2. For all n > 1,6 <2, and the series converges, so by the Comparison Test, the series converges. 3. For all n > 2, In(n) n 1 >, and the series Σ diverges, so by the Comparison Test, the series > n 6-n³ In(n) n diverges.

2. For all n > 1,6 <2, and the series converges, so by the Comparison Test, the series converges. 3. For all n > 2, In(n) n 1 >, and the series Σ diverges, so by the Comparison Test, the series > n 6-n³ In(n) n diverges.

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Question

Each of the following statements is an attempt to show that a given series is convergent or divergent using the Comparison Test (NOT the Limit Comparison Test.) For each statement, enter C (for "correct") if the argument is valid, or enter I (for "incorrect") if any part of the argument is flawed. (Note: if the conclusion is true but the argument that led to it was wrong, you must enter I.)

2. For all \( n > 1 \), \(\frac{n}{6-n^3} < \frac{1}{n^2}\), and the series \(\sum \frac{1}{n^2}\) converges, so by the Comparison Test, the series \(\sum \frac{n}{6-n^3}\) converges.

3. For all \( n > 2 \), \(\frac{\ln(n)}{n} > \frac{1}{n}\), and the series \(\sum \frac{1}{n}\) diverges, so by the Comparison Test, the series \(\sum \frac{\ln(n)}{n}\) diverges.

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