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 Statement
**Determine if the following series is absolutely convergent, conditionally convergent, or divergent. (Do not use the ratio test)**
\[
\sum_{n=1}^{\infty} \frac{\sin(n)}{n^2 + 1}
\]
### Explanation
In this problem, we are given an infinite series and must determine its convergence behavior. Specifically, we need to establish whether the series is absolutely convergent, conditionally convergent, or divergent. The series in question is:
\[
\sum_{n=1}^{\infty} \frac{\sin(n)}{n^2 + 1}
\]
#### Definitions:
- **Absolutely Convergent**: A series \(\sum a_n\) is absolutely convergent if \(\sum |a_n|\) is convergent.
- **Conditionally Convergent**: A series \(\sum a_n\) is conditionally convergent if \(\sum a_n\) is convergent, but \(\sum |a_n|\) is not convergent.
- **Divergent**: A series \(\sum a_n\) is divergent if it does not converge to a finite limit.
### Solution Outline
1. **Check for Absolute Convergence**:
- Consider the series \(\sum_{n=1}^{\infty} \left| \frac{\sin(n)}{n^2 + 1} \right|\).
- Since \(|\sin(n)| \leq 1\) for all \(n\), it follows that:
\[
\left| \frac{\sin(n)}{n^2 + 1} \right| \leq \frac{1}{n^2 + 1}
\]
- Note that \(\sum_{n=1}^{\infty} \frac{1}{n^2 + 1}\) behaves similarly to \(\sum_{n=1}^{\infty} \frac{1}{n^2}\), which is a convergent \(p\)-series with \(p=2\).
- Therefore, \(\sum_{n=1}^{\infty} \left| \frac{\sin(n)}{n^2 + 1} \right|\) converges, and so the original series is **absolutely convergent**.
2.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7dbb4ae4-0d65-4baa-9481-63f79be91eca%2F4879b545-bcca-4499-bcaa-2693c7397bad%2Fr3qsznt_reoriented.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement
**Determine if the following series is absolutely convergent, conditionally convergent, or divergent. (Do not use the ratio test)**
\[
\sum_{n=1}^{\infty} \frac{\sin(n)}{n^2 + 1}
\]
### Explanation
In this problem, we are given an infinite series and must determine its convergence behavior. Specifically, we need to establish whether the series is absolutely convergent, conditionally convergent, or divergent. The series in question is:
\[
\sum_{n=1}^{\infty} \frac{\sin(n)}{n^2 + 1}
\]
#### Definitions:
- **Absolutely Convergent**: A series \(\sum a_n\) is absolutely convergent if \(\sum |a_n|\) is convergent.
- **Conditionally Convergent**: A series \(\sum a_n\) is conditionally convergent if \(\sum a_n\) is convergent, but \(\sum |a_n|\) is not convergent.
- **Divergent**: A series \(\sum a_n\) is divergent if it does not converge to a finite limit.
### Solution Outline
1. **Check for Absolute Convergence**:
- Consider the series \(\sum_{n=1}^{\infty} \left| \frac{\sin(n)}{n^2 + 1} \right|\).
- Since \(|\sin(n)| \leq 1\) for all \(n\), it follows that:
\[
\left| \frac{\sin(n)}{n^2 + 1} \right| \leq \frac{1}{n^2 + 1}
\]
- Note that \(\sum_{n=1}^{\infty} \frac{1}{n^2 + 1}\) behaves similarly to \(\sum_{n=1}^{\infty} \frac{1}{n^2}\), which is a convergent \(p\)-series with \(p=2\).
- Therefore, \(\sum_{n=1}^{\infty} \left| \frac{\sin(n)}{n^2 + 1} \right|\) converges, and so the original series is **absolutely convergent**.
2.
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