Question

(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 1 (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, 6 <2, and the series 2 E 2. For all n > 1, nihon) < , and the series 2 3. For all n > 2, bagama > I, and the series 4. For all n > 1, < , and the series 5. For all In > 2, Ingm > , and the series & 6. For all n > 1, < , and the series converges, so by the Comparison Test, the series converges. diverges, so by the Comparison Test, the series En non diverges. converges, so by the Comparison Test, the series by converges. converges, so by the Comparison Test, the series - converges. diverges, so by the Comparison Test, the series om diverges. converges, so by the Comparison Test, the series converges.

Answer 1

on conv is correct Sola: Formula: comparison test: Let I up and son be two series of positive real numbers and there is a natural number in such that unekin for all in a mi k being a fixed positive number. Then i E un is convengent if Eunis convergent, . (u į un is divergent if Eur is divergent. Le 1. For all n32, s , and the series 2. E na converges, so by the comparison Fest, the senies Em converges. . For all nys, nom 2 , and the series 2 [ń diverges; so by the comparison Test, the series Exilmin divorges Here the con elusion that have bout the argument that is flawed. I 3. For all n>2, min) , bai, and the series Etz converges, so by the comparison Fest, the series Elnon converges. . Here the conclusion Inth) for any is correst but the argument is flawed. 2 4. For all main sme, and the series Ene converges, so by the comparison Test, the series sa ne converges. :D 5. Fon all my 2, in(n)sh, and the series 1 h diverges i so by the comparison Test the series E Incon diverges. ...02 Antoya 2 1- 3 - 1n3

I b. 03 For all nyi, are tan(n) , I, and the serica I sto con verges, so by comparison Test the series s aretan (n) converges Here the conclusion is true but the argument are tan(n) , hrallns that led to it was wrong. N3 n3 2h3