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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 10.4.65b
Chapter 10, Problem 10.4.65b

Loglog p-series Consider the series ∑ (k = 2 to ∞) 1 / (k(ln k)(ln ln k)ᵖ), where p is a real number.
b. Which of the following series converges faster? Explain.
∑ (k = 2 to ∞) 1 / (k(ln k)²) or ∑ (k = 3 to ∞) 1 / (k(ln k)(ln ln k)²)?

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1
Step 1: Understand the two series given for comparison. The first series is \( \sum_{k=2}^\infty \frac{1}{k (\ln k)^2} \) and the second series is \( \sum_{k=3}^\infty \frac{1}{k (\ln k) (\ln \ln k)^2} \). Both are infinite series with terms involving logarithmic functions in the denominator.
Step 2: Recall that the speed of convergence of a series depends on how quickly its terms approach zero. Smaller terms generally mean faster convergence. So, we want to compare the size of the terms \( \frac{1}{k (\ln k)^2} \) and \( \frac{1}{k (\ln k) (\ln \ln k)^2} \) for large \( k \).
Step 3: For large \( k \), note that \( \ln k > 0 \) and \( \ln \ln k > 0 \). Since \( (\ln k)^2 \) grows faster than \( (\ln k)(\ln \ln k)^2 \) because \( (\ln \ln k)^2 \) grows slower than \( \ln k \), the denominator \( k (\ln k)^2 \) is larger than \( k (\ln k)(\ln \ln k)^2 \) for sufficiently large \( k \).
Step 4: Since the denominator of the first series' terms is larger, its terms are smaller compared to the second series' terms for large \( k \). Smaller terms imply that the first series converges faster than the second series.
Step 5: To confirm this rigorously, one could apply the Integral Test or compare the series using the Limit Comparison Test by examining the limit of the ratio of their terms as \( k \to \infty \). This will show which series' terms decrease faster, confirming the conclusion about convergence speed.

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Key Concepts

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

Comparison Test for Series Convergence

The Comparison Test helps determine the convergence of a series by comparing it to another series with known behavior. If a series has terms smaller than those of a convergent series, it also converges. Conversely, if its terms are larger than those of a divergent series, it diverges. This test is useful for comparing the speed of convergence between series.
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Direct Comparison Test

Logarithmic and Log-Logarithmic Series

Logarithmic and log-logarithmic series involve terms with logarithmic functions in the denominator, such as ln(k) and ln(ln(k)). These series converge very slowly, and their convergence depends sensitively on the powers of these logarithmic terms. Understanding how these nested logarithms affect term size is key to analyzing convergence rates.
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Logarithms Introduction

Rate of Convergence

The rate of convergence describes how quickly the partial sums of a series approach the series limit. Series with terms that decrease faster converge more quickly. Comparing terms like 1/(k(ln k)^2) and 1/(k(ln k)(ln ln k)^2) involves analyzing which denominator grows faster, thus indicating which series converges faster.
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Intro To Related Rates
Related Practice
Textbook Question

Explain why or why not

Determine whether the following statements are true and give an explanation or counterexample.



b.If limₙ→∞aₙ = 0 and limₙ→∞bₙ = ∞, then limₙ→∞aₙbₙ = 0.

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Textbook Question

39–40. {Use of Tech} Lower and upper bounds of a series

For each convergent series and given value of n, use Theorem 10.13 to complete the following.


c. Find lower and upper bounds (Lₙ and Uₙ, respectively) for the exact value of the series.


39. ∑ (k = 1 to ∞) 1 / k⁷ ; n = 2

Textbook Question

39–40. {Use of Tech} Lower and upper bounds of a series

For each convergent series and given value of n, use Theorem 10.13 to complete the following.


b. Find an upper bound for the remainder Rₙ.


39. ∑ (k = 1 to ∞) 1 / k⁷ ; n = 2

Textbook Question

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.


c. Suppose f is a continuous, positive, decreasing function, for x ≥ 1, and aₖ = f(k), for k = 1, 2, 3, …. If ∑ (k = 1 to ∞) aₖ converges to L, then ∫ (1 to ∞) f(x) dx converges to L.

Textbook Question

57–60. Heights of bouncing balls A ball is thrown upward to a height of hₒ meters. After each bounce, the ball rebounds to a fraction r of its previous height. Let hₙ be the height after the nth bounce. Consider the following values of hₒ and r.


b. Find an explicit formula for the nth term of the sequence {hₙ}.


h₀ = 30,r = 0.25

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Textbook Question

41–44. {Use of Tech} Remainders and estimates Consider the following convergent series.


c. Find lower and upper bounds (Lₙ and Uₙ, respectively) on the exact value of the series.


41. ∑ (k = 1 to ∞) 1 / k⁶