When working with a Gamma function I ended up getting the following relation. I am wondering how to find a good (something better than Stirling's formula) asymptotic approximation of it $$ \frac{(N/2+1)!}{(N/2+1/2)!}. $$ and how to find C, such that
$$ \frac{(N/2+1)!}{(N/2+1/2)!}\leq C $$
Let $f(n) =\frac{(n/2+1)!}{(n/2+1/2)!} $.
I had come up with if $n$ is even, $f(n) \approx \sqrt{n/2+1} $ and if $n$ is odd, then $f(n) \approx \sqrt{n/2+3/2} $ and then did a search.
This is $f(n) =\dfrac{\Gamma(n/2+2)}{\Gamma(n/2+3/2)} $.
According to the result quoted below, this, with $a=2, b=3/2$, is, since $a-b = 1/2$, $f(n) = \sqrt{\dfrac{n}{2}} \left(1+\dfrac{(1/2)(-1/2)}{2(n/2)} +O\left(\dfrac1{n^2}\right) \right) = \sqrt{\dfrac{n}{2}} \left(1-\dfrac{1}{4n} +O\left(\dfrac1{n^2}\right) \right) $.
A Google search for "Ratios of Gamma functions" comes up with this (among others):
https://msp.org/pjm/1951/1-1/pjm-v1-n1-p14-s.pdf
"THE ASYMPTOTIC EXPANSION OF A RATIO OF GAMMA FUNCTIONS" by F. G.TRICOMI AND A.ERDELYI
Among the results:
As $z \to \infty$, $\dfrac{\Gamma(z+a)}{\Gamma(z+b)} =z^{a-b}\left(1+\dfrac{(a-b)(a-b-1)}{2z}+O\left(|z|^{-2}\right)\right) $.
An asymptotic expansion $\dfrac{\Gamma(z+a)}{\Gamma(z)} =z^a\sum_{n=0}^{\infty}A_n(a)z^{-n} $ where $A_0(a) = 1, A_1(a) = \binom{a}{2}, A_2(a) = \frac{3a-1}{4}\binom{a}{3}, A_3(a) = \binom{a}{2}\binom{a}{4} $ and, in general, $A_n(a) =\dfrac1{n}\sum_{m=0}^{n-1}\binom{a-m}{n-m-1}A_m(a) $.
$\dfrac{\Gamma(z+a)}{\Gamma(z+b)} \sim \sum_{n=0}^{\infty} C_n(a-b, b)z^{a-b-n} $ where
$C_0(a-b, b) =1, C_1(a-b, b) =\dfrac12(a-b)(a-b-1),\\ C_2(a-b, b) =\dfrac1{12}\binom{a-b}{2}(3(a+b-1)^2-a+b-1),\\ $
and, in general, $C_n(a-b, b) =\dfrac1{n}\sum_{m=0}^{n-1}\left[\binom{a-b-m}{n-m+1}-(-1)^{n+m}(a-b)b^{n-m} \right]C_m(a-b, b) $.