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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![**Determine whether the following series converges. Justify your answer.**
\[
\sum_{k=1}^{\infty} \frac{8^k + 10}{8^k}
\]
---
**Options:**
- **A.** The series is a geometric series with a common ratio \(\frac{1}{8}\). This is less than 1, so the series converges by the properties of a geometric series.
- **B.**
Because \(\frac{8^k + 10}{8^k} \leq \frac{8^k}{8^k} = 1\) for any positive integer \(k\) and \(\sum_{k=1}^{\infty} \frac{8^k}{8^k} = \sum_{k=1}^{\infty} \frac{1}{1}\) converges, the given series converges by the Comparison Test.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcef3f1b2-56cd-4f45-949e-fc7a38ef994c%2F2abec71d-1aa2-46da-b3cd-cc6a417cb6cf%2F3hruv4c_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Determine whether the following series converges. Justify your answer.**
\[
\sum_{k=1}^{\infty} \frac{8^k + 10}{8^k}
\]
---
**Options:**
- **A.** The series is a geometric series with a common ratio \(\frac{1}{8}\). This is less than 1, so the series converges by the properties of a geometric series.
- **B.**
Because \(\frac{8^k + 10}{8^k} \leq \frac{8^k}{8^k} = 1\) for any positive integer \(k\) and \(\sum_{k=1}^{\infty} \frac{8^k}{8^k} = \sum_{k=1}^{\infty} \frac{1}{1}\) converges, the given series converges by the Comparison Test.
![**Transcription for Educational Website**
---
**Options for Series Convergence Analysis**
**C.** Because
\[
\frac{8}{k} < \frac{8}{k} + \frac{10}{k} \text{ for any positive integer } k \text{ and } \sum \frac{8}{k}
\]
diverges, the given series diverges by the Comparison Test.
**D.** The Ratio Test yields \( r = \_\_\_ \). This is less than 1, so the series converges by the Ratio Test.
*(Type an exact answer.)*
**E.** The series is a geometric series with common ratio \( \_\_\_ \). This is greater than 1, so the series diverges by the properties of a geometric series.
**Buttons Below:**
- Clear all
- Check answer
**Note:** This exercise involves applying different convergence tests (Comparison Test, Ratio Test, and properties of geometric series) to determine the behavior of series. Each option outlines a method for assessing divergence or convergence based on mathematical principles.
---
This transcription breaks down each option methodically, explaining the reasoning behind the choice of test and the expected outcome regarding series behavior.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcef3f1b2-56cd-4f45-949e-fc7a38ef994c%2F2abec71d-1aa2-46da-b3cd-cc6a417cb6cf%2Fnoa8dul_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Transcription for Educational Website**
---
**Options for Series Convergence Analysis**
**C.** Because
\[
\frac{8}{k} < \frac{8}{k} + \frac{10}{k} \text{ for any positive integer } k \text{ and } \sum \frac{8}{k}
\]
diverges, the given series diverges by the Comparison Test.
**D.** The Ratio Test yields \( r = \_\_\_ \). This is less than 1, so the series converges by the Ratio Test.
*(Type an exact answer.)*
**E.** The series is a geometric series with common ratio \( \_\_\_ \). This is greater than 1, so the series diverges by the properties of a geometric series.
**Buttons Below:**
- Clear all
- Check answer
**Note:** This exercise involves applying different convergence tests (Comparison Test, Ratio Test, and properties of geometric series) to determine the behavior of series. Each option outlines a method for assessing divergence or convergence based on mathematical principles.
---
This transcription breaks down each option methodically, explaining the reasoning behind the choice of test and the expected outcome regarding series behavior.
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