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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![### Using the Limit Comparison Test to Determine Series Convergence
Consider the series:
\[ \sum_{k=1}^{\infty} \frac{k^2 + 14}{k(k + 4)(k - 5)(k + 2)} \]
We are tasked with using the Limit Comparison Test to determine if this series converges or diverges.
### Limit Comparison Test
To apply the Limit Comparison Test, we typically compare our given series to a known benchmark series. The goal is to find a simpler series \( b_k \) such that the limit
\[ \lim_{k \to \infty} \frac{a_k}{b_k} \]
is a positive, finite number.
In this case:
- **Series to be analyzed (\( a_k \))**: \(\frac{k^2 + 14}{k(k + 4)(k - 5)(k + 2)}\)
- **Benchmark Series (\( b_k \))**: Choose a term which simplifies the behavior of \( a_k \) as \( k \) approaches infinity. For this problem, a reasonable choice might be \( \frac{1}{k^2} \).
### Steps:
1. **Form the ratio**: \(\frac{a_k}{b_k} = \frac{\frac{k^2 + 14}{k(k + 4)(k - 5)(k + 2)}}{\frac{1}{k^2}}\).
2. **Simplify the expression**.
3. **Take the limit as \( k \) approaches infinity**.
Based on the problem setup, the solution follows these steps and determines whether the series converges or diverges.
### Conclusion
The correct answer for the convergence of the given series is indicated as:
- **\[ \bigcirc \] Diverges**
- \[ \,\_\ \] Converges
The series \(\sum_{k=1}^{\infty} \frac{k^2 + 14}{k(k + 4)(k - 5)(k + 2)}\) diverges.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F71c0ef9f-cd3d-4be0-91e6-4ca83044b87b%2F1f03055e-96db-4344-b648-9139dc934dad%2F647j7r_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Using the Limit Comparison Test to Determine Series Convergence
Consider the series:
\[ \sum_{k=1}^{\infty} \frac{k^2 + 14}{k(k + 4)(k - 5)(k + 2)} \]
We are tasked with using the Limit Comparison Test to determine if this series converges or diverges.
### Limit Comparison Test
To apply the Limit Comparison Test, we typically compare our given series to a known benchmark series. The goal is to find a simpler series \( b_k \) such that the limit
\[ \lim_{k \to \infty} \frac{a_k}{b_k} \]
is a positive, finite number.
In this case:
- **Series to be analyzed (\( a_k \))**: \(\frac{k^2 + 14}{k(k + 4)(k - 5)(k + 2)}\)
- **Benchmark Series (\( b_k \))**: Choose a term which simplifies the behavior of \( a_k \) as \( k \) approaches infinity. For this problem, a reasonable choice might be \( \frac{1}{k^2} \).
### Steps:
1. **Form the ratio**: \(\frac{a_k}{b_k} = \frac{\frac{k^2 + 14}{k(k + 4)(k - 5)(k + 2)}}{\frac{1}{k^2}}\).
2. **Simplify the expression**.
3. **Take the limit as \( k \) approaches infinity**.
Based on the problem setup, the solution follows these steps and determines whether the series converges or diverges.
### Conclusion
The correct answer for the convergence of the given series is indicated as:
- **\[ \bigcirc \] Diverges**
- \[ \,\_\ \] Converges
The series \(\sum_{k=1}^{\infty} \frac{k^2 + 14}{k(k + 4)(k - 5)(k + 2)}\) diverges.
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