In order to derive Sterling's approximation, I need to show that the following integral decays quicker than at least $\mathcal{O}(n^2)$: $\lim_{n\to\infty}\dfrac{\int_{2n}^\infty x^ne^{-x}dx}{\int_{0}^{2n} x^ne^{-x}dx}$ is at most $\mathcal{O}(n^2)$, this integral can be written as $\lim_{n\to\infty}\dfrac{\Gamma(n+1,2n)}{\gamma(n+1,2n)}$ in terms of incomplete gamma functions. I have been trying analytical methods since a month but to no use. I tried to plot the ratio of this ratio to $\frac{1}{n^6}$ i.e. $\dfrac{n^6\int_{2n}^\infty x^ne^{-x}dx}{\int_{0}^{2n} x^ne^{-x}dx}$ vs n as shown below and so I am sure that it decays at least as quick as $\mathcal{O}(n^6)$, maybe it decays exponentially, however I need to produce an analytical upper bound for ratio. enter image description here

Is there some light? or some identity on incomplete gamma functions?


1 Answer 1


We can use some fairly brutal estimates. On the one hand, \begin{align} \int_0^{2n} x^n e^{-x}\,dx &> \int_n^{2n} x^n e^{-x}\,dx \\ &> n^n \int_n^{2n} e^{-x}\,dx \\ &= n^ne^{-n}(1-e^{-n}) \\ & > \frac{n^n}{2e^n} \end{align} for $n \geqslant 1$.

On the other hand, with $g(x) = x^n e^{-x/2}$ we have $$g'(x) = \biggl(\frac{n}{x} - \frac{1}{2}\biggr)g(x) \leqslant 0$$ for $x \geqslant 2n$, whence \begin{align} \int_{2n}^{\infty} x^ne^{-x}\,dx &= \int_{2n}^{\infty} g(x) e^{-x/2}\,dx \\ &\leqslant g(2n) \int_{2n}^{\infty} e^{-x/2}\,dx \\ &= 2g(2n)e^{-n} \\ &= 4\cdot \frac{n^n}{2e^n}\cdot \biggl(\frac{2}{e}\biggr)^n\,. \end{align} Hence $$\frac{\Gamma(n+1,2n)}{\gamma(n+1,2n)} \leqslant 4\biggl(\frac{2}{e}\biggr)^n\,,$$ i.e. we have exponential decay.


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