91Ó°ÊÓ

First, let's find the absolute value of the terms in the series: $$\left| \frac{(-1)^{k}}{\sqrt{k}} \right| = \frac{1}{\sqrt{k}}$$ Now, we have a new non-alternating series: $$\sum_{k=1}^{\infty} \frac{1}{\sqrt{k}}$$ We can use the p-series test to see if the series converges. A p-series is given by the formula: $$\sum_{k=1}^{\infty} \frac{1}{k^p}$$ where p is a constant. If $$p > 1$$, the series converges, and if $$p \leq 1$$, the series diverges. In our case, we have $$p = \frac{1}{2}$$. Since $$\frac{1}{2} \leq 1$$, the new series diverges. Therefore, our original series does not converge absolutely.

Since the original series did not converge absolutely, let's test for conditional convergence using the Alternating Series Test. An alternating series is defined as: $$\sum_{k=1}^{\infty} (-1)^{k} a_k$$ where $$a_k > 0$$. The Alternating Series Test states that the series converges if: 1. $$a_{k+1} \leq a_k$$ for all k. 2. $$\lim_{k\to\infty} a_k = 0$$ Our original series is an alternating series with $$a_k = \frac{1}{\sqrt{k}}$$. Let's check the two conditions: 1. $$\frac{1}{\sqrt{k+1}} \leq \frac{1}{\sqrt{k}}$$ for all k. This condition is met because as k increases, the denominator of$$\frac{1}{\sqrt{k}}$$ also increases, resulting in a smaller fraction. 2. $$\lim_{k\to\infty} \frac{1}{\sqrt{k}} = 0$$ This limit approaches zero as k goes to infinity. Therefore, the second condition is also met. Since both conditions are met, the Alternating Series Test tells us that the original series converges conditionally.

The given series $$\sum_{k=1}^{\infty} \frac{(-1)^{k}}{\sqrt{k}}$$ converges conditionally, since it does not converge absolutely but meets the conditions for the Alternating Series Test.