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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 10.5.11
Chapter 10, Problem 10.5.11

9–36. Comparison tests Use the Comparison Test or the Limit Comparison Test to determine whether the following series converge.


∑ (k = 1 to ∞) (k² − 1) / (k³ + 4)

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1
Identify the given series: \( \sum_{k=1}^{\infty} \frac{k^{2} - 1}{k^{3} + 4} \). We want to determine if this series converges or diverges.
Analyze the behavior of the general term \( a_k = \frac{k^{2} - 1}{k^{3} + 4} \) for large \( k \). For large \( k \), the dominant terms in numerator and denominator are \( k^{2} \) and \( k^{3} \), respectively, so \( a_k \) behaves like \( \frac{k^{2}}{k^{3}} = \frac{1}{k} \).
Choose a comparison series that is simpler but has similar behavior for large \( k \). Since \( a_k \) behaves like \( \frac{1}{k} \), consider the harmonic series \( \sum_{k=1}^{\infty} \frac{1}{k} \), which is known to diverge.
Apply the Limit Comparison Test by computing the limit \( L = \lim_{k \to \infty} \frac{a_k}{b_k} = \lim_{k \to \infty} \frac{\frac{k^{2} - 1}{k^{3} + 4}}{\frac{1}{k}} = \lim_{k \to \infty} \frac{(k^{2} - 1) \cdot k}{k^{3} + 4} \). Simplify this expression to find \( L \).
Interpret the result of the limit \( L \). If \( L \) is a finite positive number, then both series either converge or diverge together. Since the harmonic series diverges, this will tell us about the convergence of the original series.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Comparison Test

The Comparison Test determines the convergence of a series by comparing it to a second series with known behavior. If the terms of the given series are smaller than those of a convergent series, it also converges. Conversely, if the terms are larger than those of a divergent series, it diverges.
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Direct Comparison Test

Limit Comparison Test

The Limit Comparison Test compares two series by taking the limit of the ratio of their terms. If this limit is a positive finite number, both series either converge or diverge together. This test is useful when direct comparison is difficult but the terms behave similarly for large indices.
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Limit Comparison Test

Behavior of Rational Functions in Series

When analyzing series with rational function terms, focus on the highest degree terms in numerator and denominator to determine the dominant behavior as k approaches infinity. This helps identify a simpler comparison series, often a p-series or geometric series, to apply convergence tests effectively.
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Geometric Series
Related Practice
Textbook Question

11–86. Applying convergence tests Determine whether the following series converge. Justify your answers.

∑ (from k = 1 to ∞) (10ᵏ + 1) / k¹⁰

Textbook Question

32–49. Choose your test Use the test of your choice to determine whether the following series converge absolutely, converge conditionally, or diverge.

∑ (from k = 1 to ∞) (−1)ᵏ k (2ᵏ⁺¹ / (9ᵏ − 1))

Textbook Question

1–10. Choosing convergence tests Identify a convergence test for each series. If necessary, explain how to simplify or rewrite the series before applying the convergence test. You do not need to carry out the convergence test.

∑ (from k = 1 to ∞) ((−1)ᵏ⁺¹) / (√2ᵏ + lnk)

Textbook Question

11–27. Alternating Series Test Determine whether the following series converge.

∑ (k = 1 to ∞) (−1)ᵏ⁺¹ k² / (k³ + 1)

Textbook Question

11–86. Applying convergence tests Determine whether the following series converge. Justify your answers.

∑ (from k = 1 to ∞) 1 / (√k × e^(√k))

Textbook Question

32–49. Choose your test Use the test of your choice to determine whether the following series converge absolutely, converge conditionally, or diverge.

∑ (from k = 1 to ∞) (√k / k − 1)²ᵏ