Answer
Divergent
Work Step by Step
The integral Test:
Let $f(x)=\frac{1}{x\sqrt {lnx}}$
$f(x)$ is continuous, positive, and decreasing on $[2,\infty)$
$\int _{2}^{\infty}f(x)dx=\int _{2}^{\infty}\frac{1}{x\sqrt {lnx}}dx$
$=\int _{2}^{\infty}\frac{1}{\sqrt {lnx}}d(lnx)$
$=2\sqrt {lnx}| _{2}^{\infty}$
$=\infty$
Thus, the series is divergent.