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Ch. 10 - Sequences and Infinite Series
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 10 - Sequences and Infinite SeriesProblem 10.8.73
Chapter 10, Problem 10.8.73

11–86. Applying convergence tests Determine whether the following series converge. Justify your answers.


∑ (from k = 0 to ∞)k² · 1.001⁻ᵏ

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Identify the given series: \( \sum_{k=0}^{\infty} k^{2} \cdot 1.001^{-k} \). Notice that the general term is \( a_k = k^{2} \cdot (1.001)^{-k} \).
Rewrite the term to see the exponential part clearly: \( a_k = k^{2} \cdot \left( \frac{1}{1.001} \right)^{k} \). Since \( \frac{1}{1.001} < 1 \), this suggests the terms involve a polynomial factor times an exponential decay.
Consider applying the Ratio Test, which is useful for series involving factorials, exponentials, or powers. The Ratio Test states to compute \( L = \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| \).
Calculate \( \frac{a_{k+1}}{a_k} = \frac{(k+1)^2 \cdot (1.001)^{-(k+1)}}{k^2 \cdot (1.001)^{-k}} = \frac{(k+1)^2}{k^2} \cdot \frac{1}{1.001} \).
Evaluate the limit \( L = \lim_{k \to \infty} \frac{(k+1)^2}{k^2} \cdot \frac{1}{1.001} = \left( \lim_{k \to \infty} \frac{(k+1)^2}{k^2} \right) \cdot \frac{1}{1.001} = 1 \cdot \frac{1}{1.001} < 1 \). Since \( L < 1 \), the Ratio Test confirms that the series converges.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Geometric Series and Its Convergence

A geometric series has terms of the form ar^k, where r is the common ratio. It converges if |r| < 1 and diverges otherwise. Understanding this helps analyze series with exponential terms like 1.001⁻ᵏ, which can be rewritten as (1/1.001)^k, a geometric term with ratio less than 1.
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Geometric Series

Comparison and Limit Comparison Tests

These tests compare a given series to a known benchmark series to determine convergence. The limit comparison test is useful when terms involve polynomials and exponentials, helping to confirm if the series behaves like a convergent geometric series despite polynomial factors like k².
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Limit Comparison Test

Effect of Polynomial Factors on Series Convergence

Polynomial terms like k² grow slower than exponential decay terms. When combined with a geometric term with ratio less than 1, the exponential decay dominates, often ensuring convergence. Recognizing this balance is key to justifying convergence in series with mixed polynomial and exponential terms.
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Convergence of Taylor & Maclaurin Series
Related Practice
Textbook Question

45–63. Absolute and conditional convergence Determine whether the following series converge absolutely, converge conditionally, or diverge.

∑ (k = 1 to ∞) (3/4)ᵏ

Textbook Question

11–86. Applying convergence tests Determine whether the following series converge. Justify your answers.

∑ (from k = 1 to ∞) (4k)! / (k!)⁴

Textbook Question

72–86. Evaluating series Evaluate each series or state that it diverges.

∑ (k = 1 to ∞) ((1/3) × (5/6)ᵏ + (3/5) × (7/9)ᵏ)

Textbook Question

55–70. More sequences

Find the limit of the following sequences or determine that the sequence diverges.


{(75n⁻¹ / 99ⁿ) + (5ⁿsinn / 8ⁿ)}

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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 ∞) 1 / (2k − √k)

Textbook Question

9–30. The Ratio and Root Tests Use the Ratio Test or the Root Test to determine whether the following series converge absolutely or diverge.

∑ (from k = 1 to ∞) ((-1)ᵏ⁺¹ × k²ᵏ) / (k! × k!)