#### Answer

Conditionally Convergent

#### Work Step by Step

A series $ \Sigma a_n$ is defined as absolutely convergent when the series $ \Sigma |a_n|$ is convergent.
From the series, we notice that
$ \Sigma_{n=1}^{\infty} |(-1)^n(\sqrt {n+1}-\sqrt n)|= \Sigma_{n=1}^{\infty} (\sqrt {n+1}-\sqrt n)=\Sigma_{n=2}^{\infty} \dfrac{1}{n^{1/2}}$
This corresponds to a p-series with common ratio $r=\dfrac{1}{2} \lt 1$.
Remember that a p-series is said to be convergent when the common ratio $ r \gt 1$.
This implies that the series is Not Absolutely Convergent.
The Alternating Series Test states that a series $\Sigma a_n$ such that $p_n=(-1)^n q_n$; $q_n \geq 0$ for all $n$
If the following conditions are satisfied then the series converges:
a) $\lim\limits_{n \to \infty} q_n=0$;
b) $q_n$ is a decreasing sequence.
We have: $q_n=\dfrac{1}{(\sqrt {n+1}-\sqrt n)}$
a) $\lim\limits_{n \to \infty} q_n=\lim\limits_{n \to \infty}\dfrac{1}{(\sqrt {n+1}-\sqrt n)}=0$
b) The series is a decreasing sequence.
This implies that the given series is Conditionally Convergent by the Alternating Series Test.