<s>[INST] Order of a binomial coefficient 
I am trying to find the Big-O order of $\binom{n-1}{\frac{n}{2}-1}$ where $n$ is even.

I know that $\binom{n}{\frac{n}{2}}=O\left(\frac{2^n}{\sqrt{n}}\right)$ using Stirling's approximation. I know that $\binom{n-1}{\frac{n}{2}-1}$ is smaller than $\binom{n}{\frac{n}{2}}$, so I'm guessing that the order of $\binom{n-1}{\frac{n}{2}-1}$ is close to $O\left(\frac{2^n}{\sqrt{n}}\right)$ like maybe $O\left(\frac{2^n}{n}\right)$?
My attempt:
\begin{align}
\binom{n-1}{n/2-1} &= \frac{(n-1)!}{(n/2)! (n/2-1)!} \\
&= \frac{(n-1)^{n-1}e^{-(n-1)}\sqrt{n-1}}{(n/2)^{n/2}e^{-n/2}\sqrt{n}\times \left(n/2-1\right)^{n/2-1}e^{-(n/2-1)}\sqrt{n/2-1}} \text{ By Stirling's } \\
&\approx\frac{(n-1)^{n-1}\sqrt{n-1}}{(n/2)^{n/2}\left(n/2-1\right)^{n/2-1}\sqrt{n(n-2)}} \\
&\approx\frac{(n-1)^{n-1}\sqrt{n-1}\times2^n}{n^{n/2}\left(n-2\right)^{n/2-1}\sqrt{n(n-2)}} \\
&\approx\frac{(n-1)^{n-1}\sqrt{n-1}\times (n-2)}{\left(n(n-2)\right)^{n/2}\sqrt{n(n-2)}}\times2^n \\
\end{align}
So I am not able to find the order the first term above. Any ideas?
 [/INST] Write this as
$$
2^n\times \frac{\sqrt{n-1}}{\sqrt{n}\cdot \sqrt{n-2}}\times \frac{(n-1)^{n-1}(n-2)}{(n(n-2))^{n/2}}\\
\\\approx 2^n\times \color{#2B2}{\frac1{\sqrt{n}}}\times { (1-\frac1n)^{n-1}(1-\frac2n)\over (1-\frac1{n/2})^{n/2}}
$$
and use the limit definition of $e$.
</s>