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.)(adsbygoogle = window.adsbygoogle || []).push({});

A.) for n > 2, [tex]\frac{n}{n^3-4}[/tex] < [tex]\frac{2}{n^2}[/tex], and the series [tex]2\sum \frac{1}{n^2}[/tex] converges, so by the comparison test, the series [tex]\sum \frac{n}{n^3-4}[/tex] converges

B.) for n > 2, [tex]\frac{ln(n)}{n^2}[/tex] > [tex]\frac{1}{n^2}[/tex], and the series [tex]\sum \frac{1}{n^2}[/tex] converges, so by the comparison test, the series [tex]\sum \frac{ln(n)}{n^2}[/tex] converges

C.) for n > 2, [tex]\frac{1}{n^2-6}[/tex] < [tex]\frac{1}{n^2}[/tex], and the series [tex]\sum \frac{1}{n^2}[/tex] converges, so by the comparison test, the series [tex]\sum \frac{1}{n^2-6}[/tex] converges

For A.) and C.) sounds true, by the p-series, both converges so that means mean the original series converges, so A.) and C.) are True right?

and B.) I also think it's true by the p-series.

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# Homework Help: Comparison Test

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