Convergence of $1 - \sqrt{ \frac{2}{n-1} } \cdot \frac{ \Gamma(n/2) }{ \Gamma( \frac{n-1}{2} ) } $ as $n \to \infty$ I stumbled upon this problem while reading about the bias of the sample standard deviation.
How to show that:
$$\bigg(1 - \sqrt{ \frac{2}{n-1} }  \cdot \frac{ \Gamma(n/2) }{ \Gamma( \frac{n-1}{2} ) } \bigg) \sim \frac{1}{4 n}$$
as $n \to \infty$ and n is positive integer.
I thought I should use Stirling's formula, or Gautschi's inequality. The following equality might also lead in the right direction:
$$\Gamma(1/2+n)=\frac{(2n)!}{4^nn!}\sqrt{\pi}$$
Can you help?

Using twice Stirling formula, we have
$$\log \left(\frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma
   \left(\frac{n-1}{2}\right)}\right)=\frac{1}{2} \log \left(\frac{n}{2}\right)-\frac{3}{4 n}-\frac{1}{2 n^2}-\frac{3}{8
   n^3}+O\left(\frac{1}{n^4}\right)\tag 1$$
Using the binomial expansion and taking the logarithm and ussing Taylor series
$$\log \left(\sqrt{\frac{2}{n-1}}\right)=-\frac{1}{2} \log \left(\frac{n}{2}\right)+\frac{1}{2 n}+\frac{1}{4 n^2}+\frac{1}{6
   n^3}+O\left(\frac{1}{n^4}\right)\tag 2$$
Adding $(1)$ to $(2)$ gives
$$\log \left(\sqrt{\frac{2}{n-1}}\frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma
   \left(\frac{n-1}{2}\right)}\right)=-\frac{1}{4 n}-\frac{1}{4 n^2}+O\left(\frac{1}{n^3}\right)\tag 3$$
$$\sqrt{\frac{2}{n-1}}\frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma
   \left(\frac{n-1}{2}\right)}=1-\frac{1}{4 n}-\frac{7}{32 n^2}+O\left(\frac{1}{n^3}\right)\tag 4$$
$$1-\sqrt{\frac{2}{n-1}}\frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma
   \left(\frac{n-1}{2}\right)}=\frac{1}{4 n}+\frac{7}{32 n^2}+O\left(\frac{1}{n^3}\right)\tag 5$$
Try for $n=10$; the exact value is
$$1-\frac{128 }{105}\sqrt{\frac{2}{\pi }}=0.0273407\cdots$$ while the truncated series gives
$$\frac{87}{3200}=0.0271875\cdots$$
