# Comparison Test

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.)

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

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

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

## Answers and Replies

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Certainly all the series converge (assuming they start at $n \geq 1$). I want you to look much more closely at the arguments they present though, and see if you can find anything wrong with any of them.

OlderDan
Homework Helper
Data said:
sure you can, for example

$$\frac{1}{n^{\frac{3}{2}}} > \frac{\ln n}{n^2} \ \forall n > 0.$$

Their particular argument is obviously wrong, though. If you look at their arguments for A and C, there are problems there too (what happens if $n=3$? Do the inequalities they claim actually hold?).
Sorry, I pulled my reply because it seemed inappropriate after yours, which I did not know was going to be there. I didn't check their claimed inequality- a serious oversight on my part.

Actually, before I deleted mine I fixed it a little - their argument for A is ok (originally I somehow read it as n^2 in the denominator, or something - but even that doesn't make sense, because I still thought it would converge! Strange, this brain of mine is).

so A is false because there is a 2 in front of the 'sum of' sign(which you cant do)? and B and C are true?

no. A is correct (of course putting the 2 in front doesn't matter! The sum itself is just a number, like any other). Look at our previous posts and then look carefully at the arguments for B and C. Why do you think they are correct?

saltydog