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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 1.) | 21/12 , and the series 2Σ <2, and the series arctan(n) converges. converges. arctan(n) <2, and the series 12³ 12³ In(n) 72² and the series 1. For all n > 2,8 < 2. For all n > 1,6 3. For all n > 1, < converges, so by the Comparison Test, the series converges, so by the Comparison Test, the series Σ Σ converges, so by the Comparison Test, the series Σ- converges, so by the Comparison Test, the series Σ ¹() 4. For all n > 1, converges. 5. For all n > 1, nln(n) < 1/12, and the series 2 Σ diverges, so by the Comparison Test, the series Σ nin(n) diverges. converges, so by the Comparison Test, the series Σ. In(n) In(n) 6. For all n > 2. and the series 7² converges. 2 converges.

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 1.) | 21/12 , and the series 2Σ <2, and the series arctan(n) converges. converges. arctan(n) <2, and the series 12³ 12³ In(n) 72² and the series 1. For all n > 2,8 < 2. For all n > 1,6 3. For all n > 1, < converges, so by the Comparison Test, the series converges, so by the Comparison Test, the series Σ Σ converges, so by the Comparison Test, the series Σ- converges, so by the Comparison Test, the series Σ ¹() 4. For all n > 1, converges. 5. For all n > 1, nln(n) < 1/12, and the series 2 Σ diverges, so by the Comparison Test, the series Σ nin(n) diverges. converges, so by the Comparison Test, the series Σ. In(n) In(n) 6. For all n > 2. and the series 7² converges. 2 converges.

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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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.) 1. For all \( n > 2 \), \(\frac{n}{n-8} < \frac{2}{n^2}\), and the series \(2 \sum \frac{1}{n^2}\) converges, so by the Comparison Test, the series \(\sum \frac{n}{n-8}\) converges. 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 \( n > 1 \), \(\frac{\arctan(n)}{n^3} < \frac{\pi}{2n^3}\), and the series \(\frac{\pi}{2} \sum \frac{1}{n^3}\) converges, so by the Comparison Test, the series \(\sum \frac{\arctan(n)}{n^3}\) converges. 4. For all \( n > 1 \), \(\frac{\ln(n)}{n^2} < \frac{1}{n^{1.5}}\), and the series \(\sum \frac{1}{n^{1.5}}\) converges, so by the Comparison Test, the series \(\sum \frac{\ln(n)}{n^2}\) converges. 5. For all \( n > 1 \), \(\frac{1}{n \ln(n)} > \frac{2}{n}\), and the series \(\sum \frac{1}{n}\) diverges, so by the Comparison Test, the series \(\sum \frac{1}{n \ln(n)}\) diverges. 6. For all \( n > 2 \), \
Transcribed Image Text: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.) 1. For all \( n > 2 \), \(\frac{n}{n-8} < \frac{2}{n^2}\), and the series \(2 \sum \frac{1}{n^2}\) converges, so by the Comparison Test, the series \(\sum \frac{n}{n-8}\) converges. 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 \( n > 1 \), \(\frac{\arctan(n)}{n^3} < \frac{\pi}{2n^3}\), and the series \(\frac{\pi}{2} \sum \frac{1}{n^3}\) converges, so by the Comparison Test, the series \(\sum \frac{\arctan(n)}{n^3}\) converges. 4. For all \( n > 1 \), \(\frac{\ln(n)}{n^2} < \frac{1}{n^{1.5}}\), and the series \(\sum \frac{1}{n^{1.5}}\) converges, so by the Comparison Test, the series \(\sum \frac{\ln(n)}{n^2}\) converges. 5. For all \( n > 1 \), \(\frac{1}{n \ln(n)} > \frac{2}{n}\), and the series \(\sum \frac{1}{n}\) diverges, so by the Comparison Test, the series \(\sum \frac{1}{n \ln(n)}\) diverges. 6. For all \( n > 2 \), \
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