Did I use Stirling formula incorrectly?

Question

I have to prove that $\frac{\sqrt{2\pi n} * e^{-n}*n^k}{k!} \to e^{-x^2/2}$ when $\frac{k-n}{\sqrt{n}} \to x$ and $n \to \infty$

I used Stirling formula and get that $\frac{\sqrt{2\pi n} * e^{-n}*n^k}{k!} = e^{k-n} * (n/k)^k \frac{\sqrt{2\pi n}}{\sqrt{2\pi k}}$

From $\frac{k-n}{\sqrt{n}} \to x$ we get that $n/k \to 1$ than $\frac{\sqrt{2\pi n}}{\sqrt{2\pi k}} \to 1$

So we get $e^{k-n} * (n/k)^k$

But $(n/k)^k = (1 + (n-k)/k)^{k/(n-k) * (n-k)} = e^{n-k}$ because $(n-k) / k \to 0$

So everything tends to $e^{k-n} * e^{n-k} = 1$

What do I do wrong?(

Answer 1

This is a mistake: $$\left(1+\frac{n-k}{k}\right)^{\frac{k}{n-k}\cdot(n-k)}\sim e^{n-k}$$Is not true, because the $(n-k)$ exponent complicates things: because it is growing while $(n-k)/k$ is decreasing, and the relative rates of growth do matter. To explain what I mean, if $a$ is constant then it is true that: $$\left(1+\frac{n-k}{k}\right)^{\frac{k}{n-k}\cdot a}\sim e^a$$But here, $a=n-k$ is not constant, and it is a priori unclear what the asymptotic behaviour is.

Let's be more careful:

$$\begin{align}\left(1+\frac{n-k}{k}\right)^k&=\exp(k\log(1+(n-k)/k))\\&=\exp(k((n-k)/k-(n-k)^2/2k^2+\psi((n-k)/k)))\\&=\exp(n-k-(n-k)^2/2k+\psi((n-k)/k))\cdot k)\end{align}$$Where $\psi(y)\in\Theta(y^3)$ as $y\to0$. Since $n/k\to1$, $(n-k)/k\to0$ as $n,k\to\infty$ and $k\cdot((n-k)/k)^{3}=(n-k)^3/k^2\sim-x^3n\sqrt{n}/k^2\sim-x^3n^{-1/2}\to0$ thus $\psi((n-k)/k)\cdot k$ vanishes as $n,k\to\infty$.

So: $$e^{k-n}(n/k)^k=\exp\left(-\frac{1}{2}\left(\frac{k-n}{\sqrt{n}}\right)^2\cdot\frac{n}{k}+o(1)\right)\to e^{-x^2/2}$$

Therefore, combining with your other correct work, we have the result.