Asymptotic of incomplete Gamma function, $\Gamma(n,n/a)$

Question

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?

Answer 1

$\Gamma(n, n/a)$ and $\Gamma(n)$ will become very close (in the sense that their ratio converges to $1$) as $n$ grow. The following computation provides a more precise error bound for their ratio:

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I will write $\alpha = 1/a$ so that $0 < \alpha < 1$. Then

$$ \Gamma(n, \alpha n) = \Gamma(n) \biggl[ 1 - \frac{1}{\Gamma(n)} \int_{0}^{\alpha n} t^{n-1}e^{-t} \, \mathrm{d}t \biggr]. $$

Since $t \mapsto t^{n-1}e^{-t}$ is increasing on $0 \leq t \leq n-1$, for large $n$ we have

$$ \int_{0}^{\alpha n} t^{n-1}e^{-t} \, \mathrm{d}t \leq (\alpha n)^n e^{-\alpha n}. $$

So by the Stirling's approximation,

$$ \frac{1}{\Gamma(n)} \int_{0}^{\alpha n} t^{n-1}e^{-t} \, \mathrm{d}t \leq \frac{(\alpha n)^n e^{-\alpha n}}{\sqrt{2\pi} \, n^{n-\frac{1}{2}} e^{-n}} = \frac{1}{\sqrt{2\pi}} n^{1/2} (\alpha e^{1-\alpha})^n . $$

Similarly, by noting that

$$ \int_{0}^{\alpha n} t^{n-1}e^{-t} \, \mathrm{d}t \geq \int_{\alpha n-1}^{\alpha n} t^{n-1}e^{-t} \, \mathrm{d}t \geq (\alpha n - 1)^{n-1} e^{-\alpha n+1}, $$

we have

$$ \frac{1}{\Gamma(n)} \int_{0}^{\alpha n} t^{n-1}e^{-t} \, \mathrm{d}t \geq c_{\alpha} n^{-1/2} (\alpha e^{1-\alpha})^{n} $$

for some constant $c_{\alpha} \in (0, 1)$ depending only on $\alpha$. This tells that

$$ \Gamma(n, \alpha n) = \Gamma(n) \left(1 - n^{\Theta(1)}(\alpha e^{1-\alpha})^n \right), $$

where $\Theta(1)$ represents a bounded sequence in $n$.

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Remarks. A less precise asymptotic formula $\Gamma(n, \alpha n) = \Gamma(n)(1 - o(1))$ can be obtained by applying the central limit theorem to

$$ \frac{\Gamma(n, \alpha n)}{\Gamma(n)} = \mathbf{P}(\tau_1 + \cdots + \tau_n \geq \alpha n), $$

where $\tau_k$'s are independent $\operatorname{Exp}(1)$ variables.

Answer 2

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).