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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 10.7.9
Chapter 10, Problem 10.7.9

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) / (k!)

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1
Identify the series given: \( \sum_{k=1}^{\infty} \frac{(-1)^k}{k!} \). Since the terms involve factorials, the Ratio Test is often convenient here.
Recall the Ratio Test formula: For the 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.
Write the general term \( a_k = \frac{(-1)^k}{k!} \). Then, \( a_{k+1} = \frac{(-1)^{k+1}}{(k+1)!} \). Compute the ratio of absolute values: \( \left| \frac{a_{k+1}}{a_k} \right| = \left| \frac{(-1)^{k+1}}{(k+1)!} \cdot \frac{k!}{(-1)^k} \right| = \frac{1}{k+1} \).
Evaluate the limit as \( k \to \infty \): \( L = \lim_{k \to \infty} \frac{1}{k+1} = 0 \). Since \( L < 1 \), the series converges absolutely.
Conclude that by the Ratio Test, the series \( \sum_{k=1}^{\infty} \frac{(-1)^k}{k!} \) converges absolutely because the limit of the ratio of consecutive terms is zero.

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Key Concepts

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

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 the nth root of the absolute value of the terms in a series. If the limit of this nth root as n approaches infinity 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 implies convergence, and it is a stronger form of convergence that allows the use of tests like the Ratio and Root Tests effectively.
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Choosing a Convergence Test
Related Practice
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

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Textbook Question

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

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Textbook Question

21–42. Geometric series Evaluate each geometric series or state that it diverges.  


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Textbook Question

46–53. Decimal expansions

Write each repeating decimal first as a geometric series and then as a fraction (a ratio of two integers).


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Textbook Question

54–69. Telescoping series

For the following telescoping series, find a formula for the nth term of the sequence of partial sums {Sₙ}. Then evaluate limₙ→∞ Sₙ to obtain the value of the series or state that the series diverges.


57. ∑ (k = 1 to ∞) 1 / ((k + 6)(k + 7))

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Textbook Question

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

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