Answer
Converges
Work Step by Step
Let $u_n=\Sigma_{n=1}^\infty\dfrac{\tan^{-1} n}{n^{1.1}}$
We compare with: $\Sigma_{n=1}^\infty\dfrac{\tan^{-1} (n)}{n^{(1.1)}} \lt \Sigma_{n=1}^\infty \dfrac{\dfrac{\pi}{2}}{n^{(1.1)}}$
Here, $\Sigma_{n=1}^\infty \dfrac{\pi/2}{n^{(1.1)}}$ shows a convergent $p$-series with $p=1.1 \gt 1$ .
Thus, the series converges by the direct comparison test.