Answer
Convergent
Work Step by Step
Plug $\tan^{-1} x= a \implies da=\dfrac{dx}{1+x^2}$
Then $\lim\limits_{x \to \infty} \int_1^\infty\dfrac{8 \tan^{-1} x}{1+x^2} dx=\lim\limits_{k \to \infty} [8a]_{(\pi/4)}^{(\pi/2)}$
so, $ (4)(\dfrac{\pi^2}{4}-\dfrac{\pi^2}{16})=\dfrac{4\pi^2}{4}-\dfrac{4\pi^2}{16}=\dfrac{3\pi^2}{4}$
As per the Integral Test, the series is convergent.