Growth of $\Gamma(n,n)$ How can you get the asymptotics for the growth of $\Gamma(n,n)$?  
$$
    \Gamma(n,n) = \int_n^\infty x^{n-1} \exp(-x) \mathrm{d}x
$$
 A: You can get it by a simple rescaling of the integral.  Let $x=n t$; then
$$\begin{align}\Gamma(n,n) &= n^n \int_1^{\infty} dt \: t^{n-1} e^{-n t}\\ &= n^n \int_1^{\infty} \frac{dt}{t} e^{n (\log{t}-t)} \\ &= n^n e^{-n} \int_0^{\infty} \frac{du}{1+u} e^{n [\log{(1+u)}-u]} \end{align}$$
The contribution to this integral is dominated by that near $u=0$ as $n \rightarrow \infty$.  We may then Taylor expand the term in the exponential and see that the leading asymptotic behavior of the integral is
$$\Gamma(n,n) \sim n^n e^{-n} \int_0^{\infty} du \: e^{-n u^2/2}  = n^n e^{-n} \sqrt{\frac{\pi}{2 n}}$$
We may find further terms in the asymptotic behavior by Taylor expanding the higher-order terms in the exponential and the term outside the exponential.  The next higher order term is
$$\left ( 1 + \frac{n u^3}{3} \right ) (1-u) \sim 1-u +\frac{n u^3}{3} $$
Evaluating the integrals that result from this expansion, we find the next higher term in the expansion:
$$\Gamma(n,n) = n^n e^{-n}  \left [ \sqrt{\frac{\pi}{2 n}} - \frac{1}{3 n} + O\left ( n^{-3/2} \right ) \right ]$$
