<s>[INST] What periodic function does $\frac{\Gamma(x+1)}{2^x \Gamma(\frac{x}{2}+1)^2}$ tend to? If one graphs $$ y= \frac{\Gamma(x+1)}{2^x \Gamma(\frac{x}{2}+1)^2} = \frac{x!}{2^x  \left( \left( \frac{x}{2}\right)!\right)^2} $$
One immediately notices that as $x \rightarrow -\infty$ the function tends to some periodic function that looks an awful lot like $\tan(x)$. Is there a way to make this idea precise and state what that function is? I was thinking something like, there exists a periodic function $f(x)$ such that $f(x) - y(x) \rightarrow 0$ as $x \rightarrow -\infty$.
See this image:
Which was produced using desmos.com/calculator

 [/INST] For negative values of $x$,  $$y= \frac{\Gamma(x+1)}{2^x\, \Gamma(\frac{x}{2}+1)^2} $$ is closer and closer to
$$\sin ^2\left(\frac{\pi  x}{2}\right) \csc (\pi  (x+1))=-\frac{1}{2} \tan \left(\frac{\pi  x}{2}\right)$$ This comes from Euler's reflection formula of the gamma function.
Edit
As @GregMartin commented, I forgot the next term of the expansion. So, give all credit to him.
Using more terms, for negative values of $x$
$$y=- \sqrt{-\frac{2}{\pi x}} \tan \left(\frac{\pi  x}{2}\right)\Bigg[ 1-\frac{1}{4 x}+\frac{1}{32 x^2}+O\left(\frac{1}{x^3}\right)\Bigg]$$
It could even be better if the series expansion is made a simple Padé approximant such as
$$\frac {64 x^2-8 x+11 }{64 x^2+8 x+11 }+O\left(\frac{1}{x^5}\right)$$
In order to check how good (or bad) is the approximation, I computed
$$I_n=\int_{-2n-0.99}^{-2n+0.99} \Bigg[\frac{\Gamma(x+1)}{2^x\, \Gamma(\frac{x}{2}+1)^2}+ \sqrt{-\frac{2}{\pi x}}\, \tan \left(\frac{\pi  x}{2}\right)\,\,\frac {64 x^2-8 x+11 }{64 x^2+8 x+11 }\Bigg]^2\, dx$$
$$\left(
\begin{array}{cc}
n & I_n \\
 10 & 5.371 \times 10^{-16} \\
 9  & 1.766\times 10^{-15} \\
 8  & 6.733\times 10^{-15} \\
 7  & 3.105\times 10^{-14} \\
 6  & 1.846\times 10^{-13}  \\
 5  & 1.567\times 10^{-12} \\
 4  & 2.271\times 10^{-11} \\
 3  & 7.961\times 10^{-10} \\
 2  & 1.562\times 10^{-07} 
\end{array}
\right)$$
What could be interesting is to see if we can approximate the roots of
$$\frac{\Gamma(x+1)}{2^x\, \Gamma(\frac{x}{2}+1)^2}=k $$
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