Puiseux Expansion of Gamma Function about Infinity In trying to find interesting proofs that Student's T Distribution converges to the Regularized Normal Distribution when $k$ (the number of desgrees of freedom) grows without bounds (i.e. $= \infty$).
One of the ways I tried involved trying to find an expansion for $\frac{\Gamma\left(\frac{v+1}{2}\right)}{\Gamma\left(\frac{v}{2}\right)}$ about $v=\infty$, though I could not make any headway on finding the expansion and I feared it would be extremely complicated. However, when I asked Wolfram Alpha I got a beautiful Puiseux Series, which is expressed as:
$$\frac{\Gamma\left(\frac{v+1}{2}\right)}{\Gamma\left(\frac{v}{2}\right)}\simeq \frac{v^{1/2}}{\sqrt2}-\frac{v^{-1/2}}{4\sqrt2}+\frac{v^{-3/2}}{32\sqrt2}+5\frac{v^{-5/2}}{128\sqrt2}-21\frac{v^{-5/2}}{2048\sqrt2}+\cdots$$
(more terms can be found in the link provided if desired). I can't help but notice that the denominators are simply $2^n \sqrt{2}$ for $n\in(0,2,5,7,11,\ldots)$ However, I can't find any pattern in $n$, nor can I explain the additional coefficients that appear starting on the fourth term, so I can't think of a way to work backward to find a proof.
Regardless, all I need is the first term for the purpose of taking a limit, as all the other terms will vanish anyway. Does anyone know how to prove the expansion? 
Edit: I should note that I have already attacked the problem using Stirlings Series,  but I found that to be somewhat brute force for such a simple summation and I hoped for a more clever argument (it does however  some light on where the coefficients come from). Again,  I only need to know the series up to $\frac{v^{1/2}}{\sqrt2} + O(v^{-1/2})$
 A: Consider $$A=\frac{\Gamma\left(\frac{v+1}{2}\right)}{\Gamma\left(\frac{v}{2}\right)}$$ and take logarithms $$\log(A)=\log\left(\Gamma\left(\frac{v+1}{2}\right)\right)-\log\left(\Gamma\left(\frac{v}{2}\right)\right)$$ and consider Stirling expansion for large values of $x$ $$\log \left(\Gamma(x)\right)=x (\log (x)-1)+\frac{1}{2} \left(\log \left(\frac{1}{x}\right)+\log (2 \pi
   )\right)+\frac{1}{12 x}-\frac{1}{360 x^3}+\frac{1}{1260
   x^5}+O\left(\frac{1}{x^{13/2}}\right)$$ So, applying this to the two different arguments, we have 
$$\log\left(\Gamma\left(\frac{v+1}{2}\right)\right)=\frac{1}{2} v (\log (v)-1-\log (2))+\frac{1}{2} \log (2 \pi )-\frac{1}{12
   v}+\frac{7}{360 v^3}-\frac{31}{1260
   v^5}+O\left(\frac{1}{v^{13/2}}\right)$$
$$\log\left(\Gamma\left(\frac{v}{2}\right)\right)=\frac{1}{2} v (\log (v)-1-\log (2))+\left(\frac{1}{2} \log
   \left(\frac{1}{v}\right)+\log \left(2 \sqrt{\pi }\right)\right)+\frac{1}{6
   v}-\frac{1}{45 v^3}+\frac{8}{315
   v^5}+O\left(\frac{1}{x^{13/2}}\right)$$ All of that leads to $$\log(A)=-\frac{1}{2} \log \left(\frac{2}{v}\right)-\frac{1}{4 v}+\frac{1}{24
   v^3}-\frac{1}{20 v^5}+O\left(\frac{1}{v^{13/2}}\right)$$ Now, using $A=e^{\log(A)}$ and Taylor series again $$A=\frac{\sqrt{v}}{\sqrt{2}}-\frac{\left(\frac{1}{v}\right)^{1/2}}{4 \sqrt{2}}+\frac{\left(\frac{1}{v}\right)^{3/2}}{32 \sqrt{2}}+\frac{5
   \left(\frac{1}{v}\right)^{5/2}}{128 \sqrt{2}}-\frac{21
   \left(\frac{1}{v}\right)^{7/2}}{2048 \sqrt{2}}-\frac{399
   \left(\frac{1}{v}\right)^{9/2}}{8192 \sqrt{2}}+\frac{869
   \left(\frac{1}{v}\right)^{11/2}}{65536
   \sqrt{2}}+O\left(\frac{1}{v^6}\right)$$
For the coefficients, you could find them in sequences $A143503$ and $A12854$ in $OEIS$.
A: Here is a method to obtain the full series of $\Gamma(a)/\Gamma(a+b)$, for large $a$, for any $b>0$ (one can then use recursion relations to access other values of $b$, for example). We have
$$ \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)} = B(a,b) = \int_0^1 x^{a-1} (1-x)^{b-1} \, dx, $$
where $B$ is the beta function, and is defined by the right-hand side. Therefore if we can find the series for the right-hand side, we can just divide by $\Gamma(b)$ to get the series we want. Putting $x=e^{-y}$, this becomes
$$ \frac{\Gamma(a)}{\Gamma(a+b)} = \frac{1}{\Gamma(b)}\int_0^{\infty} e^{-ay} (1-e^{-y})^{b-1} \, dx. $$
This integral is now in the form on which we can use Watson's lemma; the art is in obtaining the series for $(1-e^{-y})^{b-1}$. To find the first few terms, we can write $ 1-e^{-y}=y-y^2/2+\dotsb $ and use the binomial theorem, which tells us instantly that the leading order term is
$$\frac{1}{\Gamma(b)}\int_0^{\infty} y^{b-1} e^{-ay} \, dy = a^{-b}. $$
I don't think there is an easy way to obtain the rest of the series for $(1-e^{-y})^{b-1}$ when $b$ is not an integer, but it will be of the form $ \sum_{n=0}^{\infty} c_n y^{b+n-1} $, with $c_0=1$, and then Watson's lemma gives
$$ \frac{\Gamma(a)}{\Gamma(a+b)} \sim \sum_{n=0}^{\infty} c_n \frac{\Gamma(b+n)}{\Gamma(b)} a^{-b-n}. $$
