Mathematics
Mathematics, 04.11.2019 21:31, mdkester11

(1 point) 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> 2n> 2, ln(n)n> 1nln⁡(n)n> 1n, and the series ∑1n∑1n diverges, so by the comparison test, the series ∑ln(n)n∑ln⁡(n)n diverges.
2. for all n> 1n> 1, n6−n3< 1n2n6−n3< 1n2, and the series ∑1n2∑1n2 converges, so by the comparison test, the series ∑n6−n3∑n6−n3 converges.
3. for all n> 2n> 2, nn3−9< 2n2nn3−9< 2n2, and the series 2∑1n22∑1n2 converges, so by the comparison test, the series ∑nn3−9∑nn3−9 converges.
4. for all n> 2n> 2, ln(n)n2> 1n2ln⁡(n)n2> 1n2, and the series ∑1n2∑1n2 converges, so by the comparison test, the series ∑ln(n)n2∑ln⁡(n)n2 converges.
5. for all n> 1n> 1, 1nln(n)< 2n1nln⁡(n)< 2n, and the series 2∑1n2∑1n diverges, so by the comparison test, the series ∑1nln(n)∑1nln⁡(n) diverges.
6. for all n> 1n> 1, ln(n)n2< 1n1.5ln⁡(n)n2< 1n1.5, and the series ∑1n1.5∑1n1.5 converges, so by the comparison test, the series ∑ln(n)n2∑ln⁡(n)n2 converges.

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