Answer
Convergent
Work Step by Step
Since, we have $\int_3^\infty \dfrac{1/x}{(\ln x) \sqrt {(\ln x)^2-1}} dx$
Suppose $t=\ln x \implies dt=\dfrac{1}{x} dx$
Then $\int_3^\infty \dfrac{1/x}{(\ln x) \sqrt {(\ln x)^2-1}} dx=\int_{\ln 3}^\infty \dfrac{1/x}{t \sqrt {t^2-1}} dx=\lim\limits_{a \to \infty} [\sec^{-1}]_{\ln 3}^a$
or, $=\lim\limits_{a \to \infty}[(\cos^{-1} (1/a) -\sec^{-1} (\ln 3)]$
or, $\dfrac{\pi}{2}-\sec^{-1} (\ln 3) \approx 1.1439$
Hence, the series is convergent.
