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Quick Question on Limit Comparison Test

  1. Mar 1, 2010 #1
    Can the limit comparison test work on alternating series? I looked at the conditions online and it said that the series a_n and b_n must be > 0 in order to use it. To be specific it said, "Suppose that a_n > 0 and b_n>0 for all n> or = to N (N a positive integer)."
    I asked my teacher about this and he said that it does work, and yet.... I don't know. Maybe I misinterpreted the conditions.
     
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  3. Mar 2, 2010 #2

    mathman

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    When dealing with alternating series, the comparison can be carried out with absolute values of the series in question. The comparison series must have positive terms.
     
  4. Mar 2, 2010 #3

    Char. Limit

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    Isn't there an alternating series test for just such situations?
     
  5. Mar 2, 2010 #4
    To be particular, I'm curious about being able to use the Limit Comparison test for
    [tex]\sum_{n=1}^{\infty} \frac{(-1)^n}{n}[/tex]
    I understand that it's easier to do with the AST, but I tried this with the LCT and it didn't turn out nice.

    I compared it to 1/n.
    [tex]\lim_{n\rightarrow \infty} |\frac{1}{n} * \frac{n}{(-1)^n}|[/tex]
    [tex]\lim_{n\rightarrow \infty} |\frac{1}{(-1)^n}|[/tex]
    [tex] = 1[/tex]
    1/n diverges, so (-1)^n / n must diverge..? That's what I got using this test that is, and it's wrong. What did I do incorrectly with my math?
     
  6. Mar 2, 2010 #5
    The limit comparison test doesn't use absolute values. Since limnā†’āˆž 1/(-1)n doesn't exist/diverges, this comparison test doesn't help.
     
  7. Mar 2, 2010 #6

    Char. Limit

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    Also, the alternating harmonic series does converge... even if I forgot exactly why, or to what value... is it log 2?
     
  8. Mar 2, 2010 #7
    It is log(2) and comes from the http://en.wikipedia.org/wiki/Mercator_series" [Broken].
     
    Last edited by a moderator: May 4, 2017
  9. Mar 2, 2010 #8
    My first question asked if you could use the LCT on an alternating or negative series. Mathman replied that it's possible as long as you compare it to a positive function and while using absolute values. That's what I tried doing, and that's what my second post is about. Something seems off about the LCT or I made a mistake.

    I'm aware that it does converge, because it should converge conditionally by using the AST. However, I'm wondering why the LCT doesn't work for this if apparently you are able to use it on a negative or alternating series. And it's -log(2), not log(2) by the way.
     
    Last edited: Mar 2, 2010
  10. Mar 2, 2010 #9
    I also believe it will work with alternating series as long as the two series you're comparing alternate with the same sign so their limit converges to a number (since it doesn't use absolute values), unlike 1/(-1)n which oscillates between -1 and 1.
     
  11. Mar 3, 2010 #10
    The comparison tests can be applied only on the postive series.
     
  12. Mar 3, 2010 #11
    I'm getting mixed answers here.. :(

    @Bohrok: But if I use an alternating series as a comparison for another alternating series, how will I know if the series I compared converges or diverges? And if you say AST, that makes the LCT useless because you could've done AST on the original series.

    @Sweet GirL: I'm starting to believe in this. However, my teacher also said you can use it on alternating series by putting absolute values, but it doesn't seem like it works.
     
  13. Mar 3, 2010 #12

    Char. Limit

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    My math teacher agrees with Sweet Girl. LCT can only be done on positive series.

    Sorry, log(.5) then.
     
  14. Mar 3, 2010 #13
    Okay. I'll go challenge my math teacher. And by challenge, I mean ask how to do the above series using LCT and subtly hint that he's wrong.
     
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