Step 1: Concept Overview:
A conditionally convergent series converges, but its absolute value series diverges. The series in question is an alternating series: \( \sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{n^p} \).
Step 2: Method:
1. Check convergence: Use the Leibniz test (Alternating Series Test). The series converges if \( \frac{1}{n^p} \) terms are positive, decreasing, and approach 0. 2. Check absolute convergence: Examine the absolute value series: \( \sum_{n=1}^{\infty} \left| (-1)^{n-1} \frac{1}{n^p} \right| = \sum_{n=1}^{\infty} \frac{1}{n^p} \), a p-series.
Step 3: Step-by-step Explanation:
Alternating series convergence: The terms \( a_n = \frac{1}{n^p} \) are positive for \(n \ge 1\). With \(p>0\), \( n^p \) increases with \(n\), thus \( a_n \) decreases. The limit is \( \lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{1}{n^p} = 0 \) because \(p>0\). The Leibniz test confirms the alternating series converges for all \( p>0 \). Absolute convergence: The absolute value series is the p-series \( \sum_{n=1}^{\infty} \frac{1}{n^p} \). The p-series test states: - Converges if \( p>1 \). - Diverges if \( p \le 1 \). Conditional Convergence: For conditional convergence: - The alternating series must converge (which occurs for \( p>0 \)). - The absolute value series must diverge (which occurs for \( p \le 1 \)). Combining these, we need \( p>0 \) AND \( p \le 1 \), which is the interval \( (0, 1] \).
Step 4: Answer:
The series is conditionally convergent if \( p \) is in the interval \( (0, 1] \).