Answer
Converges to $\dfrac{1}{\ln 2}$
Work Step by Step
Consider $f(x)=\dfrac{1}{x(\ln x)^2}$. This is positive, continuous and decreasing for $x \geq 1$
Now, take the integral test to find the convergence and divergence for the sequence.
We have $\int_2^\infty \dfrac{1}{x(\ln x)^2} dx= \lim\limits_{a \to \infty} \int_2^a \dfrac{1}{x(\ln x)^2} dx$
or, $ \lim\limits_{a \to \infty} [\dfrac{-1}{\ln x}]_2^a=\lim\limits_{a \to \infty} [\dfrac{-1}{\ln a} +\dfrac{1}{\ln 2}]= \dfrac{1}{\ln 2}$
Hence, the sequence $\Sigma_{n=2}^\infty \dfrac{1}{n(\ln n)^2}$ converges to $\dfrac{1}{\ln 2}$
