Answer
Divergent
Work Step by Step
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.
