Answer
Divergent
Work Step by Step
Let $f(x)=\frac{1}{x\sqrt{\ln x}}$. This function is continuous and decreasing on $[2,\infty)$.
Evaluate $\int_2^\infty f(x)dx$:
$\int_2^\infty \frac{1}{x\sqrt{\ln x}}dx=\int_2^\infty \frac{1}{\sqrt{\ln x}}\cdot \frac{1}{x}dx$ (Let $u=\ln x\to du=\frac{1}{x}dx$)
$=\int_{\ln 2}^{\ln \infty}\frac{1}{\sqrt{u}}du$
$=2\sqrt{u}_{\ln 2}^{\ln \infty}$
$=2\cdot \sqrt{\ln \infty}-2\cdot \sqrt{\ln 2}$
$=\infty$
So, the integral is divergent.
Using the Integral Test for Series, $\sum_{n=2}^\infty \frac{1}{n\sqrt{\ln n}}$ is divergent.
