bound for the ratio of Gamma functions

Question

Let $x \in R$, $N$ is a natural number.

How to bound from above $$ \frac{\Gamma(1-1/x)}{\Gamma(N+1-1/x)} $$

Answer 1

Repeated application of the fact that $\Gamma(x+1)=x\cdot\Gamma(x) \text{ for } x\notin-\mathbb{N}\cup\{0\}$ yields $$\frac{\Gamma(1-1/x)}{\Gamma(N+1-1/x)}=\prod_{i=1}^N \,\left(i-\frac{1}{x}\right)^{-1}$$ Does that answer your question?

Answer 2

When $x\to 1^+$, then $$ \frac{\Gamma(1-1/x)}{\Gamma(N+1-1/x)} \sim \frac{1}{(N-1)!} \frac{x}{x-1} \underset{x\to 1^+}{\longrightarrow} +\infty $$ so it cannot be bounded above.