Show that the error terms in the asymptotic expansion of a binomial coefficient are uniformly bounded

Question

I'm interested in the asymptotic behavior of $$ a_{n,k} :=\frac{1}{2^n} \binom{n}{\frac{n}{2} +k},$$ where $k$ is within a constant times $\sqrt{n}$, say $$ |k| < C \sqrt{n }.$$

Using Stirling's approximation, I found $$ a_{n,k} = \sqrt{\frac{2}{\pi n}} e^{-\frac{2k^2}{n}}\left(1 + O\left( \frac{1}{\sqrt{n}} \right)\right).$$

In each $a_{n,k}$, let us call the relative error term $b_{n,k}$, so $$ a_{n,k} = \sqrt{\frac{2}{\pi n}} e^{-\frac{2k^2}{n}}\left(1 + b_{n,k}\right).$$

My Question: How do I show that the $\{b_{n,k}\}$ are all bounded by the same constant divided by $\sqrt{n}$.

Clarification: I'd like to claim: there's a constant $K>0$, and an index $n_0\in \mathbb{N}$, such that for all $n>n_0$, and all $k$ satisfying $|k|< C\sqrt{n}$, $$ |b_{n,k}| < \frac{K}{\sqrt{n}} .$$

Answer 1

The desired uniform bound follows from essentially the same computation that you might have used to obtain the formula for $a_{n,k}$.

Indeed, the Stirling's approximation tells that

\begin{align} n! = \sqrt{2\pi} \, n^{n+\frac{1}{2}} e^{-n} \left( 1 + \mathcal{O}\left(n^{-1}\right) \right), \end{align}

where the implicit bound is absolute. Next, fix $C > 0$. Then uniformly in $|k| \leq C\sqrt{n}$, we have

\begin{align} \left(\tfrac{n}{2}\pm k\right)!\ &= \sqrt{2\pi} \, \left(\tfrac{n}{2}\pm k\right)^{\frac{n}{2}\pm k+\frac{1}{2}} e^{-(\frac{n}{2}\pm k)} \left( 1 + \mathcal{O}\left((\tfrac{n}{2}\pm k\right)^{-1}) \right) \\ &= \sqrt{2\pi} \, \left(\tfrac{n}{2}\pm k\right)^{\frac{n}{2}\pm k+\frac{1}{2}} e^{-(\frac{n}{2}\pm k)} \left( 1 + \mathcal{O}_C\left(n^{-1}\right) \right), \end{align}

where the implicit bound for $\mathcal{O}_C\left(n^{-1}\right)$ depends only on $C$. Plugging this to $a_{n,k}$,

$$ a_{n,k} = \sqrt{\frac{2}{\pi n}} \left(1+\frac{2k}{n}\right)^{-(\frac{n}{2} + k+\frac{1}{2})} \left(1 - \frac{2k}{n}\right)^{-(\frac{n}{2}- k+\frac{1}{2})} \left( 1 + \mathcal{O}_C(n^{-1}) \right). $$

Now, by the Taylor's theorem, we have

$$ \log\left(1 \pm \frac{2k}{n} \right) = \pm\frac{2k}{n} - \frac{2k^2}{n^2} + \mathcal{O}_C(n^{-3/2}). $$

Plugging this back shows that

$$ a_{n,k} = \sqrt{\frac{2}{\pi n}} e^{-\frac{2k^2}{n}} \left( 1 + \mathcal{O}_C(n^{-1/2}) \right) $$

as desired.