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.
Find asymptotically equivalent function for $\ln \binom{n^2}{2n}$
asymptotics
Best Answer
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 $n$ $$\left( \begin{array}{ccc} n & \text{approximation} & \text{exact} \\ 3 & 4.49297 & 4.43082 \\ 4 & 9.49277 & 9.46265 \\ 5 & 15.0177 & 14.9999 \\ 6 & 20.9595 & 20.9478 \\ 7 & 27.2466 & 27.2383 \\ 8 & 33.8286 & 33.8224 \\ 9 & 40.6676 & 40.6628 \\ 10 & 47.7345 & 47.7306 \end{array} \right)$$