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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 10.7.39
Chapter 10, Problem 10.7.39

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 k³ / √(k⁸ + 1)

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1
Identify the given series: \( \sum_{k=1}^{\infty} (-1)^k \frac{k^3}{\sqrt{k^8 + 1}} \). Notice it is an alternating series because of the factor \( (-1)^k \).
To determine absolute convergence, consider the absolute value of the terms: \( \left| \frac{k^3}{\sqrt{k^8 + 1}} \right| = \frac{k^3}{\sqrt{k^8 + 1}} \). Simplify this expression to understand its behavior as \( k \to \infty \).
Analyze the limit of the absolute value of the terms as \( k \to \infty \). Since \( \sqrt{k^8 + 1} = (k^8 + 1)^{1/2} \), rewrite it as \( k^4 \sqrt{1 + \frac{1}{k^8}} \). Then the term becomes \( \frac{k^3}{k^4 \sqrt{1 + \frac{1}{k^8}}} = \frac{1}{k \sqrt{1 + \frac{1}{k^8}}} \).
Evaluate the limit \( \lim_{k \to \infty} \frac{1}{k \sqrt{1 + \frac{1}{k^8}}} \). Since \( \sqrt{1 + \frac{1}{k^8}} \to 1 \), the limit behaves like \( \frac{1}{k} \), which goes to zero.
Since the terms behave like \( \frac{1}{k} \) for large \( k \), the absolute series resembles the harmonic series \( \sum \frac{1}{k} \), which diverges. Therefore, the series does not converge absolutely. Next, apply the Alternating Series Test by checking if the terms \( b_k = \frac{k^3}{\sqrt{k^8 + 1}} \) decrease monotonically to zero to determine conditional convergence.

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Key Concepts

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

Absolute and Conditional Convergence

Absolute convergence occurs when the series of absolute values converges, ensuring the original series converges regardless of term signs. Conditional convergence happens when the series converges but not absolutely, often due to alternating signs. Understanding these distinctions helps classify the behavior of infinite series.
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Choosing a Convergence Test

Limit Comparison Test

The Limit Comparison Test compares a given series to a known benchmark series by examining the limit of their term ratios. If the limit is a positive finite number, both series share the same convergence behavior. This test is useful for series with complicated terms involving polynomials and roots.
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Alternating Series Test

The Alternating Series Test determines convergence of series whose terms alternate in sign. It requires that the absolute value of terms decreases monotonically to zero. If these conditions hold, the series converges, which helps identify conditional convergence in alternating series.
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Alternating Series Test
Related Practice
Textbook Question

Define the remainder of an infinite series.

Textbook Question

35–44. Limits of sequences Write the terms a₁, a₂, a₃, and a₄ of the following sequences. If the sequence appears to converge, make a conjecture about its limit. If the sequence diverges, explain why. 

aₙ = 1⁄10ⁿ; n = 1, 2, 3, …

Textbook Question

21–42. Geometric series Evaluate each geometric series or state that it diverges.  


21.∑ (k = 0 to ∞) (1/4)ᵏ

Textbook Question

For what values of r does the sequence {rⁿ} converge? Diverge?

1
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Textbook Question

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

∑ (k = 0 to ∞) (−1)ᵏ / (2k + 1)

Textbook Question

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


∑ (k = 4 to ∞) (1 + cos²(k)) / (k − 3)