probability theory - Show that $limlimits_{nrightarrowinfty} e^{-n}sumlimits_{k=0}^n frac{n^k}{k!}=frac{1}{2}$

Show that $\displaystyle\lim_{n\rightarrow\infty} e^{-n}\sum_{k=0}^n \frac{n^k}{k!}=\frac{1}{2}$ using the fact that if $X_j$ are independent and identically distributed as Poisson(1), and $S_n=\sum\limits_{j=1}^n X_j$ then $\displaystyle\frac{S_n-n}{\sqrt{n}}\rightarrow N(0,1)$ in distribution.

I know that $\displaystyle\frac{e^{-n} n^k}{k!}$ is the pdf of $K$ that is Poisson(n) and $k=\sum\limits_{j=1}^n X_j$ is distributed Poisson(n), but I don't know what I can do next.

Answer

Hint: $e^{-n} \sum_{k=0}^n \dfrac{n^k}{k!}$ is the probability of what?