Does the integral $\int_{2}^{\infty}\frac{dx}{\sqrt{1+x^{3}}}$ exists? I need to prove this. I need your help to verify that my proof is correct (or not) please.

Prove that this integral exists: \begin{align}
 \int_{2}^{\infty}\frac{dx}{\sqrt{1+x^{3}}} \end{align}

My attempt:
Fist we need to observe that $\frac{1}{\sqrt{1+x^{3}}}<\frac{1}{\sqrt{x^{3}}}\Longrightarrow \int_{2}^{\infty}\frac{dx}{\sqrt{1+x^{3}}}<\int_{2}^{\infty}\frac{dx}{\sqrt{x^{3}}}$
Next, we note that $\lim_{x \rightarrow \infty} \frac{\frac{1}{\sqrt{1+x^{3}}}}{\frac{1}{\sqrt{x^{3}}}}=1 \Longrightarrow$ for $f(x)=\frac{1}{\sqrt{1+x^{3}}}$, $g(x)=\frac{1}{\sqrt{x^{3}}}$ both integrals converges or both diverges.
By last, integrals of the form $\int_{1}^{\infty}\frac{dx}{x^{p}}$ converges if $p>1$, $\Longrightarrow \int_{1}^{\infty}\frac{dx}{\sqrt{x^{3}}}$  converges $\Longrightarrow \int_{2}^{\infty}\frac{dx}{\sqrt{x^{3}}}$ converges
That implies that, $\int_{2}^{\infty}\frac{dx}{\sqrt{1+x^{3}}}$ converges, therefore it exists.
Is it correct? Is there another way to prove it? Thank you very much

Your solution is correct, though you didn't really need to use the limit comparison test. You could just stop after the first line. Since $\frac{1}{\sqrt{1+x^3}}\leq\frac{1}{\sqrt{x^3}}$ and the integral $\int_2^{\infty}\frac{1}{\sqrt{x^3}}dx$ converges, we know from the usual comparison test that $\int_2^{\infty}\frac{1}{\sqrt{1+x^3}}dx$ converges as well. Of course it is important to note that these are nonnegative functions in the solution.
