(6 points) Show that the series $\sum_{n=1}^{\infty} \frac{(-1)^{n}}{\sqrt[3]{n}}$ is conditionally convergent.

Step 1 ：First, we need to check if the series is convergent. We can use the Alternating Series Test, which states that if the terms of the series are decreasing and the limit of the terms as n approaches infinity is 0, then the series is convergent.

Step 2 ：The terms of the series are $(-1)^n/\sqrt[3]{n}$. We can see that as n increases, the absolute value of the terms decreases, so the terms are decreasing.

Step 3 ：Next, we need to find the limit of the terms as n approaches infinity. We have $\lim_{n\to\infty} \frac{(-1)^n}{\sqrt[3]{n}} = 0$ because the denominator grows faster than the numerator.

Step 4 ：Since the terms are decreasing and the limit of the terms is 0, the series is convergent by the Alternating Series Test.

Step 5 ：However, to show that the series is conditionally convergent, we need to show that the series of absolute values is divergent. The series of absolute values is $\sum_{n=1}^{\infty} \frac{1}{\sqrt[3]{n}}$.

Step 6 ：We can use the p-series test to determine the convergence of this series. The p-series test states that the series $\sum_{n=1}^{\infty} \frac{1}{n^p}$ is convergent if p > 1 and divergent if p ≤ 1. In this case, p = 1/3, which is less than 1, so the series of absolute values is divergent.

Step 7 ：Therefore, the original series is conditionally convergent because it is convergent but the series of absolute values is divergent.