Answer
Convergent
Work Step by Step
Let $f(x)=\frac{1}{x(\ln x)^2}$. This function is continuous and decreasing on $[2,\infty)$.
Evaluate $\int_2^\infty f(x)dx$:
$\int_2^\infty \frac{1}{x(\ln x)^2}dx=\int_2^\infty \frac{1}{(\ln x)^2}\frac{1}{x}dx$ ($u=\ln x\to du=\frac{1}{x}dx$
$=\int_{\ln u}^{\ln \infty}\frac{1}{u^2}du$
$=-\frac{1}{u}]_{\ln 2}^{\ln \infty}$
$=-0-(-\frac{1}{\ln 2})$
$=\frac{1}{\ln 2}$
So, the integral is convergent.
Using the Integral Test for Series, $\sum_{n=2}^\infty\frac{1}{n(\ln n)^2}$ is convergent.
