Series expansion of incomplete gamma function ratio I am interested in the series expansion of:
$$S(k)=\frac{\Gamma(k+1,a)}{k!},$$
around $k=\infty$ where $\Gamma(x,z)$ is the incomplete gamma function and $a$ is some positive constant.  In particular, I would like to know how quickly this ratio converges to 1.
How to prove asymptotic limit of an incomplete Gamma function shows that:
$$1-S(k)\leq \left(\frac{e^a-1}{e^a}\right)^{k+1},$$
but is it possible to make this any tighter?
 A: If we start with a little rearranging,
$$
\begin{align}
\frac{\Gamma(k+1,a)}{k!} &= \frac{1}{k!} \int_a^\infty t^k e^{-t}\,dt \\
&= \frac{1}{k!} \left( \int_0^\infty t^k e^{-t}\,dt - \int_0^a t^k e^{-t}\,dt \right) \\
&= 1 - \frac{1}{k}\int_0^a t^k e^{-t}\,dt,
\end{align}
$$
then make the change of variables $t = ae^{-s}$, we end up with
$$
1 - \frac{\Gamma(k+1,a)}{k!} = \frac{a^{k+1}}{k!} \int_0^\infty e^{-ks} \exp\{-s-ae^{-s}\}\,ds.
$$
If we expand
$$
\begin{align}
\exp\{-s-ae^{-s}\} &= \sum_{n=0}^{\infty} c_n s^n \\
&= e^{-a} + e^{-a}(a-1) s + \tfrac{1}{2} e^{-a} (a^2-3a+1) s^2 + \cdots
\end{align}
$$
then by Watson's lemma we can exchange the order of integration and summation to obtain an asymptotic series for the integral;
$$
\begin{align}
1 - \frac{\Gamma(k+1,a)}{k!} &\approx \frac{a^{k+1}}{k!} \sum_{n=0}^{\infty} c_n \int_0^\infty e^{-ks} s^n \,ds \\
&= \frac{a^{k+1}}{k!} \sum_{n=0}^{\infty} \frac{n! c_n}{k^{n+1}} \\
&= \frac{a^{k+1} e^{-a}}{k \cdot k!} \left(1 + \frac{a-1}{k} + \frac{a^2 - 3a + 1}{k^2} + \cdots \right).
\end{align}
$$
In particular, to first order we have
$$
1 - \frac{\Gamma(k+1,a)}{k!} \sim \frac{a^{k+1} e^{-a}}{k \cdot k!}.
$$
A: Using a CAS, what it found as expansion for large values of $k$ is $$S(k)=\frac{\Gamma(k+1,a)}{k!}\simeq 1-\frac{ a^{k+1} e^{k \left(\log
   \left(\frac{1}{k}\right)+1\right)-a}}{\sqrt{2 \pi }k^{3/2}}$$ I hope this could be of some help.
