## Thomas' Calculus 13th Edition

Let $u_n=\sqrt{\dfrac{n+4}{n^4+4}}$ When the numerator increases, the value of fraction is always increasing and also, when the denominator decreases, the value of fraction is always increasing. $\sqrt{\dfrac{n+4}{n^4}} \leq \sqrt{\dfrac{n+4n}{n^4}}$ $u_n \leq \sqrt{\dfrac{n+4}{n^4}}$ $\implies u_n \leq \sqrt{\dfrac{(n+4n)}{n^4}}=\sqrt{\dfrac{n(1+4)}{n^4}}=\sqrt{(\dfrac{5}{n^3}})$ we can see that $\Sigma_{n=1}^\infty \sqrt{(\dfrac{5}{n^3})}=(\sqrt 5)\Sigma_{n=1}^\infty \dfrac{1}{n^{3/2}}$ shows a convergent p- series with $p=\dfrac{3}{2} \gt 1$. Hence, the series converges by the comparison test.