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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 10.7.19
Capitolo 10, Problema 10.7.19

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 ∞) (2ᵏ / k⁹⁹)

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Identify the series given: \( \sum_{k=1}^{\infty} \frac{2^{k}}{k^{99}} \). We want to determine if it converges absolutely or diverges using the Ratio Test or the Root Test.
Recall the Ratio Test: For a series \( \sum a_k \), compute \( L = \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| \). If \( L < 1 \), the series converges absolutely; if \( L > 1 \), it diverges; if \( L = 1 \), the test is inconclusive.
Calculate \( \frac{a_{k+1}}{a_k} \) for \( a_k = \frac{2^{k}}{k^{99}} \): \[ \frac{a_{k+1}}{a_k} = \frac{2^{k+1} / (k+1)^{99}}{2^{k} / k^{99}} = 2 \cdot \frac{k^{99}}{(k+1)^{99}} \]
Take the limit as \( k \to \infty \) of \( 2 \cdot \frac{k^{99}}{(k+1)^{99}} \). Since \( \frac{k}{k+1} \to 1 \), this limit simplifies to \( 2 \cdot 1 = 2 \).
Interpret the result: Since the limit \( L = 2 > 1 \), by the Ratio Test, the series \( \sum_{k=1}^{\infty} \frac{2^{k}}{k^{99}} \) diverges.

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Ratio Test

The Ratio Test determines the convergence of a series by examining the limit of the absolute value of the ratio of consecutive terms. If this limit is less than 1, the series converges absolutely; if greater than 1, it diverges; if equal to 1, the test is inconclusive.
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Ratio Test

Root Test

The Root Test analyzes convergence by taking the nth root of the absolute value of the nth term and evaluating its limit as n approaches infinity. If the limit is less than 1, the series converges absolutely; if greater than 1, it diverges; if equal to 1, the test is inconclusive.
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Root Test

Absolute Convergence

A series converges absolutely if the series of the absolute values of its terms converges. Absolute convergence guarantees convergence of the original series and is a stronger condition than conditional convergence.
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Choosing a Convergence Test
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