Central limit theorem with Poisson(1)

Question

Let$\ X_1,X_2,...,X_n$ be independent Poisson random variables with parameter$\ λ=1$, use the Central Limit Theorem to prove:

$\ \lim_{n→∞} \frac{1}{e^n} \sum_{k=0}^n \frac{n^k}{k!} =\frac{1}{2}$

What I've done:

Since$\ E(X_n)=1$ and$\ V(X_n)=1$

$$\frac{(X_1+\cdots+X_n)-n}{\sqrt n} \overset{} \longrightarrow N(0,1) \text{ as } n\to\infty. \tag 1$$

And for any normally distributed $Z$ with mean $\mu$, $P(Z\leq \mu) = 1/2$

How can I get to the result I'm asked for?

Answer 1

You need to note that $S_n=X_1+\cdots+X_n$ is Poisson with parameter $\lambda=n$. Then $P(S_n\le n) = e^{-n}\sum_{k=0}^n n^k/k!$, and since $0$ is a continuity point of the distribution $N(0,1)$, you have $P(S_n\le n)=P((S_n-n)/\sqrt n \le 0) \to P(Z\le 0) = 1/2$.