# $| \Gamma (x_{n})| \to 0$ as $n \to \infty$ where $x_{n} \in [-n+\delta,1-n-\delta]$

I'm trying to prove $| \Gamma (x_{n})| \to 0$ as $n \to \infty$ where $x_{n} \in [-n+\delta,1-n-\delta]$ for some $\delta > 0$.

What possible method can I use?

As a special case, I've proved the above result for $x_{n} = 1/2 - n$. Can I use this result for the proof of the above?

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From $\Gamma(x+1)=x\,\Gamma(x)$ you get $$\Gamma(-n+x)=\frac{1}{(x-n)(x-(n-1))\dots(x-1)}\Gamma(x).$$ If $x_n=-n+a_n$, so that $\delta<a_n<1-\delta$, you have $\Gamma(a_n)$ bounded and $$\frac{1}{(a_n-n)(a_n-(n-1))\dots(a_n-1)}\to 0.$$

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Don't you mean $x_n=-n+x$ and $\delta\lt x\lt1-\delta$? –  joriki Sep 27 '12 at 8:40
@user8268 Thank you very much for your very clever answer!! –  julypraise Sep 27 '12 at 8:42
@joriki I think he changed the usage of $x_{n}$ in his solution. I thknk his $a_{n}$ is the same as my $x_{n}$ –  julypraise Sep 27 '12 at 8:50
@joriki sorry for confusion, now I made my notation consistent with the question –  user8268 Sep 27 '12 at 9:29

This is false. The gamma function has poles at the negative integers, so you can choose arbitrarily large moduli for each $n$.

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Ah sorry my bad. I've modified the problem –  julypraise Sep 27 '12 at 7:51