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How can I prove: $$\lim_{z\to\infty}{\Gamma(z, x)\over\Gamma(z)} = 1$$ ? Here $\Gamma(z, x)$ is the upper incomplete gamma function and $\Gamma(z)$ is the gamma function.

This must be something trivial, but I can't figure it out. Any ideas?

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  • $\begingroup$ Nice question (+1) $\endgroup$ – user 1357113 Sep 21 '12 at 10:23
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The integrand $t^{z-1}\mathrm e^{-t}$ is unimodal and has its maximum at $t=z-1$. Thus for $z-1\gt x$ we can bound the integrand on both sides of $x$ by its value at $x$ to obtain

$$ \begin{align} \Gamma(z)-\Gamma(z,x) &= \int_0^xt^{z-1}\mathrm e^{-t}\mathrm dt \\ &\lt\int_0^xx^{z-1}\mathrm e^{-x}\mathrm dt \\ &=xx^{z-1}\mathrm e^{-x} \\ &=\frac x{z-1-x}\int_x^{z-1}x^{z-1}\mathrm e^{-x}\mathrm dt \\ &\lt\frac x{z-1-x}\int_x^{z-1}t^{z-1}\mathrm e^{-t}\mathrm dt \\ &\lt\frac x{z-1-x}\int_0^\infty t^{z-1}\mathrm e^{-t}\mathrm dt \\ &=\frac x{z-1-x}\Gamma(z)\;, \end{align} $$

so for fixed $x$ the ratio of the difference to $\Gamma(z)$ tends to $0$ for $z\to\infty$.

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  • $\begingroup$ So you proved that $\gamma(z, x) < {x\over z-1-x}\Gamma(z)$ for $z-1 > x > 0$ and from that the limit follows immediately. Nice! $\endgroup$ – Ondřej Čertík Sep 21 '12 at 18:10
  • $\begingroup$ I verified your steps, I think it's all correct. I am accepting your answer as it gives a simple proof. $\endgroup$ – Ondřej Čertík Sep 21 '12 at 21:15
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In this answer I showed that

$$ \left(\frac{e^x}{e^x-1}\right)^z \gamma(z,x) \leq \Gamma(z) $$

when $z \geq 1$ and $x > 0$, where

$$ \gamma(z,x) = \Gamma(z) - \Gamma(z,x) $$

is the lower incomplete gamma function. Rearranging we get

$$ 0 \leq 1 - \frac{\Gamma(z,x)}{\Gamma(z)} = \frac{\gamma(z,x)}{\Gamma(z)} \leq \left(\frac{e^x-1}{e^x}\right)^z \longrightarrow 0 $$

as $z \to \infty$ when $x$ is fixed.

In fact, this shows that the result is still true if $x = f(z) \geq 0$ and

$$ z \log\left(1-e^{-f(z)}\right) \longrightarrow -\infty. $$

In particular, it's still true if

$$ \frac{z}{e^{f(z)}} \longrightarrow \infty. $$

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    $\begingroup$ Thanks! As a matter of fact, I actually wanted to prove that $\gamma(z, x)\over\Gamma(z)$ goes to zero, but thought it'd be easier to do it with the upper incomplete gamma function. $\endgroup$ – Ondřej Čertík Sep 21 '12 at 18:13

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