Sample standard deviation is a biased estimator: Details in calculating the bias of $s$ In this post Why is sample standard deviation a biased estimator of $\sigma$?
the last step is shown as:
$$\sigma\left(1-\sqrt\frac{2}{n-1}\frac{\Gamma\frac{n}{2}}{\Gamma\frac{n-1}{2}}\right)
= \sigma\left(1-\sqrt\frac{2}{n-1}\frac{((n/2)-1)!}{((n-1)/2-1)!}\right)$$
How is this equal to $\frac{\sigma}{4n}$?

Making the substitution $x = \frac{n}{2}-1$, you essentially want to control
$$1 - \frac{\Gamma(x+1)}{\Gamma(x+\frac{1}{2}) \sqrt{x + \frac{1}{2}}}$$
as $x \to \infty$.
Gautschi's inequality (applied with $s=\frac{1}{2}$) implies
$$
1 - \sqrt{\frac{x+1}{x+\frac{1}{2}}}
<1 - \frac{\Gamma(x+1)}{\Gamma(x+\frac{1}{2}) \sqrt{x + \frac{1}{2}}}
< 1 - \sqrt{\frac{x}{x+\frac{1}{2}}}$$
The upper and lower bounds can be rearranged as
$$
\left|1 - \frac{\Gamma(x+1)}{\Gamma(x+\frac{1}{2}) \sqrt{x + \frac{1}{2}}}\right|
< \frac{1}{2x+1} \cdot \frac{1}{1 + \sqrt{1 - \frac{1}{2x+1}}}
\approx \frac{1}{2(2x+1)}.$$
Plugging in $x=\frac{n}{2}-1$ gives a bound of $\frac{1}{2(n-1)}$. This is weaker than the author's claim of asymptotic equivalence with $\frac{1}{4n}$, but at least it is of the same order.

Responses to comments:
When $x=\frac{n}{2}-1$ you have $x+1 = \frac{n}{2}$ and $x + \frac{1}{2} = \frac{n}{2} - 1 + \frac{1}{2} = \frac{n}{2} - \frac{1}{2} = \frac{n-1}{2}$. So $\frac{\Gamma(x+1)}{\Gamma(x+\frac{1}{2}) \sqrt{x + \frac{1}{2}}} = \frac{\Gamma(n/2)}{\Gamma((n-1)/2) \sqrt{(n-1)/2}}$.
