Answer
Converges Absolutely
Work Step by Step
The Integral Test states that the series converges iff the integral $\int_a^\infty f(x) dx$ converges.
Since, we have $\int_a^\infty f(x) dx=\int_1^\infty \dfrac{\tan^{-1} x}{1+x^2} dx$
Use u-substitution method, plug in $u =\tan^{-1} x \implies du=\dfrac{dx}{1+x^2}$
Now, $\int_{\tan^{-1}(1)}^{\tan^{-1}(\infty)} \dfrac{\tan^{-1} x}{1+x^2} dx=\int_{\pi/4}^{\pi/2} u du=[\dfrac{u^2}{2}]_{\pi/4}^{\pi/2}=\bf{Finite}$
Hence, the series Converges Absolutely by the Integral test.