#### Answer

Conditionally convergent

#### Work Step by Step

First we will check whether the series is absolutely convergent. Because If the series is absolutely convergent then the series is convergent. But if the series is not absolutely convergent, then we will check whether the series is convergent or divergent.
Remember that $\Sigma a_{n}$ is said to be absolutely convergent
if $\Sigma |a_{n}|$ is convergent.
Therefore, the given series will be absolutely convergent when the following series converges.
$\Sigma |(-1)^{n-1}n^{-1/3}|=\Sigma _{n=1}^\infty n^{-1/3}=\Sigma _{n=1}^\infty \frac{1}{n^{1/3}}$
This is a p-series with $p=\frac{1}{3} \lt 1$
Hence, the series diverges and the original series given in the question is not absolutely converges.
The Comparison Test states that the p-series $\sum_{n=1}^{\infty}\frac{1}{n^{p}}$ is convergent if $p\gt 1$ and divergent if $p\leq 1$.
Now we want to see whether the series is conditionally convergent or divergent.
Alternating Series Test
Consider the series $\Sigma a_{n}$ where, $a_{n}=(-1)^{n}b^n$ or$(-1)^{n+1}b^n$
Then the series is convergent if
1. $\lim\limits_{n \to \infty} b_{n}=0$
2. $b_{n}$ is positive.
3. $b_{n}$ is decreasing.
Note that $\lim\limits_{n \to \infty}\frac{1}{n^{1/3}}=\frac{1}{\infty}=0$
and $\frac{1}{n^{1/3}}$ is always positive for $n\geq 1$
Also, $\frac{1}{n^{1/3}}$ is decreasing because $n^{1/3}$ is increasing for $n\geq 1$.
Therefore, by AST , the given series converges conditionally.
Hence, the given series is conditionally convergent.