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

A $p$-series has the form of $\Sigma_{n=k}^{\infty}\dfrac{1}{n^p}$.It is convergent iff $p \gt 1$ otherwise diverges. From the given problem, we have $\Sigma_{n=1}^{\infty}\dfrac{1}{\sqrt n}$ The given series can be re-written as: $\Sigma_{n=1}^{\infty}\dfrac{1}{ n^{0.5}}$ Here, $p=0.5 \lt 1$ Thus, the given series divergent.