Asymptotic of incomplete Gamma function, $\Gamma(n,n/a)$ Let $a>1$. I need to approximate (for n large)
$$
\Gamma(n,n/a) \approx f(n)
$$
Approximations I found are for the case which the second argument is fixed. Is there any simple formula for this asymptotic?

If $0<\alpha<1$ and $\alpha$ is bounded away from $1$ (e.g., $1 - n^{ - 1/2} \gg \alpha$), then
$$
\Gamma (n,\alpha n) \sim \Gamma (n) + \alpha ^n \frac{{n^{n - 1} e^{ - \alpha n} }}{{1-\alpha}}\left( {1 - \frac{\alpha }{{n(1-\alpha)^2 }} + \frac{{\alpha (2\alpha  + 1)}}{{n^2 (1-\alpha  )^4 }} -  \cdots } \right)
$$
as $n\to +\infty$ (cf. http://dlmf.nist.gov/8.11.E6). If $\alpha$ is close to $1$, one may use
\begin{align*}
\frac{{\Gamma (n,n - \beta \sqrt n )}}{{\Gamma (n)}} \sim 1 & - \frac{1}{2}\operatorname{erfc}\left( {\frac{\beta }{{\sqrt 2 }}} \right) \\ &+ \frac{1}{{\sqrt {2\pi n} }}\exp \left( { - \frac{{\beta ^2 }}{2}} \right)\left( {\frac{{\beta ^2  - 1}}{3} - \frac{{\beta (2\beta ^4  - 11\beta ^2  + 3)}}{{36\sqrt n }} +  \cdots } \right)
\end{align*}
as $n\to +\infty$ with $0 < \beta \ll n^{1/6}$. Here $\operatorname{erfc}$ denotes the complementary error function (cf. https://doi.org/10.1007/s00365-018-9445-3).
