TITLE: Evaluating $\lim\limits_{n\to\infty} e^{-n} \sum\limits_{k=0}^{n} \frac{n^k}{k!}$
QUESTION [251 upvotes]: I'm supposed to calculate:
$$\lim_{n\to\infty} e^{-n} \sum_{k=0}^{n} \frac{n^k}{k!}$$
By using WolframAlpha, I might guess that the limit is $\frac{1}{2}$, which is a pretty interesting and nice result. I wonder in which ways we may approach it.

REPLY [7 votes]: I thought that it might be of instructive to post a solution to a generalization of the OP's question.  Namely, evaluate the limit
$$\lim_{n\to\infty}e^{-n}\sum_{k=0}^{N(n)}\frac{n^k}{k!}$$
where $N(n)=\lfloor Cn\rfloor$, where $C>0$ is an arbitrary constant.  To that end we now proceed.

Let $N(n)=\lfloor Cn\rfloor$, where $C>0$ is an arbitrary constant.  We denote $S(n)$ the sum of interest
$$S(n)=e^{-n}\sum_{k=0}^{N}\frac{n^k}{k!}$$
Applying the analogous methodology presented by @SangchulLee, it is straightforward to show that
$$S(n)=1-\frac{(N/e)^{N}\sqrt{N}}{N!}\int_{(N-n)/\sqrt{N}}^{\sqrt{N}}e^{\sqrt{N}x}\left(1-\frac{x}{\sqrt N}\right)^N\,dx\tag7$$
We note that the integrand is positive and bounded above by $e^{-x^2/2}$.  Therefore, we can apply the Dominated Convergence Theorem along with Stirling's Formula to evaluate the limit as $n\to\infty$.
There are three cases to examine.
Case $1$: $C>1$
If $C>1$, then both the lower and upper limits of integration on the integral in $(7)$ approach $\infty$ as $n\to \infty$.  Therefore, we find
$$\lim_{n\to \infty}e^{-n}\sum_{k=0}^{\lfloor Cn\rfloor}\frac{n^k}{k!}=1$$
Case $2$: $C=1$
If $C=1$, then the lower limit is $0$ while the upper limit approaches $\infty$ and we find
$$\lim_{n\to \infty}e^{-n}\sum_{k=0}^{n}\frac{n^k}{k!}=\frac12$$
Case $3$: $C<1$
If $C<1$, then the lower limit is approaches $-\infty$ while the upper limit approaches $\infty$ and we find
$$\lim_{n\to \infty}e^{-n}\sum_{k=0}^{n}\frac{n^k}{k!}=0$$

To summarize we have found that
$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}e^{-n}\sum_{k=0}^{\lfloor Cn\rfloor}\frac{n^k}{k!}=\begin{cases}1&,C>1\\\\\frac12&, C=1\\\\0&, C<1\end{cases}}$$