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Determining General Values of Convergence for a seqence

  1. Apr 26, 2012 #1
    Determine the values of "r" for which rn converges.


    Is there a specific procedure I should try to apply to figure this out? The only things I could intuitively come up with that will converge in this scenario are when -1 ≤ r ≤ 1......is there anything else to this?
     
  2. jcsd
  3. Apr 26, 2012 #2

    LCKurtz

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    Really? r = -1?? For |r|<1 I guess it depends on what you have to work with. You might consider what happens to ##y=\ln(|r|^n)##.
     
  4. Apr 27, 2012 #3


    I don't fully see what your getting at. From what you wrote that would become n ln(r), but in that case it would not converge.
     
  5. Apr 27, 2012 #4

    LCKurtz

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    That's "you're", not "your".

    What about ##r=-1##? Have you thought about that?

    Yes, ##y=\ln(|r|^n) = n \ln(|r|)## doesn't converge. But what does it do? Since ##y## is the logarithm of your problem, what does that tell you about your problem?
     
  6. Apr 27, 2012 #5


    In the case of r = -1, the sequence oscillates so it is divergent.

    In the second part ##y=\ln(|r|^n) = n \ln(|r|)## goes to infinity whenever r > 1, when 0 < r < 1 then it converges.

    So the main behavior to try and examine occurs between -1 < r < 1 then
     
  7. Apr 27, 2012 #6

    LCKurtz

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    Yes, that's what I'm getting at. What happens to ##y## when ##|r|<1##?
     
  8. Apr 27, 2012 #7
    In that instance it appears as if it's going to converge to 0, but when r is in the form ##ln(r)## it's tending toward negative infiniti
     
  9. Apr 27, 2012 #8

    LCKurtz

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    You can't just assert ##|r|^n\to 0##; that's what you are trying to prove if ##|r|<1##. But once you know ##\ln|r|^n\to -\infty##, doesn't that show it?
     
  10. Apr 28, 2012 #9

    After staring and thinking about it for a second is this the proper rationale behind the convergence to 0?:

    SO I know: ##\ln|r|^n\to -\infty## .

    if I raise both sides to "e": e ##ln|r|^n## --> e##-\infty##. since that would simplify to |r|n --> 1/ e##infty## and that tends to 0.
     
  11. Apr 28, 2012 #10

    After staring and thinking about it for a second is this the proper rationale behind the convergence to 0?:

    SO I know: ##\ln|r|^n\to -\infty## .

    if I raise both sides to "e": e ##ln|r|^n## --> e##-\infty##. since that would simplify to |r|n --> 1/ e##infty## and that tends to 0.

    Is that how it ends up converging to 0?
     
  12. Apr 28, 2012 #11

    LCKurtz

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    That's the idea, although you can probably find a more proper way to express it.
     
  13. Apr 29, 2012 #12
    Thanks. and thanks for having the patience to help me through it
     
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