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Ch. 10 - Sequences and Infinite Series
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243당신이 사용하는 게 아니라요?교과서 변경
10장, 문제 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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주요 개념

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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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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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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