Answer
Convergent
Work Step by Step
Since, we have $\lim\limits_{x \to \infty} \int_1^\infty\dfrac{8 \tan^{-1} x}{1+x^2}$
Suppose $tan^{-1} x= p \implies dp=\dfrac{dx}{1+x^2}$
or, $=\lim\limits_{a \to \infty} [8p]_{\pi/4}^{(\pi/2)}$
or, $ =4(\dfrac{\pi^2}{4}-\dfrac{\pi^2}{16})$
or, $=\dfrac{3\pi^2}{4}$
Hence, the series is Convergent by the Integral Test
