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Series convergence

  1. Apr 23, 2007 #1
    I'm having trouble determining whether these series converge or diverge.

    1. sigma sqrt(n/(n^4-2))

    I tried ratio test, but it gave me 1 as the answer (indeterminate)

    2. sigma sin (pi/x)

    3. sigma sin(x)

    I know that sin(x) is bounded......

    Any hints?
     
  2. jcsd
  3. Apr 23, 2007 #2

    Dick

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    Uh, what are you summing over? In the first one, if you are summing over n its a lot like n^(-3/2). In the others I'm clueless until you illuminate the first point.
     
  4. Apr 24, 2007 #3
    oh, sorry.... first one: sum over n from n=2 to infinity
    2. sum over x from x=1 to infinity
    3. sum over x from x=0 to infinity
     
  5. Apr 24, 2007 #4

    HallsofIvy

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    The first one, as Dick said, can be compared to 1/n3/2.

    For the second one, for small [itex]\theta[/itex], [itex]sin(\theta)[/itex] is approximately [itex]\theta[/itex] so that as x goes to infinity, the terms are approximately [itex]\pi/x[/itex]. Does that series converge?

    For the third one, does sin(x) go to 0?
     
  6. Apr 24, 2007 #5
    So

    1. converge
    2. no
    3. no, sin(x) doesn't go to 0, so the series diverges

    Correct?
     
  7. Apr 24, 2007 #6
    Wait, sqrt(n/(n^4-2)) > 1/n^3/2, right? So it doesn't matter that 1/n^3/2 converges?
     
  8. Apr 24, 2007 #7

    Dick

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    How about writing something like 2/n^(3/2)>sqrt(n/(n^4-2))?
     
    Last edited: Apr 24, 2007
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