Hello, how to show the integral diverges:
$\displaystyle \int_{1}^{\infty}$
I would begin with the fact that for $\displaystyle x\in(1,\infty)$ we have:
$\displaystyle \frac{1}{x^{\Large\frac{1}{6}}}< \frac{\sqrt{x}}{\sqrt[3]{x^2-1}}$
Now, let's disregard the singularity at the lower bound, and consider:
$\displaystyle I=\int_2^{\infty} \frac{\sqrt{x}}{\sqrt[3]{x^2-1}}\,dx$
What does the $\displaystyle p$-test tell us about the convergence of $\displaystyle I$?