Simplify division of two gamma functions

Question

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.

Answer 1

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.

Answer 2

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, $Γ\big(\frac{1}{2}\big)=\sqrt \pi$.