Textbook Question
Express 0.314141414… as a ratio of two integers.
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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.49
42–76. Convergence or divergence Use a convergence test of your choice to determine whether the following series converge.
∑ (from k = 1 to ∞)k⁴ / √(9k¹² + 2)
Verified step by step guidance1
First, write down the general term of the series: \(a_k = \frac{k^4}{\sqrt{9k^{12} + 2}}\).
To analyze convergence, simplify the expression inside the square root for large \(k\). Notice that \$9k^{12}\( dominates \(2\), so approximate the denominator as \(\sqrt{9k^{12}} = 3k^6\) for large \)k$.
Rewrite the term using this approximation: \(a_k \approx \frac{k^4}{3k^6} = \frac{1}{3k^2}\) for large \(k\).
Recognize that the series behaves like \(\sum \frac{1}{k^2}\) for large \(k\), which is a p-series with \(p=2\).
Since a p-series \(\sum \frac{1}{k^p}\) converges if \(p > 1\), conclude that the original series converges by the Comparison Test.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Convergence and Divergence of Infinite Series
An infinite series converges if the sum of its terms approaches a finite limit as the number of terms grows indefinitely. If the sum does not approach a finite value, the series diverges. Understanding this distinction is fundamental to analyzing series behavior.
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06:52
Convergence of an Infinite Series
Comparison Test for Series Convergence
The Comparison Test involves comparing the given series to a second series with known convergence properties. If the terms of the given series are smaller than those of a convergent series, it also converges; if larger than a divergent series, it diverges. This test is useful when terms resemble simpler series.
Recommended video:
09:25Direct Comparison Test
Asymptotic Behavior and Simplification of Terms
Analyzing the dominant terms in the numerator and denominator for large k helps simplify the general term of the series. This simplification reveals the term's growth rate, which is crucial for applying convergence tests effectively, such as comparing to p-series or geometric series.
Recommended video:
5:50Asymptotes of Hyperbolas
Related Practice
Textbook Question
Give an example (if possible) of a sequence {aₖ} that converges, while the series ∑ (from k = 1 to ∞) aₖ diverges.
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Textbook Question
Explain why or why not
Determine whether the following statements are true and give an explanation or counterexample.
e.The sequence aₙ = n² / (n² + 1) converge.
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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 ∞)k√k / k³
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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 ∞)k⁵ e⁻ᵏ
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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 / k³⁄⁷
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