Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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### Problem 13
**Determine whether**
\[ \sum_{k=1}^{\infty} \frac{(-1)^k}{k^{\frac{1}{2}}} \]
**converges absolutely, converges conditionally, or diverges.**
---
### Explanation:
This problem requires analyzing the given infinite series to conclude if it converges either absolutely, conditionally, or diverges. Here’s a step-by-step approach to solving such problems:
1. **Absolute Convergence**:
- Investigate the absolute value of the series:
\[ \sum_{k=1}^{\infty} \left|\frac{(-1)^k}{k^{\frac{1}{2}}}\right| = \sum_{k=1}^{\infty} \frac{1}{k^{\frac{1}{2}}} \]
- This transforms the given series into a p-series. Recall that a p-series \(\sum_{k=1}^{\infty} \frac{1}{k^p}\) converges if and only if \(p > 1\). Here, \(p = \frac{1}{2}\), which is less than 1, so the series \(\sum_{k=1}^{\infty} \frac{1}{k^{\frac{1}{2}}}\) diverges. Therefore, the given series does not converge absolutely.
2. **Conditional Convergence**:
- To determine conditional convergence, use the Alternating Series Test (Leibniz Test):
- Confirm that the series is in alternating form and that \(a_k = \frac{1}{k^{\frac{1}{2}}}\) is positive, monotonic decreasing, and approaches zero as \(k\) approaches infinity.
- As \(k\) increases, \(a_k = \frac{1}{k^{\frac{1}{2}}}\) decreases and \(\lim_{k \to \infty} \frac{1}{k^{\frac{1}{2}}} = 0\).
Given that these criteria are satisfied, the series converges conditionally.
---
Thus, the series \(\sum_{k=1}^{\infty} \frac{(-1)^k}{k^{\frac{1}{2}}}\) **converges conditionally**.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fff7f42fd-3734-43a2-a946-a65fb76095a7%2F79b9556b-a754-4f9f-b927-f15679a9eb87%2Favh71m4_processed.jpeg&w=3840&q=75)
Transcribed Image Text:---
### Problem 13
**Determine whether**
\[ \sum_{k=1}^{\infty} \frac{(-1)^k}{k^{\frac{1}{2}}} \]
**converges absolutely, converges conditionally, or diverges.**
---
### Explanation:
This problem requires analyzing the given infinite series to conclude if it converges either absolutely, conditionally, or diverges. Here’s a step-by-step approach to solving such problems:
1. **Absolute Convergence**:
- Investigate the absolute value of the series:
\[ \sum_{k=1}^{\infty} \left|\frac{(-1)^k}{k^{\frac{1}{2}}}\right| = \sum_{k=1}^{\infty} \frac{1}{k^{\frac{1}{2}}} \]
- This transforms the given series into a p-series. Recall that a p-series \(\sum_{k=1}^{\infty} \frac{1}{k^p}\) converges if and only if \(p > 1\). Here, \(p = \frac{1}{2}\), which is less than 1, so the series \(\sum_{k=1}^{\infty} \frac{1}{k^{\frac{1}{2}}}\) diverges. Therefore, the given series does not converge absolutely.
2. **Conditional Convergence**:
- To determine conditional convergence, use the Alternating Series Test (Leibniz Test):
- Confirm that the series is in alternating form and that \(a_k = \frac{1}{k^{\frac{1}{2}}}\) is positive, monotonic decreasing, and approaches zero as \(k\) approaches infinity.
- As \(k\) increases, \(a_k = \frac{1}{k^{\frac{1}{2}}}\) decreases and \(\lim_{k \to \infty} \frac{1}{k^{\frac{1}{2}}} = 0\).
Given that these criteria are satisfied, the series converges conditionally.
---
Thus, the series \(\sum_{k=1}^{\infty} \frac{(-1)^k}{k^{\frac{1}{2}}}\) **converges conditionally**.
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