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 ∞)5ᵏ / 2²ᵏ⁺¹
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Ch. 10 - Sequences and Infinite Series
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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Table of contents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 10, Problem 10.R.81
77–87. Absolute or conditional convergence
Determine whether the following series converge absolutely, converge conditionally, or diverge.
∑ (from k = 1 to ∞)(−1)ᵏ⁺¹(k² + 4) / (2k² + 1)
Verified step by step guidance1
Identify the given series: \( \sum_{k=1}^{\infty} (-1)^{k+1} \frac{k^2 + 4}{2k^2 + 1} \). This is an alternating series because of the factor \( (-1)^{k+1} \).
Check for absolute convergence by considering the series of absolute values: \( \sum_{k=1}^{\infty} \left| (-1)^{k+1} \frac{k^2 + 4}{2k^2 + 1} \right| = \sum_{k=1}^{\infty} \frac{k^2 + 4}{2k^2 + 1} \).
Analyze the behavior of the terms \( \frac{k^2 + 4}{2k^2 + 1} \) as \( k \to \infty \). Simplify the expression by dividing numerator and denominator by \( k^2 \) to find the limit of the terms.
Determine if the series of absolute values converges or diverges by comparing it to a known benchmark series, such as a \( p \)-series or using the Limit Comparison Test.
If the series of absolute values diverges, apply the Alternating Series Test to the original series by checking if the terms \( \frac{k^2 + 4}{2k^2 + 1} \) decrease monotonically to zero. Based on this, conclude whether the original series converges conditionally or diverges.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Absolute Convergence
A series ∑a_k converges absolutely if the series of absolute values ∑|a_k| converges. Absolute convergence implies convergence regardless of the sign of terms, and it guarantees the sum is well-defined and stable under rearrangement.
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Choosing a Convergence Test
Conditional Convergence
A series converges conditionally if it converges, but does not converge absolutely. This means ∑a_k converges, but ∑|a_k| diverges. Conditional convergence often occurs in alternating series where the terms decrease in magnitude but the absolute series diverges.
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Alternating Series Test
The Alternating Series Test states that an alternating series ∑(−1)^k b_k converges if the sequence b_k is positive, decreasing, and approaches zero. This test helps determine conditional convergence when absolute convergence fails.
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Related Practice
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
Explain why or why not
Determine whether the following statements are true and give an explanation or counterexample.
a.The terms of the sequence {aₙ} increase in magnitude, so the limit of the sequence does not exist.
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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 ∞)2ᵏ / eᵏ
49
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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!
87
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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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