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Ch. 10 - Sequences and Infinite Series
Briggs - Calculus: Early Transcendentals 3rd Edition
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Tutti i libri di testoBriggs 3rd EditionCh. 10 - Sequences and Infinite SeriesProblema 10.8.73
Capitolo 10, Problema 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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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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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
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45–63. Absolute and conditional convergence Determine whether the following series converge absolutely, converge conditionally, or diverge.

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

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13–52. Limits of sequences

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13–52. Limits of sequences

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


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