In this paper: link
It is said on the first page that we have
$$\frac{\Gamma(z+a)}{\Gamma(z+b)}=z^{a-b} \left[1+\frac{(a-b)(a+b-1)}{2z}+O(|z|^{-2})\right].$$ It is said that the conditions will be stated later.
Can this result be found in any other texts with easier proofs?
Are you able to see what conditions they talk about? It is a long proof, and I am not able to see where and what the conditions are. Is this explained easier somewhere else?
I think that the simplest way is to look at the Stirling expansion of $$\log (\Gamma (p))=p (\log (p)-1)+\frac{1}{2} \left(-\log \left({p}\right)+\log (2 \pi )\right)+\frac{1}{12 p}+O\left(\frac{1}{p^2}\right)$$ and continue the expansion using $p=z+a$. This will give $$\log (\Gamma (z+a))=z (\log (z)-1)+\frac{1}{2} \left((2 a-1) \log (z)+\log (2 \pi )\right)+\frac{6 a^2-6 a+1}{12 z}+O\left(\frac{1}{z^2}\right)$$ Finally $$\log (\Gamma (z+a))-\log (\Gamma (z+b))=(a-b) \log (z)+\frac{(a-b) (a+b-1)}{2 z}+O\left(\frac{1}{z^2}\right)$$ Just exponentiate to get the expression you wrote.