Textbook Question
77–87. Absolute or conditional convergence
Determine whether the following series converge absolutely, converge conditionally, or diverge.
∑ (from k = 1 to ∞)(−1)ᵏk·e⁻ᵏ
Ch. 10 - Sequences and Infinite Series
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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Table of contents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 10, Problem 10.R.13
12–24. Limits of sequences Evaluate the limit of the sequence or state that it does not exist.
aₙ = (–1)ⁿ (3n³ + 4n) / (6n³ + 5)
Verified step by step guidance1
Identify the given sequence: \(a_n = (-1)^n \frac{3n^3 + 4n}{6n^3 + 5}\).
Observe that the sequence has a factor \((-1)^n\) which causes the terms to alternate in sign depending on whether \(n\) is even or odd.
Focus on the rational expression \(\frac{3n^3 + 4n}{6n^3 + 5}\). To analyze its behavior as \(n \to \infty\), divide numerator and denominator by \(n^3\), the highest power of \(n\) in the expression.
After dividing, rewrite the expression as \(\frac{3 + \frac{4}{n^2}}{6 + \frac{5}{n^3}}\). As \(n\) approaches infinity, the terms with \(\frac{1}{n^k}\) approach zero, simplifying the expression to \(\frac{3}{6} = \frac{1}{2}\).
Combine this limit with the alternating factor \((-1)^n\). Since \((-1)^n\) oscillates between \(1\) and \(-1\), the sequence does not approach a single value but oscillates between \(\frac{1}{2}\) and \(-\frac{1}{2}\). Therefore, the limit of the sequence does not exist.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Limits of Sequences
The limit of a sequence is the value that the terms of the sequence approach as the index n goes to infinity. If the terms get arbitrarily close to a specific number, the sequence converges to that limit; otherwise, it diverges.
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8:22
Introduction to Sequences
Behavior of Polynomial Expressions at Infinity
When evaluating limits involving polynomials as n approaches infinity, the highest degree terms dominate the behavior. Lower degree terms become insignificant, so the limit can often be found by comparing the leading coefficients of the highest degree terms.
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07:00Taylor Polynomials
Alternating Sequences
An alternating sequence changes sign with each term, often involving factors like (-1)^n. Such sequences may oscillate and not have a limit unless the magnitude of terms approaches zero, causing the sequence to converge to zero.
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8:22Introduction to Sequences
Related Practice
Textbook Question
a.Does the sequence { k/(k + 1) } converge? Why or why not?
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Textbook Question
Explain why or why not
Determine whether the following statements are true and give an explanation or counterexample.
a.The terms of the sequence {aₙ} increase in magnitude, so the limit of the sequence does not exist.
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Textbook Question
42–76. Convergence or divergence Use a convergence test of your choice to determine whether the following series converge.
∑ (from k = 1 to ∞)2ᵏ / eᵏ
Textbook Question
42–76. Convergence or divergence Use a convergence test of your choice to determine whether the following series converge.
∑ (from k = 1 to ∞)(7 + sin k) / k²
Textbook Question
Estimate the value of the series ∑ (from k = 1 to ∞)1 / (2k + 5)³ to within 10⁻⁴ of its exact value.