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

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

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  • $\begingroup$ 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. $\endgroup$ – Greg Martin Jun 21 '20 at 17:21
  • $\begingroup$ If $x$ is any real number, $1/x$ is any real number except zero. It could be negative. $\endgroup$ – David K Jun 21 '20 at 17:26
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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?

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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.

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