Ch. 10 - Infinite Sequences and Series

Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook

Hass 15th Edition

Ch. 10 - Infinite Sequences and Series

Problem 10.PE.30

Chapter 10, Problem 10.PE.30

Determining Convergence of Series
Which of the series in Exercises 25–44 converge absolutely, which converge conditionally, and which diverge? Give reasons for your answers.

∑ (from n = 2 to ∞) 1/[n(ln n)²]

Verified step by step guidance

Identify the given series: \( \sum_{n=2}^{\infty} \frac{1}{n (\ln n)^2} \). We want to determine if it converges absolutely, conditionally, or diverges.

Since all terms \( \frac{1}{n (\ln n)^2} \) are positive for \( n \geq 2 \), absolute convergence is equivalent to convergence of the series itself.

Use the Integral Test to analyze convergence because the function \( f(x) = \frac{1}{x (\ln x)^2} \) is positive, continuous, and decreasing for \( x > 1 \).

Set up the integral \( \int_2^{\infty} \frac{1}{x (\ln x)^2} \, dx \). Use the substitution \( t = \ln x \), which implies \( dt = \frac{1}{x} dx \), so the integral becomes \( \int_{\ln 2}^{\infty} \frac{1}{t^2} \, dt \).

Evaluate the integral \( \int_{\ln 2}^{\infty} t^{-2} \, dt \) to determine if it converges. If the integral converges, then by the Integral Test, the series converges absolutely; if it diverges, the series diverges.

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Key Concepts

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

Absolute and Conditional Convergence

A series converges absolutely if the series of absolute values converges. If the original series converges but not absolutely, it is conditionally convergent. Understanding this distinction helps classify the behavior of series involving alternating or positive terms.

Integral Test for Convergence

The integral test compares a series to an improper integral of a related function. If the integral of f(x) from some point to infinity converges, then the series ∑ f(n) converges. This test is useful for series with terms like 1/[n(ln n)^p].

Behavior of Logarithmic Functions in Series

Logarithmic terms in denominators affect convergence rates. For example, series with terms 1/[n(ln n)^p] converge or diverge depending on the exponent p. Recognizing how ln n influences the series is key to applying convergence tests correctly.

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Related Practice

Textbook Question

Theory and Examples

Suppose that a₁, a₂, a₃, …, aₙ are positive numbers satisfying the following conditions:

i) a₁ ≥ a₂ ≥ a₃ ≥ …;

ii) the series a₂ + a₄ + a₈ + a₁₆ + … diverges.

Show that the series

a₁/1 + a₂/2 + a₃/3 + …

diverges.

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Textbook Question

Power Series

In Exercises 47–56, (a) find the series’ radius and interval of convergence. Then identify the values of x for which the series converges (b) absolutely and (c) conditionally.

∑ (from n = 1 to ∞) (x + 4)ⁿ/(n3ⁿ)

Textbook Question

Convergent Series

Find the sums of the series in Exercises 19–24.

∑ (from n = 2 to ∞) -2/[n(n+1)]

Textbook Question

Determining Convergence of Sequences

Which of the sequences whose nth terms appear in Exercises 1–18 converge, and which diverge? Find the limit of each convergent sequence.

aₙ = 1 + (0.9)ⁿ

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Textbook Question

30. b. By differentiating the series in part (a) term by term, show that