Answer
Divergent
Work Step by Step
Consider $f(x)=\dfrac{\ln x^2}{x}$ This is positive, continuous for $x \geq 2$ and $f(x)$ is decreasing for $x \gt 3$
Now, take the integral test to find the convergence and divergence for the sequence.
We have $\int_3^\infty \dfrac{\ln x^2}{x}dx= \lim\limits_{a \to \infty} \int_3^a \dfrac{\ln x^2}{x}dx$
or, $ \lim\limits_{a \to \infty} [2(\ln x)]_3^a=\lim\limits_{a \to \infty} [2(\ln a)-2(\ln 3)]= \infty$
Hence, the sequence $\Sigma_{n=2}^\infty \dfrac{\ln n^2}{n}=\dfrac{\ln 4}{2}+\Sigma_{n=3}^\infty \dfrac{\ln n^2}{n}$ is Divergent.
