Answer
Converges
Work Step by Step
Consider $a_n=\sqrt{\dfrac{n+4}{n^4+4}}$
When we increase the numerator, the value of the fraction will always increase:
$ \sqrt{\dfrac{n+4}{n^4}} \leq \sqrt{\dfrac{n+4n}{n^4}} $
Now, we can see that $a_n \leq \sqrt{\dfrac{n+4n}{n^4}}=\sqrt{\dfrac{n(1+4)}{n^4}}=\sqrt{\dfrac{5}{n^3}}$
Here, $\Sigma_{n=1}^\infty \sqrt{\dfrac{5}{n^3}}=\sqrt 5\Sigma_{n=1}^\infty \dfrac{1}{n^{3/2}}$ is a convergent p-series with $p=\dfrac{3}{2} \gt 1$.
Thus, the series converges by the comparison test.