# bound for the ratio of Gamma functions

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

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

• This ratio is exactly $$\big( (N-1/x)(N-1-1/x) \cdots (1-1/x) \big)^{-1},$$ by the functional equation for the Gamma function. – Greg Martin Jun 21 '20 at 17:21
• If $x$ is any real number, $1/x$ is any real number except zero. It could be negative. – David K Jun 21 '20 at 17:26

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?
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.