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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 10.8.21
Chapter 10, Problem 10.8.21

11–86. Applying convergence tests Determine whether the following series converge. Justify your answers.
∑ (from k = 1 to ∞) (−1)ᵏ × k / (k³ + 1)

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Identify the given series: \( \sum_{k=1}^{\infty} (-1)^k \frac{k}{k^3 + 1} \). Notice that it is an alternating series because of the factor \( (-1)^k \).
Consider the absolute value of the terms: \( a_k = \frac{k}{k^3 + 1} \). Simplify or analyze the behavior of \( a_k \) as \( k \to \infty \).
Check the limit of \( a_k \) as \( k \to \infty \): \( \lim_{k \to \infty} \frac{k}{k^3 + 1} \). This will help determine if the terms approach zero, which is necessary for convergence of an alternating series.
Apply the Alternating Series Test (Leibniz Test), which requires that \( a_k \) is positive, decreasing, and \( \lim_{k \to \infty} a_k = 0 \). Verify these conditions for \( a_k \).
If the Alternating Series Test confirms convergence, consider whether the series converges absolutely by testing the convergence of \( \sum_{k=1}^{\infty} \left| (-1)^k \frac{k}{k^3 + 1} \right| = \sum_{k=1}^{\infty} \frac{k}{k^3 + 1} \) using a comparison or limit comparison test with a known convergent or divergent series.

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Key Concepts

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

Alternating Series Test

The Alternating Series Test determines convergence of series whose terms alternate in sign. If the absolute value of terms decreases monotonically to zero, the series converges. This test is useful for series like ∑ (−1)^k * a_k where a_k > 0 and lim a_k = 0.
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Alternating Series Test

Limit of the General Term

For any infinite series ∑ a_k to converge, the terms a_k must approach zero as k approaches infinity. If lim (a_k) ≠ 0, the series diverges. Checking this limit is a fundamental first step in analyzing series convergence.
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Comparison Test

The Comparison Test compares a given series to a known benchmark series to determine convergence. If the terms of the given series are smaller than those of a convergent series, it converges; if larger than a divergent series, it diverges. This helps analyze the behavior of complex terms.
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Direct Comparison Test
Related Practice
Textbook Question

Suppose the sequence { aₙ} is defined by the explicit formula aₙ = 1/n, for n=1, 2, 3, .....Write out the first five terms of the sequence.

Textbook Question

54–69. Telescoping series

For the following telescoping series, find a formula for the nth term of the sequence of partial sums {Sₙ}. Then evaluate limₙ→∞ Sₙ to obtain the value of the series or state that the series diverges.


57. ∑ (k = 1 to ∞) 1 / ((k + 6)(k + 7))

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

45–63. Absolute and conditional convergence Determine whether the following series converge absolutely, converge conditionally, or diverge.

∑ (k = 1 to ∞) cos(k) / k³

Textbook Question

9–15. Geometric sums Evaluate each geometric sum.


∑ k = 0 to 83ᵏ

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Textbook Question

13–52. Limits of sequences

Find the limit of the following sequences or determine that the sequence diverges.


{(3n³ − 1)⁄(2n³ + 1)}

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