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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.R.61
Chapter 10, Problem 10.R.61

42–76. Convergence or divergence Use a convergence test of your choice to determine whether the following series converge.
∑ (from k = 3 to ∞)ln(k) / k³ᐟ²

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Identify the series given: \( \sum_{k=3}^{\infty} \frac{\ln(k)}{k^{3/2}} \). We want to determine if this series converges or diverges.
Consider the behavior of the terms \( a_k = \frac{\ln(k)}{k^{3/2}} \) as \( k \to \infty \). Since \( \ln(k) \) grows slowly and \( k^{3/2} \) grows faster, the terms approach zero, which is necessary for convergence but not sufficient.
Choose a convergence test suitable for series with positive terms and involving logarithms and powers. The Comparison Test or the Limit Comparison Test are good candidates here.
Compare \( a_k \) with a simpler series \( b_k = \frac{1}{k^{3/2}} \), which is a p-series with \( p = \frac{3}{2} > 1 \) and is known to converge.
Apply the Limit Comparison Test by evaluating \( \lim_{k \to \infty} \frac{a_k}{b_k} = \lim_{k \to \infty} \frac{\ln(k)/k^{3/2}}{1/k^{3/2}} = \lim_{k \to \infty} \ln(k) \). Since this limit diverges to infinity, refine the approach by noting that \( \ln(k) \) grows slower than any power of \( k \), so the original series behaves similarly to the convergent p-series, indicating convergence.

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Key Concepts

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

Convergence of Infinite Series

An infinite series converges if the sum of its terms approaches a finite limit as the number of terms grows indefinitely. Determining convergence involves analyzing the behavior of the terms and applying appropriate tests to see if the series sums to a finite value.
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Convergence of an Infinite Series

Comparison Test

The Comparison Test involves comparing the given series to a known benchmark series with positive terms. If the given series' terms are smaller than those of a convergent series, it also converges; if larger than a divergent series, it diverges. This test is useful when terms resemble simpler series.
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Direct Comparison Test

Behavior of Logarithmic and Power Functions in Series

Understanding how logarithmic functions like ln(k) grow compared to power functions like k^(3/2) is crucial. Since ln(k) grows slower than any positive power of k, the term ln(k)/k^(3/2) behaves similarly to 1/k^(3/2) for large k, which helps in applying convergence tests effectively.
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Representing Functions as Power Series
Related Practice
Textbook Question

42–76. Convergence or divergence Use a convergence test of your choice to determine whether the following series converge.

∑ (from k = 1 to ∞)k⁴ / √(9k¹² + 2)

Textbook Question

27–37. Evaluating series Evaluate the following infinite series or state that the series diverges.

∑ (from k = 0 to ∞)(tan⁻¹(k + 2) − tan⁻¹k)

Textbook Question

77–87. Absolute or conditional convergence

Determine whether the following series converge absolutely, converge conditionally, or diverge.

∑ (from k = 1 to ∞)(−2)ᵏ⁺¹ / k²

Textbook Question

{Use of Tech} Error in a finite alternating sum

How many terms of the series ∑ (from k = 1 to ∞)(−1)ᵏ⁺¹ / k⁴ must be summed to ensure that the approximation error is less than 10⁻⁸?

Textbook Question

Geometric sums

Evaluate the geometric sums

∑ (from k = 0 to 9) (0.2)ᵏand∑ (from k = 2 to 9) (0.2)ᵏ.

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Textbook Question

25–26. Recursively defined sequences

The following sequences {aₙ} from n = 0 to ∞ are defined by a recurrence relation. Assume each sequence is monotonic and bounded.


a.Find the first five terms a₀, a₁, ..., a₄ of each sequence.


25.aₙ₊₁ = (1 / 2) aₙ + 8;a₀ = 80

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