$\lim_{n \rightarrow \infty} \sum_{k = 0}^n \frac{e^{-n} n^k}{k!}$ =? By viewing it as the sum of Poisson random variables and using Central Limit Theorem, we might transform the formula to be the probability of a standardized gaussian random variable less or equal to 0. However, by taking the limit, the probability should be 0.5 or 0?
 A: You are asking the chance a Poisson random variable is less than or equal to the mean of the distribution.  As the Poisson parameter gets large, the distribution approaches a binomial with mean the parameter and standard deviation the square root of the parameter.  You are then asking the chance a binomial variable is below the mean, which is $\frac 12$.  You are counting the bin of $n$, while for the binomial approximation you should not, so for small $n$ the sum is somewhat greater then $\frac 12$.  I ran the numbers up to $n=20$ and eliminating the $n$ bin makes the sum somewhat too small.
A: Let $ n $ be a positive integer. First of all, we have : $$ \sum_{k=0}^{n}{\frac{n^{k}}{k!}}=\mathrm{e}^{n}-\frac{1}{n!}\int_{0}^{n}{\left(n-t\right)^{n}\mathrm{e}^{t}\,\mathrm{d}t} $$
Thus $$ \mathrm{e}^{-n}\sum_{k=0}^{n}{\frac{n^{k}}{k!}}=1-\frac{\mathrm{e}^{-n}}{n!}\int_{0}^{n}{\left(n-t\right)^{n}\mathrm{e}^{t}\,\mathrm{d}t} $$
With the substitution $ t=\sqrt{n}u $, we get the following : $$ \mathrm{e}^{-n}\sum_{k=0}^{n}{\frac{n^{k}}{k!}}=1-\frac{n^{n}\,\mathrm{e}^{-n}\sqrt{n}}{n!}\int_{0}^{\sqrt{n}}{\left(1-\frac{u}{\sqrt{n}}\right)^{n}\mathrm{e}^{\sqrt{n}u}\,\mathrm{d}u} $$
Using Stirling's formula, we have that $ \frac{n^{n}\,\mathrm{e}^{-n}\sqrt{n}}{n!}\underset{n\to +\infty}{\sim}\frac{1}{\sqrt{2\pi}} $, and notice that $ \lim\limits_{n\to +\infty}{\left(1-\frac{u}{\sqrt{n}}\right)^{n}\mathrm{e}^{\sqrt{n}u}}=\mathrm{e}^{-\frac{u^{2}}{2}} $, for any positive real $ u \cdot $ So by applying the dominated convergence theorem on the function $ f_{n} $ defined on $ \mathbb{R}_{+} $ as follows : $$ \left(\forall u\in\mathbb{R}_{+}\right),\ f_{n}\left(u\right)=\left\lbrace\begin{aligned}\left(1-\frac{u}{\sqrt{n}}\right)^{n}\mathrm{e}^{\sqrt{n}u}\ \ \ \ \ \ \ & \textrm{If }0\leq u\leq\sqrt{n}\\ 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ &\textrm{If }\ \ \ \ u\geq \sqrt{n}\end{aligned}\right. $$, we get that : $$ \lim_{n\to +\infty}{\int_{0}^{\sqrt{n}}{\left(1-\frac{u}{\sqrt{n}}\right)^{n}\mathrm{e}^{\sqrt{n}u}\,\mathrm{d}u}}=\lim_{n\to +\infty}{\int_{0}^{+\infty}{f_{n}\left(u\right)\mathrm{d}u}}=\int_{0}^{+\infty}{\lim_{n\to +\infty}{f_{n}\left(u\right)}\,\mathrm{d}u}=\int_{0}^{+\infty}{\mathrm{e}^{-\frac{u^{2}}{2}}\,\mathrm{d}u} $$ And thus : $$ \lim_{n\to +\infty}{\mathrm{e}^{-n}\sum_{k=0}^{n}{\frac{n^{k}}{k!}}}=1-\frac{1}{\sqrt{2\pi}}\int_{0}^{+\infty}{\mathrm{e}^{-\frac{u^{2}}{2}}\,\mathrm{d}u}=\frac{1}{2} $$
