Expand $\frac{\Gamma\left(\frac{x}{2}\right)}{\Gamma\left(\frac{x-1}{2}\right)}$ at $x=\infty$ I used Wolfram Alpha for this problem and it gives me a "Puiseux expansion": $$\sqrt{x}-\frac{3 \sqrt{\frac{1}{x}}}{8}-\frac{7}{128}\left(\frac{1}{x}\right)^{3 / 2}-\frac{9\left(\frac{1}{x}\right)^{5 / 2}}{1024}+O\left(\left(\frac{1}{x}\right)^{3}\right)$$
which is exactly what I need.
My only question is how one would be able to obtain this expansion manually.
 A: Note that the expansion you gave is for $\Gamma(x)/\Gamma(x-1/2)$. Using the beta function and an appropriate change of integration variables, we obtain
\begin{align*}
&\frac{{\Gamma \left( {\frac{x}{2}} \right)}}{{\Gamma \left( {\frac{{x - 1}}{2}} \right)}} = \frac{{x - 1}}{{2\sqrt \pi  }}\frac{{\Gamma \left( {\frac{x}{2}} \right)\Gamma \left( {\frac{1}{2}} \right)}}{{\Gamma \left( {\frac{{x + 1}}{2}} \right)}} = \frac{{x - 1}}{{2\sqrt \pi  }}B\left( {\frac{x}{2},\frac{1}{2}} \right) = \frac{{x - 1}}{{2\sqrt \pi  }}\int_0^1 {\frac{{t^{x/2 - 1} }}{{\sqrt {1 - t} }}dt} 
\\ &
 = \frac{{x - 1}}{{\sqrt \pi  }}\int_0^{ + \infty } {\frac{{e^{ - xs} }}{{\sqrt {1 - e^{ - 2s} } }}ds}  = \frac{{x - 1}}{{\sqrt {2\pi } }}\int_0^{ + \infty } {e^{ - xs} s^{ - 1/2} \sqrt {\frac{{2s}}{{1 - e^{ - 2s} }}} ds} .
\end{align*}
By the definition of the Nörlund numbers (cf. https://mathworld.wolfram.com/NorlundPolynomial.html),
$$
\sqrt {\frac{{2s}}{{1 - e^{ - 2s} }}}  = \sum\limits_{n = 0}^\infty  {( - 1)^n \frac{{2^n B_n^{(1/2)} }}{{n!}}s^n } 
$$
for $|s|<\pi$. Thus, by Watson's lemma (cf. http://dlmf.nist.gov/2.3.ii),
\begin{align*}
\frac{{\Gamma \left( {\frac{x}{2}} \right)}}{{\Gamma \left( {\frac{{x - 1}}{2}} \right)}} & \sim \frac{{x - 1}}{{\sqrt {2\pi } }}\sum\limits_{n = 0}^\infty  {( - 1)^n \frac{{2^n B_n^{(1/2)} }}{{n!}}\int_0^{ + \infty } {e^{ - xs} s^{n - 1/2} ds} } 
\\ &
 = \frac{{x - 1}}{{\sqrt {2\pi } }}\sum\limits_{n = 0}^\infty  {( - 1)^n \frac{{2^n B_n^{(1/2)} }}{{n!}}\Gamma \left( {n + \frac{1}{2}} \right)\frac{1}{{x^{n + 1/2} }}} 
\\ &
 = \sqrt {\frac{x}{2}}  + \sum\limits_{n = 0}^\infty  {\frac{{( - 1)^{n + 1} }}{{2\sqrt \pi  n!}}\Gamma \left( {n + \frac{1}{2}} \right)\left( {\frac{{2n + 1}}{{n + 1}}B_{n + 1}^{(1/2)}  + B_n^{(1/2)} } \right)\left(\frac{2}{{x}}\right)^{n+1/2}} 
\end{align*}
as $x\to +\infty$. It is possible to simplify this result to
$$
\frac{{\Gamma \left( {\frac{x}{2}} \right)}}{{\Gamma \left( {\frac{{x - 1}}{2}} \right)}} \sim \sqrt {\frac{x}{2}}  + \sum\limits_{n = 0}^\infty  {  
   \binom{ 1/2}{n + 1}B_{n + 1}^{(3/2)} \left(\frac{2}{{x}}\right)^{n+1/2}} .
$$
Note that this is a divergent asymptotic expansion. See http://dlmf.nist.gov/5.11.iii for more general results and references.
