Answer
Convergent
Work Step by Step
Define $f(x)=\frac{\ln x}{x^3}$. This function is continuous and decreasing on $[2,\infty)$.
Evaluate $\int_2^\infty f(x) dx$:
$\int_2^\infty \frac{\ln x}{x^3}dx=\int_2^\infty \ln x\cdot \frac{1}{x^3}dx$ (Let $u=\ln x$ and $dv=\frac{1}{x^3}dx$ $\to du=\frac{1}{x}dx$ and $v=-\frac{1}{2x^2}$)
$=\ln x\cdot (-\frac{1}{2x^2})]_2^\infty -\int_2^\infty -\frac{1}{2x^2}\cdot \frac{1}{x}dx$ (By Integration by Parts)
$=-\frac{\ln x}{2x^2}]_2^\infty -\int_2^\infty -\frac{1}{2x^3}dx$
$=0+\frac{\ln 2}{8}-[\frac{1}{4x^2}]_2^\infty$
$=\frac{\ln 2}{8}-(0-\frac{1}{8})$
$=\frac{\ln 2+1}{8}$
It means that the integral $\int_2^\infty f(x)dx$ is convergent.
Using The Integral Test for Series, $\sum_{n=1}^\infty \frac{\ln n}{n^3}=\sum_{n=2}^\infty \frac{\ln n}{n^3}$ is convergent.
