Approximate asymptotic behavior

Question

Let be $n$ and $k, (k\le \frac{n}{2}) $ positive integers. I am trying to show that the expression: $$ p(k) = \frac{n!}{2^n(\frac{n}{2} + k)! \hspace{0.1cm} (\frac{n}{2} - k)!} $$ reduces to: $$ p(0) \hspace{0.1cm} e^{-\frac{2 \hspace{0.1cm} k^2}{n}} $$ when $n$ tends to infinity. I lost myself trying to use Stirling approximation... have you any idea about the direction I should go?

Answer 1

If you get lost in Stirling's approximation, it can be helpful to instead use the logarithmic form $$ \ln n! \approx n\ln n - n+\frac{\ln n}{2} +\frac{\ln 2\pi}{2} + O(n^{-1}), $$ then look at each of the three terms separately to find cancellations. Anyways, now let's look at a variation on the original equation, $$ -\ln\frac{p(2n,k)}{p(2n,0)} = \ln (n-k)!+\ln (n+k)!-2\ln n! $$ We then use the log form of Stirling's approximation and simplify to get \begin{eqnarray} -\ln\frac{p(2n,k)}{p(2n,0)} &\approx& \left[ (n-k)\ln(n-k)+(n+k)\ln(n+k)-2n \ln n\right]+ \left[ \frac{\ln(n-k)+\ln(n+k)-2\ln n}{2}\right]\\ &\approx& n\left[\left(1-\frac{k}{n}\right)\ln\left(1-\frac{k}{n}\right)+\left(1+\frac{k}{n}\right)\ln\left(1+\frac{k}{n}\right) \right] + \frac{1}{2}\ln\left(1 -\frac{k^2}{n^2}\right) \end{eqnarray} Now, before we continue, we need to make sure we didn't make the error term relevant. The error terms are all of the form $c n^{-m}$, and $(n-k)^{-m}+(n+k)^{-m}-2n^{-m}\sim m(m+1)k^2/n^{m+2} = O(n^{-m-2})$. So the leading term is $O(n^{-3})$ and we'll be fine as long as we don't have cancellations down to that order. Using the small argument approximation for the logs gives $$ -\ln\frac{p(2n,k)}{p(2n,0)} = \frac{k^2}{n}-\frac{k^2}{2n^2}+O(n^{-3}), $$ and indeed we have terms above $O(n^{-3})$. Now, to get this as an approximation of $p(n,k)$, we drop the $n^{-2}$ term and do a bit of algebra to get $$ p(n,k) \approx p(n,0)e^{-2k^2/n} $$

Answer 2

Some hand-waving that can probably be formalized more rigorously:

For fixed $k$ while $n \to \infty$, \begin{align} \frac{p(k)}{p(0)} &= \frac{(n/2)! (n/2)!}{(n/2 + k)! (n/2 - k)!} \\ &\sim \frac{(n/2)^{n+1}}{(n/2 + k)^{n/2 + k + 1/2} (n/2 - k)^{n/2 - k + 1/2}} & \text{Stirling} \\ &= \frac{1}{(1 + 2k/n)^{n/2 + k + 1/2} (1 - 2k/n)^{n/2 - k + 1/2}} \\ &\sim e^{-(2k/n)(n/2 + k + 1/2)} e^{(2k/n)(n/2 - k + 1/2)} & \text{$(1+1/m)^m \to e$} \\ &= e^{-4k^2/n} \end{align}

Not sure why I have a $4$ instead of $2$, maybe I made an error somewhere.