If you apply the direct comparison test to assess convergence of arctan ²k, k¹.01 to which of these series would you compareit? And, what is the outcome? A. the series converges k=1 T 00 B.Σk ¹.01; the series diverges k=1 C.k 1.01; the seriesconverges k=1 D.arctank; the series diverges k=1 B
If you apply the direct comparison test to assess convergence of arctan ²k, k¹.01 to which of these series would you compareit? And, what is the outcome? A. the series converges k=1 T 00 B.Σk ¹.01; the series diverges k=1 C.k 1.01; the seriesconverges k=1 D.arctank; the series diverges k=1 B
Calculus: Early Transcendentals
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Author:James Stewart
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Chapter1: Functions And Models
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![### Assessing Series Convergence using Direct Comparison Test
**Problem Statement:**
If you apply the direct comparison test to assess the convergence of the following series:
\[ \sum_{k=1}^{\infty} \frac{\arctan^2 k}{k^{1.01}} \]
to which of these series would you compare it? And, what is the outcome?
**Answer Choices:**
A.
\[ \sum_{k=1}^{\infty} \frac{1}{k} \]
- The series converges
B.
\[ \sum_{k=1}^{\infty} \frac{\pi^2 / 4}{k^{1.01}} \]
- The series diverges
C.
\[ \sum_{k=1}^{\infty} \frac{\pi^2 / 4}{k^{1.01}} \]
- The series converges
D.
\[ \sum_{k=1}^{\infty} \arctan^2 k \]
- The series diverges
**Options for Answer:**
- ☐ A
- ☐ B
- ☐ C
- ☐ D
### Explanation:
**Direct Comparison Test:**
To apply the direct comparison test, you must find a series with terms that are always greater than or less than the terms of the original series and have a known convergence or divergence behavior.
Consider the given series:
\[ \sum_{k=1}^{\infty} \frac{\arctan^2 k}{k^{1.01}} \]
#### Comparison Series:
Let's consider a known convergent series:
\[ \sum_{k=1}^{\infty} \frac{1}{k^{p}} \]
where \( p > 1 \). Notably, if \( p \leq 1 \), the series diverges.
Given that \( \arctan(k) \) is bounded and \( \arctan(k) \leq \frac{\pi}{2} \) for all \( k \geq 1 \), it follows that:
\[ \arctan^2(k) \leq \left(\frac{\pi}{2}\right)^2 \]
Thus,
\[ \frac{\arctan^2 k}{k^{1.01}} \leq \frac{\left(\frac{\pi](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F98ea0ed6-be6f-4556-80a5-46c7ae10c3b9%2F2cf157f4-212b-4606-8361-a5e4a07011e3%2Frqqenca_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Assessing Series Convergence using Direct Comparison Test
**Problem Statement:**
If you apply the direct comparison test to assess the convergence of the following series:
\[ \sum_{k=1}^{\infty} \frac{\arctan^2 k}{k^{1.01}} \]
to which of these series would you compare it? And, what is the outcome?
**Answer Choices:**
A.
\[ \sum_{k=1}^{\infty} \frac{1}{k} \]
- The series converges
B.
\[ \sum_{k=1}^{\infty} \frac{\pi^2 / 4}{k^{1.01}} \]
- The series diverges
C.
\[ \sum_{k=1}^{\infty} \frac{\pi^2 / 4}{k^{1.01}} \]
- The series converges
D.
\[ \sum_{k=1}^{\infty} \arctan^2 k \]
- The series diverges
**Options for Answer:**
- ☐ A
- ☐ B
- ☐ C
- ☐ D
### Explanation:
**Direct Comparison Test:**
To apply the direct comparison test, you must find a series with terms that are always greater than or less than the terms of the original series and have a known convergence or divergence behavior.
Consider the given series:
\[ \sum_{k=1}^{\infty} \frac{\arctan^2 k}{k^{1.01}} \]
#### Comparison Series:
Let's consider a known convergent series:
\[ \sum_{k=1}^{\infty} \frac{1}{k^{p}} \]
where \( p > 1 \). Notably, if \( p \leq 1 \), the series diverges.
Given that \( \arctan(k) \) is bounded and \( \arctan(k) \leq \frac{\pi}{2} \) for all \( k \geq 1 \), it follows that:
\[ \arctan^2(k) \leq \left(\frac{\pi}{2}\right)^2 \]
Thus,
\[ \frac{\arctan^2 k}{k^{1.01}} \leq \frac{\left(\frac{\pi
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