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.
 A: 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 $n\to \infty$.
A: $$\lim_{n\rightarrow \infty} e^{-n}\displaystyle\sum_{k=0}^{n-1}\frac{n^{k-1}}{k!}=\lim_{n\rightarrow \infty} \frac{e^{-n}}n\displaystyle\sum_{k=0}^{n-1}\frac{n^{k}}{k!}$$

Lemma : $\lim_{n\rightarrow \infty}
 e^{-n}\displaystyle\sum_{k=0}^{n}\frac{n^{k}}{k!}=\frac12$
Proof : Let $(X_i)$ a sequence of random variables independant of same distribution $P(\lambda)$ (Poisson of parameter $\lambda$). Then
   $X_1+\dots+X_n$ is of distribution $P(n\lambda)$, thus
   $E(X_1+\dots+X_n)=n\lambda$ and $Var(X_1+\dots+X_n)=n\lambda$. We take
  $\lambda=1$, so with central limit theorem :
   $$P\left\{\frac{X_1+\dots+X_n}{\sqrt n}\leq
 0\right\}\rightarrow_{n\to\infty}\frac1{2\pi}\int_{-\infty}^0e^{-t^2}dt=\frac12$$
   But : $$P\left\{\frac{X_1+\dots+X_n}{\sqrt n}\leq
 0\right\}=P\left\{X_1+\dots+X_n\leq
 n\right\}=e^{-n}\sum_{k=0}^{n}\frac{n^{k}}{k!}\rightarrow_{n\to\infty}\frac12$$

Finally :
$$\lim_{n\rightarrow \infty} e^{-n}\displaystyle\sum_{k=0}^{n-1}\frac{n^{k-1}}{k!}=\lim_{n\rightarrow \infty} \frac{e^{-n}}n\displaystyle\sum_{k=0}^{n-1}\frac{n^{k}}{k!}=\frac12\lim_{n\rightarrow \infty} \frac1n=0$$
