Asymptotic for binomial coefficient with square root I'm looking for asymptotic estimate for the binomial coefficient:
$$
\ln{\binom{n}{[\sqrt{n}]}}
$$
I assume Stirling's approximation can help, but I'm not sure I will get any good estimation with this approach. Is there any good way to make an estimation for this coefficient? Thanks in advance.

I shall derive a simple asymptotic formula for
$$\binom{n}{\sqrt{n}+\xi}$$
where $\xi$ is bounded and $n\to+\infty$. It follows from the main result of this paper that
$$
\log ((n + a)!) = \left( {n + a + \frac{1}{2}} \right)\log n - n + \frac{1}{2}\log (2\pi ) + \frac{{6a^2  + 6a + 1}}{{12n}} + \mathcal{O}\!\left( {\frac{{\max (\left| a \right|^3 ,1)}}{{n^2 }}} \right)
$$
provided $n + a+1 \ge 0$ and $\left| a+1 \right| < \frac{3}{5}n$, where the implied constant does not depend on $n$ or $a$. Then
$$
\log (n!) = \left( {n + \frac{1}{2}} \right)\log n - n + \frac{1}{2}\log (2\pi ) + \mathcal{O}\!\left( {\frac{1}{n}} \right),
$$
$$
\log ((\sqrt n  + \xi )!) = \left( {\sqrt n  + \xi  + \frac{1}{2}} \right)\log \sqrt n  - \sqrt n  + \frac{1}{2}\log (2\pi ) + \mathcal{O}\!\left( {\frac{1}{{\sqrt n }}} \right)
$$
and
$$
\log ((n - \sqrt n  - \xi )!) = \left( {n - \sqrt n  - \xi  + \frac{1}{2}} \right)\log n - n + \frac{1}{2}\log (2\pi ) + \frac{1}{2} + \mathcal{O}\!\left( {\frac{1}{{\sqrt n }}} \right),
$$
as $n\to+\infty$, uniformly with respect to bounded values of $\xi$. Accordingly,
$$
\log \binom{n}{\sqrt{n}+\xi} = \left( {\sqrt n  + \xi  - \frac{1}{2}} \right)\log \sqrt n  + \sqrt n  - \frac{1}{2}\log (2\pi {\rm e}) + \mathcal{O}\!\left( {\frac{1}{{\sqrt n }}} \right)
$$
or
$$
\binom{n}{\sqrt{n}+\xi}= \frac{1}{{\sqrt {2\pi {\rm e}} }}\sqrt n ^{\sqrt n  + \xi  - \frac{1}{2}} {\rm e}^{\sqrt n } \left( {1 + \mathcal{O}\!\left( {\frac{1}{{\sqrt n }}} \right)} \right),
$$
as $n\to+\infty$, uniformly with respect to bounded values of $\xi$. In particular,
$$
\binom{n}{[\sqrt{n}]}= \frac{1}{{\sqrt {2\pi {\rm e}} }}\sqrt n ^{[\sqrt{n}]- \frac{1}{2}} {\rm e}^{\sqrt n } \left( {1 + \mathcal{O}\!\left( {\frac{1}{{\sqrt n }}} \right)} \right),
$$
as $n\to+\infty$.
