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Can't figure out how to evaluate a sequence as it goes to infinity.

  1. Jan 30, 2012 #1
    1. The problem statement, all variables and given/known data

    An = (((-1)^(n-1))n)/(n^2 + 1)

    I need to know if it converges or diverges and if it converges the limit.

    2. Relevant equations



    3. The attempt at a solution

    I know it converges to 0. But I don't know how to show it when evaluating. I tried evaluation An| in the absolute value but I keep ending up with oo/oo
     
  2. jcsd
  3. Jan 30, 2012 #2

    Dick

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    Try dividing numerator and denominator by n.
     
  4. Jan 30, 2012 #3
    lin n -> oo abs (((n(-1)^(n-1))/(n + 1/n)) * (1/n)) So it ends up being (infinity/infinity) * 0 = 0. Is that really a valid way to evaluate it?
     
  5. Jan 30, 2012 #4

    Dick

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    No, it's not. But n*(1/n)=1. Take the absolute value and you've got 1/(n+1/n). What's that limit?
     
  6. Jan 30, 2012 #5
    The limit is to infinity.
    http://www4d.wolframalpha.com/Calculate/MSP/MSP44701a03g21d31b6h65500003b8813i95da1dcbe?MSPStoreType=image/gif&s=25&w=119&h=40 [Broken]
     
    Last edited by a moderator: May 5, 2017
  7. Jan 30, 2012 #6

    Dick

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    Why would you think that?
     
  8. Jan 30, 2012 #7
    Well the problem text says Determine whether the sequences converges or diverges. If it converges find the limit.
     
  9. Jan 30, 2012 #8

    Dick

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    I don't think limit n->infinity of 1/(n+1/n) is infinity.
     
  10. Jan 30, 2012 #9
    Not sure if I wrote the problem bad. Here is what the problem looks like the text. an= 0aikG.gif
     
  11. Jan 30, 2012 #10

    Dick

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    That's what I thought it was. I'm going to just have to repeat what I said to start with. Divide numerator and denominator by n. That doesn't change the limit. Now you don't have infinity/infinity. What form do you have?
     
  12. Jan 30, 2012 #11
    ZFAub.gif

    I don't see how it helps unless you take the absolute value of it.
     
  13. Jan 30, 2012 #12

    Dick

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    Doesn't the denominator go to infinity and the numerator not go to infinity? The numerator is bounded. Taking the absolute value wouldn't hurt. But it's not infinity/infinity anymore.
     
  14. Jan 30, 2012 #13
    Numerator will be -1 or 1. Divided by infinity it evaluates to 0. But I am not sure if that is a valid way to show that it evaluates to 0. In the book I am using (Stewart) it gives a therom if Lim n-> infinity |an = o than Lim n-> infinity |an = 0.
     
  15. Jan 30, 2012 #14

    Dick

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    I think you mean lim n->infinity |an|=0 then lim n->infinity an=0. |an|=1/(n+1/n). Surely that approaches 0. It has the form 1/infinity doesn't it? You probably don't need a epsilon style proof here.
     
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