Textbook Question
77–87. Absolute or conditional convergence
Determine whether the following series converge absolutely, converge conditionally, or diverge.
∑ (from k = 1 to ∞)(−1)ᵏk·e⁻ᵏ
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.11b
b.Does the series ∑ (from k = 1 to ∞) k/(k + 1) converge? Why or why not?
Verified step by step guidance1
Identify the general term of the series: \(a_k = \frac{k}{k+1}\).
Examine the behavior of the terms \(a_k\) as \(k\) approaches infinity by finding \(\lim_{k \to \infty} a_k\).
Calculate the limit: \(\lim_{k \to \infty} \frac{k}{k+1} = \lim_{k \to \infty} \frac{k}{k(1 + \frac{1}{k})} = \lim_{k \to \infty} \frac{1}{1 + \frac{1}{k}}\).
Since \(\lim_{k \to \infty} a_k = 1 \neq 0\), recall the necessary condition for series convergence: if the terms do not approach zero, the series cannot converge.
Conclude that because the terms do not approach zero, the series \(\sum_{k=1}^\infty \frac{k}{k+1}\) diverges.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Infinite Series and Convergence
An infinite series is the sum of infinitely many terms. Convergence means the series approaches a finite limit as more terms are added. Determining convergence involves analyzing the behavior of the partial sums or applying convergence tests.
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Convergence of an Infinite Series
Term Test for Divergence
If the terms of a series do not approach zero as k approaches infinity, the series cannot converge. This is a quick initial test to rule out convergence by examining the limit of the general term.
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05:44Divergence Test (nth Term Test)
Behavior of the General Term k/(k+1)
The term k/(k+1) simplifies to 1 - 1/(k+1), which approaches 1 as k grows large. Since the terms do not approach zero, the series ∑ k/(k+1) diverges by the term test.
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06:11Introduction to Riemann Sums
Related Practice
Textbook Question
a.Does the sequence { k/(k + 1) } converge? Why or why not?
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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 = (2ⁿ + 5ⁿ⁺¹) / 5ⁿ
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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!
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Textbook Question
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)
Textbook Question
Estimate the value of the series ∑ (from k = 1 to ∞)1 / (2k + 5)³ to within 10⁻⁴ of its exact value.