Evaluate $\frac{\Gamma(n/2)}{\Gamma(n/2+h)},$ $n,h \in \mathbb{N}$ Suppose $n,h \in \mathbb{N}$. How can I evaluate $$\frac{\Gamma(n/2)}{\Gamma(n/2+h)},$$
where $\Gamma$ is the Euler gamma function.
All I know is that $$\Gamma(n+1) = n!$$
 A: If you consider the case where $n$ is large (and much larger than $h$), take logarithms and use Stirling approximation. This would give
$$\log \left(\frac{\Gamma \left(\frac{n}{2}\right)}{\Gamma
   \left(\frac{n}{2}+h\right)}\right)=\log
   \left(2^h\right)-h \log ({n})+\frac{h-h^2}{n}+\frac{2 h^3-3 h^2+h}{3
   n^2}+O\left(\frac{1}{n^{3}}\right)$$
For example, using $n=100$ and $h=10$, the exact value would be $\approx 4.38613\times 10^{-18}$ while the truncated expression would lead to  $\approx 4.40747\times 10^{-18}$.
Edit
Continuing from the previous expansion and Taylor series, usind $A=e^{\log(A)}$, we can avoid exponentiation and get (as a slightly worse approximation)
$$\frac{\Gamma \left(\frac{n}{2}\right)}{\Gamma
   \left(\frac{n}{2}+h\right)}=\left(\frac{2}{n}\right)^h\left(1-\frac{(h-1) h}{n}+\frac{(h-1) h (h+1) (3 h-2)}{6
   n^2}+O\left(\frac{1}{n^3}\right)\right)$$
A: In fact you have the relationship $\Gamma(x+1) = x\Gamma(x)$ holds for any $x \in \mathbb{R^{+}}$, which is in fact one of the reason why mathematician came up with the Gamme function, i.e. a function that generalizes the factorial over the positive real numbers.
Now by succesivelly applying it you should get: 
$$\Gamma\left(\frac n2 + h\right) = \left(\frac n2 + h - 1\right)\Gamma\left(\frac n2 + (h-1)\right) = \dots $$
$$= \left(\frac n2 + h - 1\right)\left(\frac n2 + h - 2\right) \cdots \left(\frac n2 \right)\Gamma\left(\frac n2\right)$$
A: The well known recurrence
$$n!=n(n-1)!$$ extends to the Gamma function, so that
$$\Gamma(n)=(n-1)\Gamma(n-1).$$
Then
$$\frac{\Gamma(n/2)}{\Gamma(n/2+h)}=\frac{\Gamma(n/2)}{(n/2+h-1)\Gamma(n/2+h-1)}=\frac{\Gamma(n/2)}{(n/2+h-2)(n/2+h-1)\Gamma(n/2+h-2)}\\=\cdots=\frac{\Gamma(n/2)}{(n/2+h-2)(n/2+h-1)\cdots(n/2+h-h)\Gamma(n/2+h-h)}.$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
{\Gamma\pars{n/2} \over \Gamma\pars{n/2 + h}} & =
{1 \over \Gamma\pars{n/2 + h}/\Gamma\pars{n/2}} =
{1 \over \pars{n/2}^{\large\overline{h}}} =
{1 \over \prod_{k = 0}^{h - 1}\pars{n/2 + k}}
\\[5mm] & =
\bbx{1 \over \pars{n/2}\pars{n/2 + 1}\cdots\pars{n/2 + h - 1}}
\end{align}
A: H.M. Edwards gives the expression in the opening of Riemann's Zeta Function (1974, reprint 2001)
$$ \Gamma(s) = \prod_{n=1}^\infty\bigg(1+\frac{s-1}{n}\bigg)^{-1}\bigg(1+\frac{1}{n}\bigg)^{s-1}, $$
which implies 
$$\frac{\Gamma(s/2)}{\Gamma(s/2+h)}= \prod_{n=1}^\infty\frac{\bigg(1+\frac{(s/2)-1}{n}\bigg)^{-1}\bigg(1+\frac{1}{n}\bigg)^{(s/2)-1}}{\bigg(1+\frac{(s/2)+h-1}{n}\bigg)^{-1}\bigg(1+\frac{1}{n}\bigg)^{(s/2)+h-1}} = $$
$$ = \prod_{n=1}^\infty \frac{\bigg(1+\frac{(s/2)-1}{n}\bigg)^{-1}}{\bigg(1+\frac{(s/2)+h-1}{n}\bigg)^{-1}\bigg(1+\frac{1}{n}\bigg)^{h}} = \prod_{n=1}^\infty \frac{\bigg(1+\frac{(s/2)+h-1}{n}\bigg)}{\bigg(1+\frac{(s/2)-1}{n}\bigg)\bigg(1+\frac{1}{n}\bigg)^{h}}$$
Another expression is given by Titchmarsh for when Re$(s)$ > 1.
$$ \Gamma(s) = \frac{1}{\zeta(s)} \int_0^{\infty}\frac{x^{s-1}}{e^x-1}dx $$
which also implies an expression
$$\frac{\Gamma(s/2)}{\Gamma(s/2+h)} = \frac{\zeta(s/2 + h)\int_0^{\infty}\frac{n^{(s/2)-1}}{e^n-1}dn}{\zeta(s/2)\int_0^{\infty}\frac{m^{(s/2)+h-1}}{e^m-1}dm} $$
And the integral over the first quadrant $(m,n) \in Q_1 = \bigg(\mathbb{R} \cap [0,\infty) \bigg)^2 $ 
$$ \frac{\int_0^{\infty}\frac{n^{(s/2)-1}}{e^n-1}dn}{\int_0^{\infty}\frac{m^{(s/2)+h-1}}{e^m-1}dm} = \int_{Q_1}\bigg( \frac{e^m -1}{e^n-1} \bigg) \bigg( \frac{1}{m^h} \bigg) \bigg( \frac{n}{m} \bigg)^{(s/2)-1} {dm} \ {dn},  $$
implying
$$ \frac{\Gamma(s/2)}{\Gamma(s/2+h)} = \frac{\zeta( (s/2) + h)}{\zeta(s/2)} \int_{Q_1}\bigg( \frac{e^m -1}{e^n-1} \bigg) \bigg( \frac{1}{m^h} \bigg) \bigg( \frac{n}{m} \bigg)^{(s/2)-1} {dm} \ {dn} $$
A: This is more a comment
than an answer,
though it can
give some 
possibly useful asymptotics.
$\lim_{x \to \infty} \dfrac{(x+(n-1)/2)^n-x(x+1)...(x+n-1)}{(x+(n-1)/2)^{n-2}}
=\dfrac{n^3-n}{24}
$.
As a consequence,
$x(x+1)...(x+n-1)
= (x+(n-1)/2)^n-(x+(n-1)/2)^{n-2}\dfrac{n^3-n}{24}
+O(x^{n-3})
$.
