Checking Answers for Comparison Test

[ProBasket]

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.

 

[Data]

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]

sure you can, for example

\frac{1}{n^{\frac{3}{2}}} &gt; \frac{\ln n}{n^2} \ \forall n &gt; 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.

 

[Data]

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).

 

[ProBasket]

so A is false because there is a 2 in front of the 'sum of' sign(which you can't do)? and B and C are true?

 

[Data]

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]

If a given series A, converges and you compare it against another series say B, in which each member of B is LARGER than A then nothing can be said for the series B.