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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 Limil 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 1 converges, so by the Comparison Test, the series > n n 1. For all n > 2, and the series 2 >÷ converges. n3 – 7 n2 1 n3 – 7 n n converges, so by the Comparison Test, the series >. n2 2. For all n > 1, < -, and the series > converges. 5 – n3 sin (n) n2 5 - n3 3. For all n > 1, n2 1 4. For all n > 1, 1 , and the series ). n2 sin (n) converges, so by the Comparison Test, the series >. n? 1 < converges. and the series 2 ) diverges, so by the Comparison Test, the series >. diverges. W WW 曲 田

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 Limil 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 1 converges, so by the Comparison Test, the series > n n 1. For all n > 2, and the series 2 >÷ converges. n3 – 7 n2 1 n3 – 7 n n converges, so by the Comparison Test, the series >. n2 2. For all n > 1, < -, and the series > converges. 5 – n3 sin (n) n2 5 - n3 3. For all n > 1, n2 1 4. For all n > 1, 1 , and the series ). n2 sin (n) converges, so by the Comparison Test, the series >. n? 1 < converges. and the series 2 ) diverges, so by the Comparison Test, the series >. diverges. W WW 曲 田

Calculus: Early Transcendentals
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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čach of the following statements is an attempt to show that a given series is convergent or divergent not using the Comparison Test (NOT the Limil uomparison 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 and the series 2 ). n- 1 converges, so by the Comparison Test, the series >. n2 1 converges, so by the Comparison Test, the series n2 n 1. For all n > 2, n3 converges. n3 – 7 n n 2. For all n > 1, Σ ... and the series n2 converges. 5 — п3 sin? (n) 5 — п3 sin?(n) 1 and the series n2 2 and the series 2 converges, so by the Comparison Test, the series >. n2 3. For all n > 1, n2 1 4. For all n > 1, ... ... converges. ... ... diverges, so by the Comparison Test, the series diverges. n In(n) In(n) converges, so by the Comparison Test, the series >. n In(n) In(n) 5. For all n > 2, n2 1 6. For all n > 2, n2 n n and the series n2 converges. n2 1 converges, so by the Comparison Test, the series n2 n2 1 and the series n- Σ converges. 4 n? – 4 V V
Transcribed Image Text:čach of the following statements is an attempt to show that a given series is convergent or divergent not using the Comparison Test (NOT the Limil uomparison 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 and the series 2 ). n- 1 converges, so by the Comparison Test, the series >. n2 1 converges, so by the Comparison Test, the series n2 n 1. For all n > 2, n3 converges. n3 – 7 n n 2. For all n > 1, Σ ... and the series n2 converges. 5 — п3 sin? (n) 5 — п3 sin?(n) 1 and the series n2 2 and the series 2 converges, so by the Comparison Test, the series >. n2 3. For all n > 1, n2 1 4. For all n > 1, ... ... converges. ... ... diverges, so by the Comparison Test, the series diverges. n In(n) In(n) converges, so by the Comparison Test, the series >. n In(n) In(n) 5. For all n > 2, n2 1 6. For all n > 2, n2 n n and the series n2 converges. n2 1 converges, so by the Comparison Test, the series n2 n2 1 and the series n- Σ converges. 4 n? – 4 V V
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