Answer
The series $\Sigma_{k=2}^{\infty} \dfrac{1}{k \ln k}$ also diverges by using integral test.
Work Step by Step
Here, we have
$\int_2^\infty \dfrac{1}{x \ln x} \ dx=\lim\limits_{b \to \infty} \int_2^b (\ln x)^{-1} d (\ln x)=\lim\limits_{b \to \infty} \ln (\ln x)|_2^b$
or, $=\lim\limits_{b \to \infty} [\ln (\ln b)-\ln (\ln 2)]$
or, $=\infty$
Thus, it has been proved that the series $\Sigma_{k=2}^{\infty} \dfrac{1}{k \ln k}$ also diverges by using integral test.
