Comparison Test

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  • #1
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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, [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.
 

Answers and Replies

  • #2
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Certainly all the series converge (assuming they start at [itex]n \geq 1[/itex]). 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.
 
  • #3
OlderDan
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Data said:
sure you can, for example

[tex]\frac{1}{n^{\frac{3}{2}}} > \frac{\ln n}{n^2} \ \forall n > 0.[/tex]

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 [itex]n=3[/itex]? 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.
 
  • #4
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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).
 
  • #5
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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?
 
  • #6
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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?
 
  • #7
saltydog
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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.
 

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