Taking limit of gamma functions

Question

$$\lim_{n\rightarrow\infty}\frac{\Gamma(n)\Gamma(n+1)\Gamma(n-a)\Gamma(n-a+1)\Gamma(2n+2)}{\Gamma(2n-a+2)\Gamma(2n-a)\Gamma(2n-2a+1)}$$ where $1<a<n.$ Any help is appreciated.I have tried to use Stirling approximation but it did not help me or I am doing something wrong. Thanks.

Answer 1

Using $\;\Gamma(m)=(m-1)!\;$ for $\;m\in\Bbb N\;$ , and $\;\Gamma(s+1)=s\Gamma(s)\;$ :$${}$$

$$\frac{\Gamma(n)\Gamma(n+1)\Gamma(n-a)\Gamma(n-a+1)\Gamma(2n+2)}{\Gamma(2n-a+2)\Gamma(2n-a)\Gamma(2n-2a+1)}=$$

$$=\frac{(n-1)!n!(2n+1)!(n-a)\Gamma(n-a)^2}{(2n-a+1)(2n-a)\Gamma(2n-a)^2(2n-2a)\Gamma(2n-2a)}$$

This limit looks like $\;\infty\;$ . Perhaps someone else can finish the argument.