4
$\begingroup$

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?

$\endgroup$
4
$\begingroup$

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

$$ 1 - \frac{\Gamma(k+1,a)}{k!} \sim \frac{a^{k+1} e^{-a}}{k \cdot k!}. $$

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Can you show how you did this with CAS? $\endgroup$ – ablmf Feb 11 '17 at 7:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.