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