- #1
ani9890
- 11
- 0
Series Comparison Test, URGENT help?
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.
1. For all n>2, ∑ 1/ (n^2−1) < 1/n^2 so converges
2. For all n> 1, ∑ arctan(n) / n^3 < pi / 2n^3 so converges
3. For all n>1, ∑ ln(n)/n^2 < 1/n^1.5 so converges
4. For all n>1, ∑ 1/nln(n) < 2/n so diverges
5. For all n>2, ∑ n/(n^3 - 8) < 2/n^2 so converges
6. For all n>2, ∑ ln(n)/n > 1/n so diverges
I believe (1) is incorrect.
(2), (3), and (6) I believe are correct.
But I can't figure out 4 and 5 ? I think 4 is incorrect but I'm not sure, and for 5 shouldn't it be compared to 1/n^2 ?
Please help!
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.
1. For all n>2, ∑ 1/ (n^2−1) < 1/n^2 so converges
2. For all n> 1, ∑ arctan(n) / n^3 < pi / 2n^3 so converges
3. For all n>1, ∑ ln(n)/n^2 < 1/n^1.5 so converges
4. For all n>1, ∑ 1/nln(n) < 2/n so diverges
5. For all n>2, ∑ n/(n^3 - 8) < 2/n^2 so converges
6. For all n>2, ∑ ln(n)/n > 1/n so diverges
I believe (1) is incorrect.
(2), (3), and (6) I believe are correct.
But I can't figure out 4 and 5 ? I think 4 is incorrect but I'm not sure, and for 5 shouldn't it be compared to 1/n^2 ?
Please help!
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