91Ó°ÊÓ

First, let's consider the absolute value of the terms in the series: $$ a_k = \left|\frac{(-1)^k}{k^{2/3}}\right| = \frac{1}{k^{2/3}} $$

Now we will test for absolute convergence. To do so, we want to examine the series formed by the absolute values of the terms in our given series: $$ \sum_{k=1}^{\infty} a_k = \sum_{k=1}^{\infty} \frac{1}{k^{2/3}} $$ To determine if the series converges or diverges, we can apply the p-series test, as it looks like a p-series. A p-series converges if the p value is greater than 1: $$ \text{p-series}: \sum_{k=1}^{\infty} \frac{1}{k^p} $$ In our case, p = 2/3. Since p < 1, the series diverges. Thus, the given series does not converge absolutely.

Since the series does not converge absolutely, we will check for conditional convergence using the Alternating Series Test. The AST states that a series converges if it meets two criteria: 1. The sequence of absolute values of terms, \(\{a_k\}\), is decreasing. 2. The limit as n goes to infinity of the absolute values of terms is 0: \(\lim_{k\rightarrow\infty} a_k = 0\). We already have our sequence of absolute values from Step 1. To check the first criterion, let's find \(a_{k+1}\): $$ a_{k+1} = \frac{1}{(k+1)^{2/3}} $$ Since the exponent 2/3 is positive, we can say that the sequence is decreasing. Therefore, the first criterion is satisfied. Now, let's check the second criterion by finding the limit of \(a_k\) as \(k\rightarrow\infty\): $$ \lim_{k\rightarrow\infty} a_k = \lim_{k\rightarrow\infty} \frac{1}{k^{2/3}} = 0 $$ The second criterion is also satisfied. Therefore, our given series converges conditionally by the Alternating Series Test.

The given series \(\sum_{k=1}^{\infty} \frac{(-1)^{k}}{k^{2 / 3}}\) does not converge absolutely but converges conditionally.