An approximation by Stirling's formula Let $0<\alpha <1$ be a real number and $\alpha n$ be an integer. I want to prove the following fact
$${n\choose \alpha n}=\frac{1+o(1)}{\sqrt{2\pi \alpha (1-\alpha)n}}2^{nH(\alpha)},$$ where $H(\alpha)=-\alpha \log_2 \alpha -(1-\alpha)\log_2 (1-\alpha)$. To do this, I've used the Stirling's formula for the factorial: $$n!=(\frac{n}{e})^n \sqrt{2\pi n}e^{r(n)},$$ where $1/(12n+1)<r(n)<1/12n$.
What I've tried:
$${n\choose \alpha n}=\frac{n!}{(\alpha n)! ((1-\alpha)n)!}=\frac{(\frac{n}{e})^n \sqrt{2\pi n}e^{r(n)}}{(\frac{\alpha n}{e})^{\alpha n} \sqrt{2\pi \alpha n}e^{r(\alpha n)}(\frac{(1-\alpha)n}{e})^{(1-\alpha)n} \sqrt{2\pi (1-\alpha)n}e^{r((1-\alpha)n)}}\\ \frac{e^{r(n)-r(\alpha n)-r((1-\alpha)n)}}{(\alpha^{\alpha}(1-\alpha)^{(1-\alpha)})^n\sqrt{2\pi \alpha(1-\alpha)n}}=\frac{e^{r(n)-r(\alpha n)-r((1-\alpha)n)}}{\sqrt{2\pi \alpha (1-\alpha)n}}2^{nH(\alpha)}.$$
Now how can I show that $e^{r(n)-r(\alpha n)-r((1-\alpha)n)}=1+o(1)$?
Thanks in advance.
 A: Since $f(n)=o(1)$ just means $\frac{f(n)}{1}\rightarrow 0$ as $n\rightarrow\infty$ (i.e. $f(n)=1+o(1)$ means $f(n)\rightarrow 1$), our goal is to show that $e^{r(n)-r(\alpha n)-r((1-\alpha)n)}\rightarrow 1$, which can be done by showing $r(n)-r(\alpha n)-r((1-\alpha)n)\rightarrow 0$. We can use the bounds on $r(n)$ to show that
$$r(n)-r(\alpha n)-r((1-\alpha)n)<\frac{1}{12n}-\frac{1}{12\alpha n+1}-\frac{1}{12(1-\alpha)n+1}\underset{n\rightarrow\infty}{\rightarrow}0$$
and
$$r(n)-r(\alpha n)-r((1-\alpha)n)>\frac{1}{12n+1}-\frac{1}{12\alpha n}-\frac{1}{12(1-\alpha)n}\underset{n\rightarrow\infty}{\rightarrow}0$$
So since $r(n)-r(\alpha n)-r((1-\alpha)n)$ is squeezed between $0$ on both sides asymptotically, indeed we have $r(n)-r(\alpha n)-r((1-\alpha)n)\rightarrow 0$, and we are done.
A: You are only having difficulty because the version of Stirling you're using is too precise. Instead, use $n! = (1+o(1)) \sqrt{2\pi n} (\frac ne)^n$ - replacing all of your $e^{r(n)}$ factors by $1+o(1)$.
The step you're stuck on, where you're trying to show that $e^{r(n)-r(\alpha n)-r((1-\alpha)n)}=1+o(1)$, becomes $\frac{1+o(1)}{(1+o(1))(1+o(1))} = 1+o(1)$, which is straightforward.
(Technically, the $1+o(1)$ factor we get from Stirling's approximation is $1+o_{n\to \infty}(1)$, so we also have to observe that $1+o_{\alpha n \to \infty}(1)$ and $1 + o_{(1-\alpha)n \to \infty}(1)$ are the same asymptotically. This is fine as long as $\alpha$ is a constant, or even more generally than that.)
