23. 8 ω k=1 sin*k k?
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
use the Direct Comparison Test to determine whether the infinite series is convergent.
![### Infinite Series Summation
#### Problem Statement
Evaluate the infinite series given by the summation:
\[ \sum_{k=1}^{\infty} \frac{\sin^2(k)}{k^2} \]
This series involves summing the terms \(\frac{\sin^2(k)}{k^2}\) for \(k\) starting from 1 and increasing to infinity.
#### Explanation:
- **Summation Symbol (\(\Sigma\)**):
Indicates that we are summing a sequence of terms.
- **Sinusoidal Function (\(\sin(k)\))**:
The sine function, commonly used in trigonometry, provides periodic results based on angle \(k\).
- **Squared Term (\(\sin^2(k)\))**:
The square of the sine function ensures that the values are non-negative since \(\sin^2(k) \geq 0\).
- **Denominator (\(k^2\))**:
The squared term in the denominator acts to diminish the size of each term as \(k\) increases, promoting convergence of the series.
In this series, the role of \(\sin^2(k)\) divided by \(k^2\) ensures that each term contributes less to the sum as \(k\) grows, facilitating the convergence of the series.
Educational Tip: When dealing with infinite series involving trigonometric functions, consider investigating the convergence by applying tests like the Comparison Test, Ratio Test, or analyzing the behaviors of individual components.
For instance, comparing \(\sin^2(k)\) to its bounded maximum (1), it becomes evident that:
\[ \frac{\sin^2(k)}{k^2} \leq \frac{1}{k^2} \]
The series \(\sum_{k=1}^{\infty} \frac{1}{k^2}\) is a known convergent p-series (where \(p = 2 > 1\)). Thus, by comparison, the given series also converges.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F37708386-d0d8-46c3-a2d8-8e337c754037%2F4d8769ef-512e-4828-a702-14be2fc2108c%2F3n83si_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Infinite Series Summation
#### Problem Statement
Evaluate the infinite series given by the summation:
\[ \sum_{k=1}^{\infty} \frac{\sin^2(k)}{k^2} \]
This series involves summing the terms \(\frac{\sin^2(k)}{k^2}\) for \(k\) starting from 1 and increasing to infinity.
#### Explanation:
- **Summation Symbol (\(\Sigma\)**):
Indicates that we are summing a sequence of terms.
- **Sinusoidal Function (\(\sin(k)\))**:
The sine function, commonly used in trigonometry, provides periodic results based on angle \(k\).
- **Squared Term (\(\sin^2(k)\))**:
The square of the sine function ensures that the values are non-negative since \(\sin^2(k) \geq 0\).
- **Denominator (\(k^2\))**:
The squared term in the denominator acts to diminish the size of each term as \(k\) increases, promoting convergence of the series.
In this series, the role of \(\sin^2(k)\) divided by \(k^2\) ensures that each term contributes less to the sum as \(k\) grows, facilitating the convergence of the series.
Educational Tip: When dealing with infinite series involving trigonometric functions, consider investigating the convergence by applying tests like the Comparison Test, Ratio Test, or analyzing the behaviors of individual components.
For instance, comparing \(\sin^2(k)\) to its bounded maximum (1), it becomes evident that:
\[ \frac{\sin^2(k)}{k^2} \leq \frac{1}{k^2} \]
The series \(\sum_{k=1}^{\infty} \frac{1}{k^2}\) is a known convergent p-series (where \(p = 2 > 1\)). Thus, by comparison, the given series also converges.
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