-k, 2-k, .....-2,-1, 0, 1, 2, .....\infty\tag3$
It means that we can devide the sum (2) into two parts:
$e^{-\lambda} \frac{{z}^k}{k!}\sum\limits_{m=1}^{k-1}\frac{\left[ \lambda(1-z) \right]^{-m}}{(-m)!}+e^{-\lambda} \frac{{z}^k}{k!}\sum\limits_{m=0}^\infty \frac{\left[ \lambda(1-z) \right]^{m}}{m!}\tag4$
The first sum of (4) equal to zero, because $ (-m)!\rightarrow \infty$ because m integer.
Finally we get:
$e^{-\lambda} \dfrac{{z}^k}{k!}e^{\left[ \lambda(1-z) \right]}= \dfrac{{z}^k}{k!}e^{-\lambda z}\tag5$