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Ch. 10 - Sequences and Infinite Series
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 10 - Sequences and Infinite SeriesProblem 10.R.13
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)

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1
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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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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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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Related Practice
Textbook Question

Explain why or why not

Determine whether the following statements are true and give an explanation or counterexample.


g.The series ∑ (from k = 1 to ∞) (k² / (k² + 1)) converges.

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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 ∞)3 / (2 + eᵏ)

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Textbook Question

12–24. Limits of sequences Evaluate the limit of the sequence or state that it does not exist.

aₙ = 8ⁿ / n!

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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 ∞)(7 + sin k) / k²

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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 j = 0 to ∞)2 ⋅ 4ʲ / (2j + 1)!

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Textbook Question

Building a tunnel — first scenario

A crew of workers is constructing a tunnel through a mountain. Understandably, the rate of construction decreases because rocks and earth must be removed a greater distance as the tunnel gets longer. Suppose each week the crew digs 0.95 of the distance it dug the previous week. In the first week, the crew constructed 100 m of tunnel.


b.What is the longest tunnel the crew can build at this rate?

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