$\frac{\Gamma(\frac{n_T + n_C - 2}{2})}{\sqrt{\frac{n_T+n_C-2}{2}}\Gamma\left(\frac{n_T+n_C-3}{2}\right)} \approx 1 - \frac{3}{4(n_T+n_C+2)-1}$ In this answer to a question I asked (which derives the variance of Cohen's $d$), the approximation
$$\frac{\Gamma\left(\frac{n_T + n_C - 2}{2}\right)}{\sqrt{\frac{n_T+n_C-2}{2}}\Gamma\left(\frac{n_T+n_C-3}{2}\right)} \approx 1 - \frac{3}{4(n_T+n_C+2)-1}$$
is used. We can reasonably assume that $n_T, n_C> 0$ are integers.
How is this approximation derived? The answerer states:

I pulled it from the Hedges paper -- don't know its derivation at the moment but will think about it some more.

I wish I had more to contribute to this question than that, but the removal of the $\Gamma$ function I find completely baffling, and I wouldn't even know where to start.
Edit: Currently trying out Stirling's approximation, seeing if that leads me anywhere. And so far, I'm quite lost as to how to deal with the division by $2$ in the $\Gamma$ functions.
 A: Using Stirling's Approximation
$$
\log(\Gamma(n))=n\log(n)-n-\frac12\log(n)+\frac12\log(2\pi)+\frac1{12n}+O\left(\frac1{n^3}\right)
$$
we can compute
$$
\log\left[\frac{\Gamma(n-1)}{\sqrt{n-1}\,\Gamma\left(n-\frac32\right)}\right]=-\frac3{8n}-\frac1{2n^2}+O\left(\frac1{n^3}\right)
$$
and therefore,
$$
\begin{align}
\frac{\Gamma(n-1)}{\sqrt{n-1}\,\Gamma\left(n-\frac32\right)}
&=1-\frac3{8n}-\frac{55}{128n^2}+O\left(\frac1{n^3}\right)\\
&=1-\frac3{8n-\frac{55}6}+O\left(\frac1{n^3}\right)
\end{align}
$$
Set $n=\frac{n_T+n_C}2$ and we get
$$
\begin{align}
\frac{\Gamma(n-1)}{\sqrt{n-1}\,\Gamma\left(n-\frac32\right)}
&=1-\frac3{8n-\frac{55}6}+O\left(\frac1{n^3}\right)\\
&=1-\frac3{4\left(n_T+n_C-2\right)-\frac76}+O\left(\frac1{\left(n_T+n_C\right)^3}\right)\\
\end{align}
$$
If I am correct, there seems to be a sign problem in $n_T+n_C\color{#FF0000}{-}2$, and $\frac76$ has been simplified to $1$.
A: Here is a simple, elementary derivation, using only the recursion relation $\Gamma(x+1)=x\ \Gamma(x)$ instead of Stirling's approximation. 
We define 
$$f(x) = \frac{\Gamma(x)}{\sqrt{x}\,\Gamma(x-\frac12)}\ ,$$ 
where $x = \frac{n_T+n_C-2}2$. Then what we want to prove is the following asymptotic behavior for large $x$: 
$$f(x)=1-\frac3{8x-\frac76}+O(x^{-3})\ .$$
The trick is to consider the product $f(x)f(x+\frac12)$. The above definition of $f(x)$ leads to 
$$f(x)f(x+{\small\frac12}) 
= \frac{\Gamma(x)}{\sqrt{x}\,\Gamma(x-\frac12)}\frac{\Gamma(x+\frac12)}{\sqrt{x+\frac12}\,\Gamma(x)}
= \frac{x-\frac12}{\sqrt{x}\sqrt{x+\frac12}}
= \frac{1-\frac1{2x}}{\sqrt{1+\frac1{2x}}}
\ ,  $$
where in the 2nd equality we have used the recursion for $\Gamma$. Expanding the squareroot in the denominator yields
$$f(x)f(x+{\small\frac12}) = 1-\frac3{4x}+\frac7{32x^2}+O(x^{-3})\ .$$
On the other hand, setting 
$$f(x)=1-\frac1{ax+b}+O(x^{-3})\ ,$$
we obtain another expansion 
$$f(x)f(x+{\small\frac12}) 
= \left( 1-\frac1{ax}\frac1{1+\frac b{ax}} \right)
  \left( 1-\frac1{ax}\frac1{1+\frac{\frac a2 +b}{ax}} \right)+O(x^{-3}) $$ 
$$ = 1-\frac2{ax}+\frac{1+\frac a2 +2b}{a^2x^2}+O(x^{-3})\ .$$
Comparison of the coefficients of the $x^{-1}$ and $x^{-2}$ terms in the two expansions finally gives $a=\frac83$ and $b=-\frac7{18}$, which concludes the proof.
