For $n \to \infty$, prove $\int_{2n}^{\infty} t^n e^{-t}\ dt = o(\Gamma(n+1))$.
I have no clue to resolve this.
I started like this : $\Gamma(n+1) =\int_{0}^{\infty} t^n e^{-t} dt$ and $\Gamma(n+1)= n\Gamma(n)$ but now I am stuck. I am not allowed to use Stirling's Approximation.
Could anyone help me?
In the interval $[2n, +\infty)$ we have (by splitting $e^{-t} = e^{-t/2} e^{-t/2}$) that $e^{-t} \leq e^{-n} e^{-t/2}$. Thus, by a change of variables $u = t/2$, $$ \int_{2n}^{+\infty} t^n e^{-t} dt \leq e^{-n} \int_{2n}^{+\infty} t^n e^{-t/2} dt = 2 \cdot 2^n e^{-n} \int_{n}^{+\infty} u^n e^{-u} du \leq 2\cdot 2^n e^{-n} \Gamma(n+1). $$
Hence, $$ \lim_{n \to \infty} \frac{\int_{2n}^{+\infty} t^n e^{-t} dt}{\Gamma(n+1)} = 0 $$
because $e > 2$.