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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Question
![### Problem Statement
**Determine convergence or divergence by any method.**
\[
\sum_{n=0}^{\infty} \frac{(-1)^n n}{\sqrt{n^2 + 8}}
\]
### Explanation
The problem presented is to determine whether the given infinite series converges or diverges. The series is as follows:
\[
\sum_{n=0}^{\infty} \frac{(-1)^n n}{\sqrt{n^2 + 8}}
\]
### Steps to Solve
1. **Identify the general term:**
\[
a_n = \frac{(-1)^n n}{\sqrt{n^2 + 8}}
\]
2. **Apply the necessary tests for convergence or divergence.** Given the alternating sign \((-1)^n\), the Alternating Series Test might be appropriate, and we might also consider other tests if needed.
### Key Points of the Series
- **Alternating Series Test** can be applied if the series has the form \(\sum (-1)^n b_n\) where \(b_n\) is positive, decreasing, and approaches zero. In this case, \(b_n = \frac{n}{\sqrt{n^2 + 8}}\).
- **Behavior of \(b_n\)**:
- As \(n\) increases, \(n^2\) will dominate over the constant 8, making the fraction \(\frac{n}{\sqrt{n^2 + 8}}\) approach 1.
- Thus, \(\lim_{n \to \infty} \frac{n}{\sqrt{n^2 + 8}} = 1\).
- If the limit of \(b_n\) as \(n\) approaches infinity is not zero, the series does not satisfy all the conditions for the Alternating Series Test, suggesting divergence.
### Conclusion
Using these steps and observations, it can be evaluated whether the given series converges or diverges. In this particular example, we observe that as \(n\) approaches infinity, the term \(\frac{n}{\sqrt{n^2 + 8}}\) does not approach zero but rather approaches 1. Hence, this suggests that the series does not converge and therefore diverges.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1542e5de-e392-4bdb-9d95-e5abdf8e6267%2Fe6eb5fb6-f262-4574-8200-f712dbb808ef%2Fyp72diq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement
**Determine convergence or divergence by any method.**
\[
\sum_{n=0}^{\infty} \frac{(-1)^n n}{\sqrt{n^2 + 8}}
\]
### Explanation
The problem presented is to determine whether the given infinite series converges or diverges. The series is as follows:
\[
\sum_{n=0}^{\infty} \frac{(-1)^n n}{\sqrt{n^2 + 8}}
\]
### Steps to Solve
1. **Identify the general term:**
\[
a_n = \frac{(-1)^n n}{\sqrt{n^2 + 8}}
\]
2. **Apply the necessary tests for convergence or divergence.** Given the alternating sign \((-1)^n\), the Alternating Series Test might be appropriate, and we might also consider other tests if needed.
### Key Points of the Series
- **Alternating Series Test** can be applied if the series has the form \(\sum (-1)^n b_n\) where \(b_n\) is positive, decreasing, and approaches zero. In this case, \(b_n = \frac{n}{\sqrt{n^2 + 8}}\).
- **Behavior of \(b_n\)**:
- As \(n\) increases, \(n^2\) will dominate over the constant 8, making the fraction \(\frac{n}{\sqrt{n^2 + 8}}\) approach 1.
- Thus, \(\lim_{n \to \infty} \frac{n}{\sqrt{n^2 + 8}} = 1\).
- If the limit of \(b_n\) as \(n\) approaches infinity is not zero, the series does not satisfy all the conditions for the Alternating Series Test, suggesting divergence.
### Conclusion
Using these steps and observations, it can be evaluated whether the given series converges or diverges. In this particular example, we observe that as \(n\) approaches infinity, the term \(\frac{n}{\sqrt{n^2 + 8}}\) does not approach zero but rather approaches 1. Hence, this suggests that the series does not converge and therefore diverges.
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