Answer
converges to $\dfrac{1}{\ln 2}$
Work Step by Step
Let us consider $f(x)=\dfrac{1}{x(\ln x)^2}$.
Here, the function $f(x)$ is positive, continuous and decreasing for $x \geq 1$
Then $\int_2^\infty \dfrac{1}{x(\ln x)^2} dx= \lim\limits_{k \to \infty} \int_2^k \dfrac{1}{x(\ln x)^2} dx=\lim\limits_{k \to \infty} [\dfrac{-1}{\ln x}]_2^k$
and $\lim\limits_{k \to \infty} [\dfrac{-1}{\ln k} +\dfrac{1}{\ln 2}]= \dfrac{1}{\ln 2}$
Thus, the sequence $\Sigma_{n=2}^\infty \dfrac{1}{n(\ln n)^2}$ converges to $\dfrac{1}{\ln 2}$.
