Find the limit of the infinite series $\lim_{n\rightarrow \infty} e^{-n}\displaystyle\sum_{k=0}^{n-1}\frac{n^{k-1}}{k!}$. what is the value of it ?
This is a problem of a competitive examination. I try it, butI get does not exists, is it right? otherwise please solve it.

Considering the other answers are much more complicated, I might have an error here. Feel free to point it out. :)
For $n > 0$:
\begin{align}
a_n 
&= e^{-n} \sum_{k=0}^{n-1} \frac{n^{k-1}}{k!} \\
&< e^{-n} \frac{1}{n} \sum_{k=0}^\infty \frac{n^k}{k!}
= e^{-n} \frac{1}{n} e^n = \frac{1}{n}
\end{align}
So this vanishes in the limit $n\to \infty$.
