Thomas' Calculus 13th Edition

Consider the series $\Sigma_{n=1}^\infty \dfrac{3}{\sqrt n}$ It can be re-written as: $3\Sigma_{n=1}^\infty \dfrac{1}{\sqrt n}=3\Sigma_{n=1}^\infty \dfrac{1}{n^{1/2}}$ This shows a $p$-series with $p=\dfrac{1}{2} \leq 1$ Hence, the series is divergent.