Textbook Question
46–53. Decimal expansions
Write each repeating decimal first as a geometric series and then as a fraction (a ratio of two integers).
49.0.037̅ = 0.037037…
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 10.2.15
13–52. Limits of sequences
Find the limit of the following sequences or determine that the sequence diverges.
{(3n³ − 1)⁄(2n³ + 1)}
Verified step by step guidance1
Identify the given sequence: \(a_n = \frac{3n^3 - 1}{2n^3 + 1}\).
To find the limit as \(n\) approaches infinity, observe the highest power of \(n\) in the numerator and denominator, which is \(n^3\).
Divide both the numerator and the denominator by \(n^3\) to simplify the expression: \(\frac{\frac{3n^3}{n^3} - \frac{1}{n^3}}{\frac{2n^3}{n^3} + \frac{1}{n^3}} = \frac{3 - \frac{1}{n^3}}{2 + \frac{1}{n^3}}\).
As \(n\) approaches infinity, the terms \(\frac{1}{n^3}\) approach 0, so the expression simplifies to \(\frac{3 - 0}{2 + 0} = \frac{3}{2}\).
Conclude that the limit of the sequence is \(\frac{3}{2}\).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Limits of Sequences
The limit of a sequence is the value that the terms of the sequence approach as the index n goes to infinity. If the terms get arbitrarily close to a specific number, the sequence converges to that limit; otherwise, it diverges.
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Guided course
8:22
Introduction to Sequences
Dominant Term in Polynomial Expressions
When evaluating limits of sequences involving polynomials, the highest degree terms dominate the behavior as n becomes large. Lower degree terms become insignificant, so the limit can often be found by comparing the leading terms.
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07:00Taylor Polynomials
Limit Laws and Simplification Techniques
Limit laws allow the separation and simplification of complex expressions. For rational sequences, dividing numerator and denominator by the highest power of n helps simplify the expression and find the limit more easily.
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05:50One-Sided Limits
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.
57. ∑ (k = 1 to ∞) 1 / ((k + 6)(k + 7))
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
21–42. Geometric series Evaluate each geometric series or state that it diverges.
27.1 + 1.01 + 1.01² + 1.01³ + ⋯
Textbook Question
48–63. Choose your test Determine whether the following series converge or diverge using the properties and tests introduced in Sections 10.3 and 10.4.
∑ (k = 1 to ∞) k / √(k² + 1)
Textbook Question
11–86. Applying convergence tests Determine whether the following series converge. Justify your answers.
∑ (from k = 1 to ∞) (−1)ᵏ × k / (k³ + 1)