Since $\arctan \left(n\right)\to \mathrm{\π}\/2$ as $n\to \infty$, it would seem that comparing this series to the convergent $p$-series $\sum _{n\=1}^{\infty }\frac{1}{n^2}$ might be a useful ploy, in which case consider

$\underset{n\to \infty }{lim}\frac{\left(\frac{\arctan \left(n\right)}{n^2\+4}\right)}{1\/n^2}\=\mathrm{\π}\/2\>0$ 

By part (1) of the Limit-Comparison test (Table 8.3.1), both series either converge or diverge. But the $p$-series converges, so the given series converges. Since the given series has only positive terms, the convergence is absolute.

As per the Mathematical Solution above, the absolute convergence of the given series can be established by the Limit-Comparison test, by a comparison to the convergent $p$-series $\sum _{n\=1}^{\infty }\frac{1}{n^2}$. Absolute convergence can also be established by the Integral test, but the integration is not elementary.

The "0.I" appended to $0.53317$ is an artifact of the numeric approximation of the exact answer Maple finds for this integral. In essence, the value of the integral is a real, positive number, and by the terms of the Integral test, the given series converges absolutely.