Textbook Question
11–86. Applying convergence tests Determine whether the following series converge. Justify your answers.
∑ (from k = 1 to ∞)1 / ln(eᵏ + 1)
Ch. 10 - Sequences and Infinite Series
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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Table of contents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 10, Problem 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)
Verified step by step guidance1
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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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 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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07:15Root 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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07:51Choosing a Convergence Test
Related Practice
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Is it possible for a series of positive terms to converge conditionally? Explain.
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21–42. Geometric series Evaluate each geometric series or state that it diverges.
39.∑ (k = 2 to ∞) (–0.15)ᵏ
Textbook Question
35–44. Limits of sequences Write the terms a₁, a₂, a₃, and a₄ of the following sequences. If the sequence appears to converge, make a conjecture about its limit. If the sequence diverges, explain why.
aₙ = 3 + cos(π*ⁿ) ; n = 1, 2, 3, …
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Limits of sequences
Find the limit of the following sequences or determine that the sequence diverges.
{nsin(6 / n)}
1
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Textbook Question
Does a geometric series always have a finite value?