Does $\sqrt{2\pi n} P(S_n =k)\to \exp(-x^2/2)$?

Question

This is exercise 3.1.2 from the 4th edition of Prof. Durrett's probability:

Suppose $X_i$ have a Poisson distribution with mean $1$, $S_n=\sum\limits_{i=1}^n X_i$, and $(k-n)/\sqrt{n}\to x$, then prove $$\sqrt{2\pi n} P(S_n =k)\to \exp(-x^2/2)\,.$$

Here's my attempt:

\begin{align} \sqrt{2\pi n} P(S_n =k)&=\sqrt{2\pi n} e^{-n}n^{k}/k! \\&\sim \sqrt{2\pi n}e^{-n}n^{k}/ \sqrt{2\pi k}e^{-k}k^{k} \\&=e^{k-n}(n/k)^{1/2+k} \\&=e^{k-n}(1+(k-n)/n)^{-1/2-k} \\&\to e^{k-n}(1+x/\sqrt{n})^{-1/2-k}\to e^{-x^2} \end{align}

My last step is problematic. How should I continue from the second to last step?

Answer 1

Instead write $(n/k)^{1/2+k}=(1+(n-k)/k)^{1/2+k}=(1-x\sqrt{n}/k)^{1/2+k}$. The Stirling-approximated log-probability is$$\begin{align}k-n+(1/2+k)\ln(1-x\sqrt{n}/k)&=x\sqrt{n}+(1/2+k)\ln(1-x\sqrt{n}/k)\\&=x\sqrt{n}+(1/2+k)(-x\sqrt{n}/k-x^2n/(2k^2)+o(x^2))\\&=-x\sqrt{n}/(2k)-x^2n/(2k)-x^2n/(4k^2).\end{align}$$With $k\approx n$, the $x$ coefficient $\to0$ as $n\to\infty$, while the $x^2$ coefficient $\to-1/2$.