Find asymptotically equivalent function for $\ln \binom{n^2}{2n}$ How can I find an asymptotically equivalent function for $\ln \binom{n^2}{2n}$?
I think I need to use the definition of the binomial coefficient and Stirling's approximation, but I'm not sure what to do after this.
 A: By Stirling approximation $n! \sim \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n$ we have that
$$ \binom{n^2}{2n}=\frac{n^2!}{(2n)!(n^2-2n)!}\sim \frac{\sqrt{2 \pi n^2}\left(\frac{n^2}{e}\right)^{n^2}}{\sqrt{2 \pi 2n}\left(\frac{2n}{e}\right)^{2n}\sqrt{2 \pi (n^2-2n)}\left(\frac{n^2-2n}{e}\right)^{n^2-2n}}$$
$$\sim\frac{\sqrt{2 \pi }n^{2n^2+1}e^{-n^2}}{2\sqrt \pi n^\frac12 2^{2n}n^{2n}e^{-2n}\cdot \sqrt{2 \pi}nn^{2n^2-4n}e^{-n^2+2n}}=$$
$$=\frac1{2\sqrt \pi} \cdot {n^{2n-\frac12}} \cdot 2^{-2n} $$
and therefore
$$\ln \binom{n^2}{2n}\sim n\ln n$$
A: In the same spirit an in answer and comment, we can even get a very good approximation.
Writing
$$\binom{n^2}{2 n}=\frac{(n^2)!}{(2n)! \, (n^2-2n)!}$$
$$\log \left(\binom{n^2}{2 n}\right)=\log\left((n^2)!\right)-\log\left((2n)!\right)-\log\left((n^2-2n)!\right)$$
Using Stirling approximation 
$$\log(p!)=p (\log (p)-1)+\frac{1}{2} \left(\log (2 \pi )+\log
   \left({p}\right)\right)+\frac{1}{12 p}-\frac{1}{360
   p^3}+O\left(\frac{1}{p^5}\right)$$ apply to each term and continue with Taylor series for large values of $n$. This should give
$$\log \left(\binom{n^2}{2 n}\right)=2 n \left(\log (n)+\log \left(\frac{e}{2}\right)\right)-\log \left(2 e^2 \sqrt{\pi }\right)-\log{(\sqrt n)}-\frac 3 {8n}+O\left(\frac{1}{n^2}\right)$$
As shown below, this is even quite good for very small values of $n$
$$\left(
\begin{array}{ccc}
n & \text{approximation} & \text{exact} \\
 3 & 4.49297 & 4.43082 \\
 4 & 9.49277 & 9.46265 \\
 5 & 15.0177 & 14.9999 \\
 6 & 20.9595 & 20.9478 \\
 7 & 27.2466 & 27.2383 \\
 8 & 33.8286 & 33.8224 \\
 9 & 40.6676 & 40.6628 \\
 10 & 47.7345 & 47.7306
\end{array}
\right)$$
