Question

The given series \( 1 - \frac{1}{2^p} + \frac{1}{3^p} - \frac{1}{4^p} + ... (p>0) \) is conditionally convergent, if 'p' lies in the interval:

• A. \( (0, 1] \)
• B. \( [0, 1] \)
• C. \( (1, \infty) \)
• D. \( [1, \infty) \)

Hint:

For an alternating p-series \( \sum (-1)^{n-1}/n^p \):
\(p>1\): Absolutely convergent.
\(0<p \le 1\): Conditionally convergent.
\(p \le 0\): Divergent (as the terms don't go to zero). Memorizing these conditions is very useful for exam questions on series.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A series is conditionally convergent if it converges, but the series of its absolute values diverges. The given series is an alternating series of the form \( \sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{n^p} \). 

Step 2: Key Formula or Approach: 
1. Check for convergence: We use the Leibniz test (Alternating Series Test). The series converges if the terms \( \frac{1}{n^p} \) are positive, decreasing, and have a limit of 0. 2. Check for absolute convergence: We examine the series of absolute values, which is \( \sum_{n=1}^{\infty} \left| (-1)^{n-1} \frac{1}{n^p} \right| = \sum_{n=1}^{\infty} \frac{1}{n^p} \). This is a p-series. 

Step 3: Detailed Explanation: 
Convergence of the alternating series: The terms \( a_n = \frac{1}{n^p} \) are positive for all \(n \ge 1\). Since \(p>0\), the denominator \( n^p \) increases as \(n\) increases, so the terms \( a_n \) are decreasing. The limit is \( \lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{1}{n^p} = 0 \) because \(p>0\). By the Leibniz test, the alternating series converges for all \( p>0 \). Absolute convergence: The series of absolute values is the p-series \( \sum_{n=1}^{\infty} \frac{1}{n^p} \). The p-series test states that this series: - Converges if \( p>1 \). - Diverges if \( p \le 1 \). Conditional Convergence: For the original series to be conditionally convergent, it must converge but not converge absolutely. - We need the alternating series to converge, which happens for \( p>0 \). - We need the series of absolute values to diverge, which happens for \( p \le 1 \). Combining these two conditions, we need \( p>0 \) AND \( p \le 1 \). This corresponds to the interval \( (0, 1] \). 

Step 4: Final Answer: 
The series is conditionally convergent if p lies in the interval \( (0, 1] \).