## University Calculus: Early Transcendentals (3rd Edition)

Consider $a_n=\dfrac{1}{\sqrt{n^3+2}}$ and $b_n=\dfrac{1}{ \sqrt{n^3}}$ Now, $\lim\limits_{n \to \infty}\dfrac{a_n}{b_n} =\lim\limits_{n \to \infty}\dfrac{\dfrac{1}{\sqrt{n^3+2}}}{\dfrac{1}{ \sqrt{n^3}}}$ Thus, we have $=\lim\limits_{n \to \infty} \sqrt{\dfrac{n^3}{n^3+2}}$ or, $=1$ Hence, the series converges due to the limit comparison test because $\Sigma_{n=1}^\infty \dfrac{1}{ \sqrt{n^3}}=\dfrac{1}{ n^{3/2}}$ has a p-series with $p=\dfrac{3}{2}$, which also converges.