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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 10.5.31
Chapter 10, Problem 10.5.31

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


∑ (k = 1 to ∞) 20 / (∛k + √k)

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1
Identify the given series: \( \sum_{k=1}^{\infty} \frac{20}{\sqrt[3]{k} + \sqrt{k}} \). We want to determine if this series converges or diverges.
Analyze the behavior of the terms for large \( k \). Notice that \( \sqrt[3]{k} = k^{1/3} \) and \( \sqrt{k} = k^{1/2} \). Since \( k^{1/2} \) grows faster than \( k^{1/3} \), the denominator behaves roughly like \( k^{1/2} \) for large \( k \).
Simplify the general term for large \( k \) to compare it with a simpler series. The term behaves like \( \frac{20}{k^{1/2}} \) because \( \sqrt{k} \) dominates \( \sqrt[3]{k} \). So, consider the comparison series \( \sum_{k=1}^{\infty} \frac{1}{k^{1/2}} \).
Recall that the p-series \( \sum \frac{1}{k^p} \) converges if and only if \( p > 1 \). Here, \( p = \frac{1}{2} < 1 \), so the comparison series diverges.
Use the Comparison Test or Limit Comparison Test: Since the original terms behave like \( \frac{1}{k^{1/2}} \) and this comparison series diverges, conclude that the original series also diverges.

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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 Series Terms Involving Roots

Understanding how terms with roots like cube roots and square roots behave as the index grows large is crucial. For example, ∛k grows slower than √k, so the dominant term in the denominator affects the term's size. Analyzing this helps in choosing an appropriate comparison series.
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Intro to Series: Partial Sums
Related Practice
Textbook Question

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

∑ (from k = 1 to ∞)k! / (kᵏ + 3)

Textbook Question

21–26. Recurrence relations Write the first four terms of the sequence {aₙ} defined by the following recurrence relations.

aₙ₊₁ = 2aₙ; a₁ = 2

Textbook Question

48–63. Choose your test Determine whether the following series converge or diverge using the properties and tests introduced in Sections 10.3 and 10.4.

∑ (k = 1 to ∞) 1 / ( (3k + 1)(3k + 4) )

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.

1 / 1! + 4 / 2! + 9 / 3! + 16 / 4! + ⋯

Textbook Question

48–63. Choose your test Determine whether the following series converge or diverge using the properties and tests introduced in Sections 10.3 and 10.4.

∑ (k = 1 to ∞) 3ᵏ / (k² + 1)

Textbook Question

For what values of p does the series ∑ (k = 10 to ∞) 1 / kᵖ converge (initial index is 10)? For what values of p does it diverge?