#### Answer

conditionally convergent.

#### Work Step by Step

The alternating series test states that if the following conditions are met then the series is convergent.
1. $\lim\limits_{ n\to \infty} b_n=0$
2.$b_n$ is a decreasing sequence.
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 $\Sigma_{n=1}^{\infty}|\dfrac{(-1)^n}{\sqrt n}|$
The given series can be re-written as: $a_n=\Sigma_{n=1}^{\infty}\dfrac{1}{ n^{1/2}}$
Here, $p=1/2=0.5 \lt 1$
Thus, the given series is divergent. So, it can not be absolutely convergent.
From the given series, we get
1. $\lim\limits_{ n\to \infty} b_n=\lim\limits_{ n\to \infty} \dfrac{1}{2}\dfrac{1}{ n^{1/2}}=0$
2.$b_n=\dfrac{1}{\sqrt n}$is a decreasing sequence.
Therefore, the given series is conditionally convergent.