Calculus: Early Transcendentals 8th Edition

Use Limit Comparison Test with $a_{n}=\frac{5+2n}{{(1+n^{2})^{2}}}$ and $b_{n}=\frac{1}{n^{3}}$ $\lim\limits_{n \to \infty}\frac{a_{n}}{b_{n}}=\lim\limits_{n \to \infty}\frac{n^{3}(5+2n)}{(1+n^{2})^{2}}$ $=\lim\limits_{n \to \infty}\frac{5n^{3}+2n^{4}}{(1+n^{2})^{2}}$ $=\lim\limits_{n \to \infty}\frac{\frac{5}{n}+2}{(\frac{1}{n^{2}}+1)^{2}}$ $=2\gt 0$ $\Sigma_{n=1}^{\infty}\frac{1}{n^{3}}$ is convergent because a $p-$series with $p=3 \gt 1$ is convergent; thus the series $\Sigma_{n=1}^{\infty}\frac{5+2n}{{(1+n^{2})^{2}}}$ also converges.