Textbook Question
Explain why the magnitude of the remainder in an alternating series (with terms that are nonincreasing in magnitude) is less than or equal to the magnitude of the first neglected term.
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 10.2.35
13–52. Limits of sequences
Find the limit of the following sequences or determine that the sequence diverges.
{tan⁻¹(n)}
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Recognize that the sequence is given by \( a_n = \tan^{-1}(n) \), where \( \tan^{-1} \) is the inverse tangent function, also known as arctangent.
Recall the behavior of the arctangent function: as its input \( x \) approaches infinity, \( \tan^{-1}(x) \) approaches a horizontal asymptote at \( \frac{\pi}{2} \).
Since \( n \) is a sequence that increases without bound (\( n \to \infty \)), consider the limit \( \lim_{n \to \infty} \tan^{-1}(n) \).
Use the known limit property: \( \lim_{x \to \infty} \tan^{-1}(x) = \frac{\pi}{2} \). This implies the sequence \( a_n = \tan^{-1}(n) \) converges to \( \frac{\pi}{2} \).
Conclude that the limit of the sequence \( \{ \tan^{-1}(n) \} \) as \( n \to \infty \) is \( \frac{\pi}{2} \).
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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 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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Introduction to Sequences
Inverse Tangent Function (arctan)
The inverse tangent function, arctan(x), returns the angle whose tangent is x. It is defined for all real numbers and has horizontal asymptotes at ±π/2, meaning as x approaches ±∞, arctan(x) approaches ±π/2.
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Behavior of arctan(n) as n → ∞
As the input n grows without bound, arctan(n) approaches its horizontal asymptote π/2. This means the sequence {arctan(n)} converges to π/2, since the values get closer and closer to this limit for large n.
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Related Practice
Textbook Question
45–63. Absolute and conditional convergence Determine whether the following series converge absolutely, converge conditionally, or diverge.
∑ (k = 1 to ∞) (−1)ᵏ⁺¹ · (k!) / (kᵏ)(Hint: Show that k! / kᵏ ≤ 2 / k², for k ≥ 3.)
Textbook Question
23–38. Divergence, Integral, and p-series Tests Use the Divergence Test, the Integral Test, or the p-series test to determine whether the following series converge.
∑ (k = 1 to ∞) (k / (k + 10))ᵏ
Textbook Question
45–63. Absolute and conditional convergence Determine whether the following series converge absolutely, converge conditionally, or diverge.
∑ (k = 1 to ∞) (−1)ᵏ / k^(2/3)
Textbook Question
13–52. Limits of sequences
Find the limit of the following sequences or determine that the sequence diverges.
{ⁿ√(e³ⁿ⁺⁴)}
1
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Textbook Question
9–36. Comparison tests Use the Comparison Test or the Limit Comparison Test to determine whether the following series converge.
∑ (k = 1 to ∞) sin(1 / k) / k²