How to prove that: $\lim_{n \rightarrow\infty} e^{-n}\sum_{k=0}^{n}\frac{n^k}{k!} = \frac{1}{2}$
It's about the Normal Approximation to Poisson distribution. But it only calculates part of the Poisson distribution (only under expectation).
When we try to prove Normal distribution is an approximation of Poisson we use Characteristic Function. But I cannot use it here (because it's only part of a distribution).
Let $N_n \sim \operatorname{Poisson}(n)$ and write $N_n = n + \sqrt{n}Z_n$. We know from CLT that $Z_n \Rightarrow \mathcal{N}(0,1)$. Then
$$ \sum_{k=0}^{n} \frac{n^k}{k!}e^{-n} = \mathbb{P}[ N_n \leq n] = \mathbb{P}[Z_n \leq 0]. $$
Since $0$ is the continuity point of the CDF $\Phi$ of the standard normal distribution $\mathcal{N}(0, 1)$, it follows that
$$\mathbb{P}[Z_n \leq 0] \xrightarrow[n\to\infty]{} \Phi(0) = \frac{1}{2}. $$