How can i compute this limit related with poisson distribution?

Question

let $\alpha \in(0,\infty)$

What is the limit of $\sum_{0}^{\alpha n}{(e^{-n}n^{k})\over k!}$ as n goes $\infty$?

it depends on $\alpha$ and looks like somewhat related to the possion distribution, but don't know how to compute.

I guess it is $e^{-{1\over \alpha}}$ but not sure. How can i compute this?

Answer 1

If $X_n$ has Poisson distribution wit parameter $n$ then a well known result says $\frac {X_n-n} {\sqrt n} \to Y$ where $Y$ has standard normal distribution. What you are seeking is the limit of $P\{X_n \leq \alpha n\}=P\{\frac {X_n-n} {\sqrt n} \leq \frac {(\alpha -1)n} {\sqrt n}\}$. The limit is 1/2 if $\alpha =1$, 1 if $\alpha >1$ and 0 if $\alpha <1$.