Limit of exponentials Why is $n^n (n+m)^{-{\left(n+m\over 2\right)}}(n-m)^{-{\left(n-m\over 2\right)}}$ asymptotically equal to $\exp\left(-{m^2\over 2n}\right)$ as $n,m\to \infty$?
 A: Let $m = n x$.
Take the logarithm:
$$
  n \log(n) - n \frac{1+x}{2} \left(\log n + \log\left(1+x\right) \right) - n \frac{1-x}{2} \left( \log n + \log\left(1-x\right) \right) 
$$
Notice that all the terms with $\log(n)$ cancel out, so we are left with
$$
   -\frac{n}{2} \left( (1+x) \log(1+x) - (1-x) \log(1-x) \right) 
$$
It seems like the you need to assume that $x$ is small here, meaning that $ m \ll n$. Then, using Taylor series expansion of the logarithm:
$$
  (1+x) \log(1+x) + (1-x) \log(1-x) = (1+x) \left( x - \frac{x^2}{2} + \mathcal{o}(x^2) \right) + (1-x) \left(-x - \frac{x^2}{2} + \mathcal{o}(x^2)\right) = x^2 + \mathcal{o}(x^3)
$$ 
Hence the original expression, asymptotically, equals
$$
    \exp\left( -\frac{n}{2} x^2 + \mathcal{o}(n x^3)\right) = \exp\left(- \frac{m^2}{2n} + \mathcal{o}\left(\frac{m^3}{n^2}\right) \right)
$$
A: By Stirling's approximation we have $$ \binom{2n}{n+m}= \frac{(2n)!}{(n+m)!(n-m)!} \sim \frac{\sqrt{2\pi n} (2n/e)^{2n}}{\sqrt{2\pi(n+m)} \left( \frac{n+m}{e} \right)^{n+m} \sqrt{2\pi(n-m)} \left( \frac{n-m}{e} \right)^{n-m} }= \frac{1}{\sqrt{2\pi (n^2-m^2)}} \cdot \frac{(2n)^{2n} }{ (n+m)^{n+m} (n-m)^{n-m}} .$$
Now if $m$ is "small" compared to "n" then $$n^n (n+m)^{-(n+m)/2} (n-m)^{-(n-m)/2} \sim \frac{1}{2^n} \sqrt{ \sqrt{2\pi (n^2-m^2)}\binom{2n}{n+m} }.$$
We make the assumption that $m$ is small compared to $n$ precise in the sense that we take $m=\mathcal{o}(n^{2/3})$ so that we can apply the refined entropy formula found at equation 8 of the link, which gives the result.
