statistics - Show that $lim_{nrightarrow infty} sum_{i=0}^{n}frac{e^{-n}n^{i}}{i!}rightarrow frac{1}{2}$

$\lim_{n\rightarrow \infty} \sum_{i=0}^{n}\frac{e^{-n}n^{i}}{i!}\rightarrow \frac{1}{2}$

Tried:
here suppose N is poission distribution with parameter n
$\lim_{n\rightarrow \infty} \sum_{i=0}^{n} P(N\geq i) $
$= \lim_{n\rightarrow \infty} \sum_{i=0}^{n}(P(N=i)+P(N>i) + P(N
$= \lim_{n\rightarrow \infty} \sum_{i=0}^{n}(P(N= i)+P(i$= \lim_{n\rightarrow \infty} \sum_{i=0}^{n}(P(N= i)-P(N$=\sum_{i=0}^{\infty}(P(N= i)-\sum_{i=0}^{\infty}P(N$=1-\frac{1}{2}$ $=\frac{1}{2}$