Answer
Convergent
Work Step by Step
Use u-substitution method .
Plug $u=\ln x $ and $du=\dfrac{1}{x} dx$
So, $\int_3^\infty \dfrac{1/x}{(\ln x) \sqrt {(\ln x)^2-1}} dx=\int_{\ln 3}^\infty \dfrac{\dfrac{1}{x}}{u \sqrt {u^2-1}} dx=\lim\limits_{k \to \infty} [\sec^{-1} x]_{\ln 3}^k$
and $\lim\limits_{k \to \infty}[(\sec^{-1} (\dfrac{1}{k}) -\sec^{-1} (\ln 3)]=(\dfrac{\pi}{2})-\sec^{-1} (\ln 3) \approx 1.1439$
Thus, the series is convergent.