Can we determinine the convergence of $\int_0^\infty \frac{x^{2n - 1}}{(x^2 + 1)^{n + 3}}\,dx$ without evaluating it? 
Can we determine convergence without evaluating this improper integral? $$\int_0^\infty  {\frac{x^{2n - 1}}{{\left( x^2 + 1 \right)}^{n + 3}}\,dx}\quad\quad n\geq 1\;,\; n\in\mathbb{Z}$$

When trying to bound the integrand, relating a known function, the integral does not converge.
But when I use the software package Maple I see that the integral converges to:
$$\int_0^\infty  {\frac{x^{2n - 1}}{{\left( x^2 + 1 \right)}^{n + 3}}\,dx}=\frac{1}{{\left( {n + 1} \right)\left( {n + 2} \right)n}}$$
Evaluating it is not difficult, but I want to know if you can determine convergence (delimiting the integrand) without evaluating such an improper integral. 
 A: The numerator is of degree $2n - 1$. The denominator looks like a polynomial of degree $2n + 6$, and is bounded away from $0$. As a general rule, integrals of the form
$$ \int_0^\infty \frac{1}{(1 + x)^n}dx$$
converge for $n > 1$ and diverge for $n \leq 1$. Your integral looks like this integral, but with $n = 7$. So it converges. You can make this argument more formal through limit comparison, if that's something you're familiar with.
A: By equivalents:
$$\frac{x^{2\;n - 1}}{(x^2 + 1)^{n + 3}}\sim_{\infty}\frac{x^{2\;n - 1}}{x^{2(n + 3)}}=\frac1{x^7}$$
and the latter is convergent since the exponent of $x$ in the denominator is $>1$.
A: The naive comparison would simply drop the $1$ in the denominator, but the result is not bounded, whereas your integrand is bounded. To fix this, keep the $1$ on, say, $[0,1]$ and then drop it on $[1,\infty)$. The effect is that on $[0,1]$ the integrand is bounded while on $[1,\infty)$ the integrand is bounded, and in particular bounded by $Cx^{-7}$ for some $C>0$.
A: Setting $x=\tan(t)$, we obtain $$\dfrac{x^{2n-1}}{\left(x^2+1\right)^{n+3}} = \dfrac{\tan^{2n-1}(t)}{\sec^{2n+6}(t)} = \sin^{2n-1}(t)\cos^7(t)$$
Hence, we have
$$\int_0^{\infty}\dfrac{x^{2n-1}}{\left(x^2+1\right)^{n+3}} dx = \int_0^{\pi/2} \sin^{2n-1}(t)\cos^5(t)dt$$
which clearly exists, since the integrand is continuous and bounded on $[0,\pi/2]$.
Also, what better way to prove convergence than evaluating the integral. The integral can be easily evaluated using the identity here and the definition of the $\beta$ function. We have
$$\beta(x,y) = 2 \int_0^{\pi/2}\sin^{2x-1}(t)\cos^{2y-1}(t)dt$$
which gives us that
$$\int_0^{\pi/2} \sin^{2n-1}(t)\cos^5(t)dt = \dfrac{\beta(n,4)}2 = \dfrac{\Gamma(n)\Gamma(3)}{2\Gamma(n+3)} = \dfrac1{(n+2)(n+1)n}$$
