## Thomas' Calculus 13th Edition

Let $a_n=\dfrac{1}{\sqrt {n(n+1)(n+2)}}$ Applying comparison, we have $\Sigma_{n=1}^\infty \dfrac{1}{\sqrt {n(n+1)(n+2)}} \lt \Sigma_{n=1}^\infty \dfrac{1}{\sqrt {(n)(n)(n)}}=\Sigma_{n=1}^\infty \dfrac{1}{{\sqrt [3] {n}}}$ $\implies a_n=\Sigma_{n=1}^\infty \dfrac{1}{{\sqrt [3] {n}}}=\Sigma_{n=1}^\infty \dfrac{1}{n^{(3/2)}}$ Here, the series $\Sigma_{n=1}^\infty \dfrac{1}{n^{3/2}}$ shows a convergent p-series with $p=\dfrac{3}{2} \gt 1$ . Thus, the series Converges Absolutely by the comparison test.