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? 1. For all n ≥ 2, the Comparison Test, the series ? 2. For all n > 3, n³ by the Comparison Test, the series ? 5-n³ n 5-n³ 3. For all n ≥ 3, n 1 n² 8 1 the Comparison Test, the series Σ - 1 - n - < n² - 8 2 converges. - n² " 1 n² and the series " and the series 2 converges. and the series converges. Σ=1/12 n² converges, so b converges, sa converges, so b

? 1. For all n ≥ 2, the Comparison Test, the series ? 2. For all n > 3, n³ by the Comparison Test, the series ? 5-n³ n 5-n³ 3. For all n ≥ 3, n 1 n² 8 1 the Comparison Test, the series Σ - 1 - n - < n² - 8 2 converges. - n² " 1 n² and the series " and the series 2 converges. and the series converges. Σ=1/12 n² converges, so b converges, sa converges, so b

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Below are statements to show if a series is convergent or divergent only using the comparison test. Determine whether each statement is correct or incorrect:
 

1. For all \( n \geq 2 \), \(\frac{n}{5 - n^3} < \frac{1}{n^2}\), and the series \(\sum \frac{1}{n^2}\) converges, so by the Comparison Test, the series \(\sum \frac{n}{5 - n^3}\) converges. 2. For all \( n \geq 3 \), \(\frac{n}{n^3 - 1} < \frac{2}{n^2}\), and the series \(2 \sum \frac{1}{n^2}\) converges, so by the Comparison Test, the series \(\sum \frac{n}{n^3 - 1}\) converges. 3. For all \( n \geq 3 \), \(\frac{1}{n^2 - 8} < \frac{1}{n^2}\), and the series \(\sum \frac{1}{n^2}\) converges, so by the Comparison Test, the series \(\sum \frac{1}{n^2 - 8}\) converges.
Transcribed Image Text:1. For all \( n \geq 2 \), \(\frac{n}{5 - n^3} < \frac{1}{n^2}\), and the series \(\sum \frac{1}{n^2}\) converges, so by the Comparison Test, the series \(\sum \frac{n}{5 - n^3}\) converges. 2. For all \( n \geq 3 \), \(\frac{n}{n^3 - 1} < \frac{2}{n^2}\), and the series \(2 \sum \frac{1}{n^2}\) converges, so by the Comparison Test, the series \(\sum \frac{n}{n^3 - 1}\) converges. 3. For all \( n \geq 3 \), \(\frac{1}{n^2 - 8} < \frac{1}{n^2}\), and the series \(\sum \frac{1}{n^2}\) converges, so by the Comparison Test, the series \(\sum \frac{1}{n^2 - 8}\) converges.
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