TITLE: Why does $\lim\limits_{n\to\infty} e^{-n}\sum_{i=1}^{n}\frac{n^i}{i!} = \frac{1}{2}$ and not 1?
QUESTION [5 upvotes]: The limit
$$\lim_{n\to\infty} e^{-n}\sum_{i=1}^{n}\frac{n^i}{i!}$$
can be seen to be $\frac{1}{2}$, yet isn't the sum in this expression just going to be $\lim\limits_{n\to\infty}e^{n}$, making the limit 1?
I'm having trouble wrapping my head around why this isn't the case. Any help would be much appreciated.

REPLY [2 votes]: A probabilistic approach. Let $X_n\sim \text{Poi}(n)$ be a Poisson random variable with rate $\lambda=n$. By the central limit theorem, it follows that
$$
Z_n=\frac{X_n-n}{\sqrt{n}}\stackrel{d}{\to} Z\sim N(0,1).
$$ 
Your limit equals
$$
\lim_{n\to \infty} P(X_n\leq n)=\lim_{n\to \infty}P(Z_n\leq 0)=P(Z\leq 0)=1/2.
$$