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

Consider $a_n=\dfrac{1}{\sqrt {n(n+1)(n+2)}}$ Since, we see that $\Sigma_{n=1}^\infty \dfrac{1}{\sqrt {n(n+1)(n+2)}} \lt \Sigma_{n=1}^\infty \dfrac{1}{\sqrt {(n)(n)(n)}}=\Sigma_{n=1}^\infty \dfrac{1}{{\sqrt [3] {n}}}$ Now, $\Sigma_{n=1}^\infty \dfrac{1}{{\sqrt [3] {n}}}=\Sigma_{n=1}^\infty \dfrac{1}{n^{3/2}}$ Thus, the series $\Sigma_{n=1}^\infty \dfrac{1}{n^{3/2}}$ is a convergent p-series. Hence, the given series Converges Absolutely by the direct comparison test.