Simplify division of two gamma functions I got the following division of two gamma functions with $n>0$:
$$
\frac{Γ(\frac{1}{2}(n+2))}{ Γ(\frac{1}{2}(n+3))}.
$$
Is there any way to further simplify this expression? I do have the feeling that there must be a way - at least to get rid of the $\frac{1}{2}$ but I am running a bit blank at the moment.
 A: One way is to use the duplication formula:
$$\Gamma(x)\Gamma(x+{1\over 2}) = 2^{1-2x}\sqrt{\pi}\Gamma(2x)$$
Revised as a response for clarification as to why this helps.
I have been down-voted by presenting just the formula above without providing any clarification. Mea Culpa indeed. Here are some details.
There is nothing really deep here. A direct application of the duplication formula gives the following:
$$
{\sqrt{\pi}(2k+1)!\over {2^{2k+1} k!}}=
\begin{cases}
\Gamma({n+2\over 2}),\enspace\hbox{if $n=2k+1$} \\
\\
\Gamma({n+3\over 2}),\enspace\hbox{if $n=2k$} \\
\end{cases}
$$
For $n=2k+1$ we have $\Gamma({n+3\over 2})=(k+1)!$ and for
$n=2k$ we have $\Gamma({n+2\over 2})=k!$, so combining these one gets:
$${\Gamma({n+2\over 2})\over \Gamma({n+3\over 2})}=
\begin{cases}
{\sqrt{\pi}\,n!\over {2^n ({n-1\over 2})!({n+1\over 2})!}},\enspace\hbox{if $n$ is odd} \\
\\
{2^{n+1}\cdot ({n\over 2})!({n\over 2})!\over \sqrt{\pi}(n+1)!},\enspace\hbox{if $n$ is even} \\
\end{cases}
$$
Assuming $n$ to be an integer, it seems to me that factorial functions and powers of constants is conceptually simpler than the $\Gamma$ function, at least in the sense that they can be computed numerically much easier. 
A: Sorry I have done a big mistake previously. Thanks to @gammatester for pointing it out.
Please check this one out. I hope that this time I have done no mistake. If I have done any, please let me know about it.
Consider two cases:-
Case-$1$
When $n$ is even,
$$\frac{Γ\big(\frac{1}{2}(n+2)\big)}{ Γ\big(\frac{1}{2}(n+3)\big)}$$
$$=\frac{\big(\frac{n}{2}\big)!}{Γ\big(\frac{1}{2}(n+3)\big)}\dots (1).$$
Now,
$$Γ\big(\frac{1}{2}(n+3)\big)=\big(\frac{n+1}{2}\big)Γ\big(\frac{1}{2}(n+1)\big)=\big(\frac{n+1}{2}\big)\big(\frac{n-1}{2}\big)\dots \frac{1}{2}Γ\big(\frac{1}{2}\big)$$
$$=\frac{\big(\frac{n+1}{2}\big)\big(\frac{n}{2}\big)\dots \frac{1}{2}Γ\big(\frac{1}{2}\big)}{\big(\frac{n}{2}\big)\dots 1}=\frac{\frac{(n+1)!}{2^{(n+1)}}Γ\big(\frac{1}{2}\big)}{\big(\frac{n}{2}\big)!}\dots(2).$$
Substitute the value of $Γ\big(\frac{1}{2}(n+3)\big)$ in equation $(1)$ from equation $(2)$.
Case-$2$
When $n$ is odd,
$$Γ\big(\frac{1}{2}(n+3)\big)=\big(\frac{n+1}{2}\big)!\dots (3).$$
$$Γ\big(\frac{1}{2}(n+2)\big)=\big(\frac{n}{2}\big)Γ\big(\frac{n}{2}\big)=\big(\frac{n}{2}\big)\big(\frac{n-2}{2}\big)\dots \frac{1}{2}Γ\big(\frac{1}{2}\big)$$
$$=\frac{\big(\frac{n}{2}\big)\big(\frac{n-2}{2}\big)\dots \frac{1}{2}Γ\big(\frac{1}{2}\big)}{\big(\frac{n-1}{2}\big)\dots 1}=\frac{\frac{n!}{2^n}Γ\big(\frac{1}{2}\big)}{\big(\frac{n-1}{2}\big)!}\dots(4).$$
From equation $(3) \ \& \ (4)$, you can easily evaluate the desired expression, and put the value of, $Γ\big(\frac{1}{2}\big)=\sqrt \pi$.
