Answer
Converges Absolutely
Work Step by Step
Consider $\int_a^\infty f(x) dx=\int_1^\infty \dfrac{\tan^{-1} x}{1+x^2} dx$
In order to solve the given integral we will have to take the help of u-substitution method, then plug in $u =\tan^{-1} x$ and $ du=\dfrac{dx}{1+x^2}$
Thus, we have $\int_{\tan^{-1}(1)}^{\tan^{-1}(\infty)} \dfrac{\tan^{-1} x}{(1+x^2)} dx=\int_{(\pi/4)}^{(\pi/2)} (u) du$
or, $[\dfrac{u^2}{2}]_{(\pi/4)}^{(\pi/2)}=$ Finite value
Hence, the series Converges Absolutely by Integral test.
