# Upper bound for quotient of Gamma functions

Yesterday I posted a question about the Gamma function (see here: Lower bound for combined Gamma functions)

It helped me very much in solving all my remaining exercises concerning the Gamma function, except one last. It is the following. Let $$0 and $$b\ge 2, b\in\mathbb{N}$$. Prove that there exists a positive constant $$c$$ which does not depend on $$a$$ (but it would be fine it would depend on $$b$$) such that $$\frac{\Gamma\big((b+2a)/2\big)}{\Gamma(2-a)}\le c.$$

Here $$\Gamma$$ is defined, for $$Re(z)>0$$, as $$\Gamma(z) = \int_0^\infty t^{z-1} e^{-t} dt$$ (here $$(b+2a)/2$$ and $$2-a$$ are all real and nonnegative numbers).

I tried by using the same strategy in the linked question. It worked with all the other exercise, but in this case I can not get rid of the dependence on $$a$$.

Anyone could help?

This has nothing to do with the Gamma function. You only need that $$\Gamma$$ is continuous and does not have any zeros. Then for every fixed $$b$$, the function $$f : [0,1] \to \mathbb{R}, \quad f(a) = \frac{\Gamma((b+2a)/2)}{\Gamma(2-a)}$$ is continuous and hence bounded.