Approximate variance of sample standard deviation based on the Gamma function Sorry for the long question title. I observed something interesting about the Gamma function but am not sure if this observation is true in general and whether it can be proven. Any help is highly appreciated!
I am interested in the distribution of the standard deviation of a sample normal random variables, i.e.,
$s = \sqrt{\frac{1}{N}\sum_{i=1}^N(X_i-\bar{X})^2}$
where $X_i$'s are i.i.d. normals and $\bar{X}$ is the sample mean.
In particular, I am interested in the variance of $s$. According to this link
the variance of $s$ is given by 
I want to somehow simplify the expression in the square bracket, may be an approximation that works for large $N$ but not $N\rightarrow\infty$. So I used R (and other software) to see if this is possible. For example:
> N = 100 
> N - 1 - 2*(gamma(N/2)/gamma(N/2-0.5))^2
[1] 0.498731

Trying out a few values for $N$ (both odd and even), the result is always slightly less than $0.5.$ The larger $N$ is, the closer the result is to $0.5.$ 
Can this result be proven mathematically, perhaps using some properties of the Gamma function?
According to this Wikipedia page one of the Gamma function properties is as follow:

Setting $n=\frac{N-1}{2}$ and $\alpha=\frac{1}{2}$, this seems to suggest that, for large $N$,
$$
N-1- \frac{2\Gamma^2 ( \frac{N}{2})}{\Gamma^2(\frac{N-1}{2})} \approx N-1 - 2 \left( \left (\frac{N-1} 2 \right)^{0.5}\right)^2 =N-1 - (N-1) = 0.
$$
This seems to be contradicting with the R outputs. Could anyone offer some insights to reconcile this puzzle? Thanks! 
 A: You have to apply a slightly better formula. Using the first two term of the asymptotic expansion from http://dlmf.nist.gov/5.11.E13
$$f(z,a,b)=\frac{\Gamma(z+a)}{\Gamma(z+b)} 
=z^{a-b} \left(1 + \frac{G_2(a,b)}{z} \right), \quad G_2(a,b)=\tfrac{1}{2}(a-b)(a+b-1)$$
your expression can be written as
$$N-1- 2f\left(\frac{N}{2},0,-\frac{1}{2}\right)^2
= N-1-N\left(1-\frac{3}{4N}\right)^2$$
Therefore the expression you search is asymptotically 
$$ \frac{1}{2}- \frac{9}{16N} + O\left(\frac{1}{N^2}\right)$$
If you keep one more term of the AE you get
$$\frac{1}{2}-\frac{1}{8N}-\frac{21}{64N^2} +  O\left(\frac{1}{N^3}\right)$$
A: Your limit formula shows that $\frac{2\Gamma^2(N/2)}{\Gamma^2((N-1)/2)}\sim N-1,$ but that is it. Note that it is also asymptotic to $N$, $N+1$ or $N+37:$ any of those has the property that $\lim\frac{f(N)}{N+1}=1.$ In other words, this tells you nothing about the constant term and it's just a coincidence that it cancelled out to zero. In fact, using the Stirling series, you can obtain $$ \frac{2\Gamma^2(N/2)}{\Gamma^2((N-1)/2)} = N-3/2+ O(1/N)$$ which agrees with your calculation.
