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Determining convergence or divergence

  1. Dec 18, 2011 #1
    1. The problem statement, all variables and given/known data
    Use a valid convergence test to see if the sum converges.
    Ʃ(n^(2))/(n^(2)+1)


    2. Relevant equations

    Well, according to p-series, I'd assume this sum diverges, but I don't know which test to use.

    3. The attempt at a solution

    I probably can't do this, but I was think about comparing it to (n^(2))/(n^(2)), which diverges, thus, the original sum diverges. This is one of those problems where it seems so easy, but it's hard. Please help.
     
  2. jcsd
  3. Dec 18, 2011 #2

    SammyS

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    Use the limit test, a.k.a. the nth term test.
     
  4. Dec 18, 2011 #3
    Okay. So I take the limit of this sum as n→∞. Thus, I'm left with ∞/∞, which is undefined, meaning the series diverges. Correct?
     
  5. Dec 18, 2011 #4

    SammyS

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    The limit is of the form ∞/∞, but such a limit can potentially be anything -- maybe convergent, maybe not.

    To find the limit: [itex]\displaystyle\lim_{n\to\infty}\frac{n^2}{n^2+1}\,,[/itex] multiply the numerator & denominator by 1/n2 .
     
  6. Dec 18, 2011 #5
    Okay, so we're using limit comparison. When I multiply that out, I get (n^(4))/(n^(2)+1), and the limit as n goes to infinity of that is just ∞. So the series diverges.

    EDIT: I forgot I could use l'hopital's rule, in this case. So, take the derivative of the top and bottom and simplify, and I'm left with 2n^(2). Take the limit of that as n goes to infinity, and you're left with infinity. So, the series diverges.
     
  7. Dec 18, 2011 #6
    Correct me if I'm wrong, but, the limit as n approaches infinity shouldn't be 1? Applying L'Hopital's rule you should get 2n/2n, which would give you 1.
     
  8. Dec 18, 2011 #7

    LCKurtz

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    No, you aren't using the limit comparison test. You are checking whether the nth term goes to zero which, if it didn't, would imply divergence. But you haven't simplified the expression correctly. It doesn't go to ∞. Recheck your implementation of SammyS's suggestion. What is the correct limit of that expression?
     
  9. Dec 18, 2011 #8
    Well, this sum diverges. So, if I'm doing just a regular limit test, I have to get a limit that equals to zero, or is undefined, in order to prove divergence.

    Another thing. If I'm not doing limit comparison test, then why is he making me multiply the sum by (1)/(n^(2))?
     
  10. Dec 18, 2011 #9
    Timebomb3750:
    So you can solve the limit without L'Hopital's rule.
     
  11. Dec 18, 2011 #10
    Okay. But, since I'm doing a plain old limit test, I have to get a limit equal to 0, or is undefined in order to prove divergence. If I get a limit of 1, then the test didn't work, or the series converges. The answer is that the series diverges. So, I'm stuck here.
     
  12. Dec 18, 2011 #11

    Dick

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    No you don't. If the limit isn't 0 (and it isn't) then the series diverges. You could also just do a comparison test with the constant series a_n=1/2.
     
  13. Dec 18, 2011 #12

    LCKurtz

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    @Timebomb -- You seem to be confusing the limit comparison test with the nth term test. For a series Ʃan, a necessary condition for the series to converge is that an → 0 as n → ∞. So if that fails, the series must diverge. This is not any kind of comparison or ratio test.

    You could also use the limit comparison test with Ʃ1, which is a little, but not much, more complicated. Your main problem seems to be your inability to multiply the numerator and denominator of an by 1/n2 correctly. If you do that algebra correctly you will get an→ 1.
     
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