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.15
12–24. Limits of sequences Evaluate the limit of the sequence or state that it does not exist.
aₙ = (2ⁿ + 5ⁿ⁺¹) / 5ⁿ
Verified step by step guidance1
Identify the given sequence: \(a_n = \frac{2^n + 5^{n+1}}{5^n}\).
Rewrite the sequence by separating the terms in the numerator over the denominator: \(a_n = \frac{2^n}{5^n} + \frac{5^{n+1}}{5^n}\).
Simplify each term: \(\frac{2^n}{5^n} = \left(\frac{2}{5}\right)^n\) and \(\frac{5^{n+1}}{5^n} = 5^{(n+1)-n} = 5\).
Express the sequence as \(a_n = \left(\frac{2}{5}\right)^n + 5\).
Analyze the limit as \(n\) approaches infinity: since \(\left(\frac{2}{5}\right)^n\) tends to 0 (because \(\frac{2}{5} < 1\)), the limit of \(a_n\) is the limit of \(0 + 5\), which is 5.
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2mKey Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Limits of Sequences
The limit of a sequence describes the value that the terms of the sequence approach as the index n becomes very large. If the terms get arbitrarily close to a specific number, the sequence converges to that limit; otherwise, it diverges.
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Introduction to Sequences
Properties of Exponential Functions
Exponential functions with different bases grow at different rates. When comparing terms like 2ⁿ and 5ⁿ, the term with the larger base dominates as n increases, which helps simplify the limit by focusing on the dominant term.
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06:21Properties of Functions
Algebraic Manipulation of Sequences
Simplifying sequences often involves factoring and dividing numerator and denominator by the highest power term to reveal dominant behavior. This technique helps in evaluating limits by reducing complex expressions to simpler forms.
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Related Practice
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.
59. ∑ (k = –3 to ∞) 4 / ((4k – 3)(4k + 1))
Textbook Question
b.Does the series ∑ (from k = 1 to ∞) k/(k + 1) converge? Why or why not?
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.