Find the limit of $\lim_{n\to\infty} e^{-n} \sum_{k=0}^{n+\sqrt{n}} \frac{n^k}{k!}$

Question

This question has been asked so many times but for $k$ from $0$ to $n$, what would be different if $k$ goes to $n+\sqrt{n}$ instead. Any hints?

$$\lim_{n\to\infty} e^{-n} \sum_{k=0}^{n+\sqrt{n}} \frac{n^k}{k!}$$

Answer 1

Observe that your sum can be written in terms of the Poisson distribution as follows. Let $X_n$ be a Poisson random variable with mean $n$. Then $$ e^{-n}\sum_{k=0}^{n+\sqrt n}\frac{n^k}{k!}=\mathbb P(X_n\leq n+\sqrt n)=\mathbb P\left(\frac{X_n-\mathbb EX_n}{\sqrt{\mathbb EX_n^2}}\leq 1\right). $$ Since $X_n$ can be written as a sum of $n$ independent and identically distributed random variables (each of which is Poisson with mean $1$) the central limit theorem applies, and therefore $$ \frac{X_n-\mathbb EX_n}{\sqrt{\mathbb EX_n^2}} $$ converges in distribution to a standard normal random variable as $n\to\infty$. (See here for an explanation.) Thus your limit equals $$ \lim_{n\to\infty}\mathbb P\left(\frac{X_n-\mathbb EX_n}{\sqrt{\mathbb EX_n^2}}\leq 1\right)=\mathbb P(Z\leq 1)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^1e^{-x^2/2}\ dx, $$ where $Z$ is standard normal.