Absolute convergence, conditional or divergence? $\sum_1^\infty \frac{(-1)^{n+1}}{n^\frac{1}{3}}$ $$\sum_1^\infty \frac{(-1)^{n-1}}{\sqrt[3]{n}}$$
This seems fairly simple to me, I know I can look at the absolute value and if that diverges than I am done.
$$\sum_1^\infty \frac{(-1)^{n-1}}{\sqrt[3]{n}}$$
$p-$ series says that this diverges. Done.
The answer is wrong though, why?

The series is NOT absolutely convergent, as you found using the p-series test.
Recall, your series would be absolutely convergent if and only if the following series were convergent: $$\sum_{n = 1}^\infty \Big|\frac{(-1)^{n-1}}{\sqrt[3]{n}}\Big| = \sum_{n = 1}^\infty \frac{1}{n^{1/3}}$$ But as you observe, this series does not converge.
But our given series is an alternating series, and as such, we can check to see if it satisfies the alternating series test: it does; i.e., it satisfies the Leibniz test. Hence it is conditionally convergent.
