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Ch. 10 - Infinite Sequences and Series
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 10, Problem 10.3.8

Applying the Integral Test
Use the Integral Test to determine if the series in Exercises 1–12 converge or diverge. Be sure to check that the conditions of the Integral Test are satisfied.
∑ (from n = 2 to ∞) ln(n²) / n

Verified step by step guidance
1
First, identify the function corresponding to the terms of the series: define \( f(x) = \frac{\ln(x^2)}{x} \) for \( x \geq 2 \).
Check the conditions for the Integral Test: verify that \( f(x) \) is continuous, positive, and decreasing on the interval \([2, \infty)\).
Simplify the function \( f(x) \) using logarithm properties: \( \ln(x^2) = 2 \ln(x) \), so \( f(x) = \frac{2 \ln(x)}{x} \).
Set up the improper integral to apply the Integral Test: \( \int_2^{\infty} \frac{2 \ln(x)}{x} \, dx \).
Evaluate the integral (or analyze its behavior) to determine if it converges or diverges, which will tell you whether the series converges or diverges by the Integral Test.

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Key Concepts

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

Integral Test for Series Convergence

The Integral Test determines the convergence or divergence of an infinite series by comparing it to an improper integral. If the function corresponding to the series terms is positive, continuous, and decreasing for x ≥ N, then the series and the integral either both converge or both diverge.
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Conditions for Applying the Integral Test

Before using the Integral Test, verify that the function f(x) representing the series terms is positive, continuous, and decreasing on the interval [N, ∞). These conditions ensure the integral comparison is valid and the test can be applied correctly.
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Evaluating the Improper Integral

To apply the Integral Test, evaluate the improper integral of f(x) from N to infinity. This often involves techniques like substitution or integration by parts. The convergence or divergence of this integral directly determines the behavior of the original series.
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