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arctan(n) T C 1. For all n > 1, and the n3 2n3 Σ 1 converges, so by the Comparison Test, the series 2 n3 arctan(n) series > converges. n3 In(n) 2. For all n > 2, n2 1 and the series C n2 1 converges, so by the Comparison Test, the series In(n) converges. n2 sin? (n) 1 and the C 3. For all n > 1, n2 n2 1 >- converges, so by the Comparison Test, the series n2 sin? (n) series converges. n2 1 4. For all n > 1, and the - n In(n) series 2 > 1 diverges, so by the Comparison Test, the series >. 1 diverges. n In(n)

arctan(n) T C 1. For all n > 1, and the n3 2n3 Σ 1 converges, so by the Comparison Test, the series 2 n3 arctan(n) series > converges. n3 In(n) 2. For all n > 2, n2 1 and the series C n2 1 converges, so by the Comparison Test, the series In(n) converges. n2 sin? (n) 1 and the C 3. For all n > 1, n2 n2 1 >- converges, so by the Comparison Test, the series n2 sin? (n) series converges. n2 1 4. For all n > 1, and the - n In(n) series 2 > 1 diverges, so by the Comparison Test, the series >. 1 diverges. n In(n)

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 not 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.) arctan(n) C 1. For all n > 1, and the n3 2n3 1 series 2 n3 converges, so by the Comparison Test, the arctan(n) series converges. n3 In(n) 2. For all n > 2, n2 1 and the series C n2 1 converges, so by the Comparison Test, the series In(n) converges. n2 sin? (n) 1 and the n2 C 3. For all n > 1, n2 1 converges, so by the Comparison Test, the n2 series sin (n) series > converges. n2 1 4. For all n 1, 2 and the n In(n) 1 series 2 >= diverges, so by the Comparison Test, the n 1 series >. diverges. n In(n) 1 5. For all n> 2, 1 and the C
Transcribed Image Text:Each of the following statements is an attempt to show that a given series is convergent or divergent not 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.) arctan(n) C 1. For all n > 1, and the n3 2n3 1 series 2 n3 converges, so by the Comparison Test, the arctan(n) series converges. n3 In(n) 2. For all n > 2, n2 1 and the series C n2 1 converges, so by the Comparison Test, the series In(n) converges. n2 sin? (n) 1 and the n2 C 3. For all n > 1, n2 1 converges, so by the Comparison Test, the n2 series sin (n) series > converges. n2 1 4. For all n 1, 2 and the n In(n) 1 series 2 >= diverges, so by the Comparison Test, the n 1 series >. diverges. n In(n) 1 5. For all n> 2, 1 and the C
n° 2n° 1 23 2 converges, so by the Comparison Test, the 2 series arctan(n) series > converges. n3 In(n) 2. For all n > 2, n2 1 and the series C n2' 1 converges, so by the Comparison Test, the series n2 In(n) converges. n2 sin? (n) 1 and the n2 C 3. For all n > 1, n2 1 converges, so by the Comparison Test, the series n2 sin? (n) series > converges. n2 1 4. For all n > 1, 2 and the n In(n) n Σ Σ 1 diverges, so by the Comparison Test, the n 1 series 2 ) series > diverges. n In(n) 1 5. For all n > 2, n2 1 and the C 1 n2 1 series >5 converges, so by the Comparison Test, the n2 1 Σ series converges. n2 – 1 Vn +1 1 > –, and the 6. For all n > 2, 1 series >- diverges, so by the Comparison Test, the Vn + 1 series > diverges.
Transcribed Image Text:n° 2n° 1 23 2 converges, so by the Comparison Test, the 2 series arctan(n) series > converges. n3 In(n) 2. For all n > 2, n2 1 and the series C n2' 1 converges, so by the Comparison Test, the series n2 In(n) converges. n2 sin? (n) 1 and the n2 C 3. For all n > 1, n2 1 converges, so by the Comparison Test, the series n2 sin? (n) series > converges. n2 1 4. For all n > 1, 2 and the n In(n) n Σ Σ 1 diverges, so by the Comparison Test, the n 1 series 2 ) series > diverges. n In(n) 1 5. For all n > 2, n2 1 and the C 1 n2 1 series >5 converges, so by the Comparison Test, the n2 1 Σ series converges. n2 – 1 Vn +1 1 > –, and the 6. For all n > 2, 1 series >- diverges, so by the Comparison Test, the Vn + 1 series > diverges.
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