Show if the integral converge or diverge For this integral $$ \int_0^{\infty} x^3 e^{-x} dx$$
I need to show if it converges or not!
I calculated it by integration by parts and I found that it converges (this is the only way I know to prove the convergence) but it is needed to prove it  using the convergence tests  without calculating it and I don't know how to apply them here!
 A: We have that the function is continuous and


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*$x\to 0\implies x^3 e^{-x}\to 0$
then we have
$$\int_0^{\infty} x^3 e^{-x} dx=\int_0^{1} x^3 e^{-x} dx+\int_1^{\infty} x^3 e^{-x} dx$$
and the first one is a proper integral while the second one converges by limit comparison test with $\int_1^\infty \frac1{x^2}$, indeed 
$$\frac{x^3 e^{-x}}{\frac1{x^2}}=\frac{x^5}{e^x}\to 0$$
Refer also to the related


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*Limit comparison test for Improper Integrals
A: Alternative approach: since $g(x)=x^3 e^{-x/2}$ is non-negative and bounded by $6^3 e^{-3}$ on $\mathbb{R}^+$ (since $g'(x)$ only vanishes at $x=0$ and $x=6$) we have $\int_{0}^{+\infty}x^3 e^{-x}\,dx = \int_{0}^{+\infty} g(x) e^{-x/2}\,dx \leq 2\cdot 6^3 e^{-3}$.
We may also interpolate this inequality: for any $a\in(0,1)$ we have
$$ \int_{0}^{+\infty}x^3 e^{-x}\,dx \leq \frac{27 e^{-3n/a}}{a^3(1-a)} $$
and by picking the optimal $a=\frac{3}{4}$ we get $\int_{0}^{+\infty}x^3 e^{-x}\,dx\leq\frac{256}{e^3}.$
A further generalization leads to
$$ n! \leq (n+1)^{n+1}e^{-n}. $$
A: Recall:
$e^n =1+n+n^2/2!+...n^5/5!..> n^5/5!$
1) $0 < a_n:=n^3e^{-n} < 5!n^{-2}:=b_n;$
$s_n:=\sum_{k=1}^{n} a_k$ converges
since $r_n:=\sum_{k=1}^{n}b_k$ converges (comparison test). 
2) Integral comparison test for series ;
Let $f: [1,\infty] \rightarrow \mathbb{R^+}$ a monotonically decreasing  function then
$\sum_{n=1}^{\infty}f(n)$ converges $\iff $
$\int_{1}^{\infty}f(x)dx$ converges,
$f(x):=x^3e^{-x}$;
$f(n)=a_n$ in 1) and we are done.
