TITLE: How to prove asymptotic limit of an incomplete Gamma function
QUESTION [11 upvotes]: 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?

REPLY [3 votes]: 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
$$
\frac{z}{e^{f(z)}} \longrightarrow \infty.
$$