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P-Series or Comparison Test Question

  1. Oct 31, 2014 #1

    RJLiberator

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    K≥0 ∑ ((sqrt(k)+2)/(k+5))

    I am trying to prove that this diverges. The divergence test is inconclusive.
    Now I am left with a great option of a comparison test. I'm not quite sure what to compare it with, but I know I need to compare it with something smaller (denominator is larger) that diverges to prove divergence.

    While looking at it, I am wondering if this is as simple as using the p-series test and taking off the smaller quantities of 2 and 5 and just using k^(1/2)/k where p would then = 1/2 and prove divergence.

    Can I do this with the p-series?

    IF not, I need to find something that is in comparison with this to prove divergence. Would k^(1/2)/2k be ok to use?
     
  2. jcsd
  3. Oct 31, 2014 #2

    vela

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    This last line of reasoning is good. You care about what the terms do as k gets large, so looking at the dominant terms in the numerator and denominator is a good idea.

    For large k, the series looks like ##k^{-1/2}##, so using the limit comparison test with ##k^{-1/2}## might work.

    You could also make an argument that the terms of the series are larger than a series of the form ##c/\sqrt{k}## and use the direct comparison test.

    You can't say it's a p-series because it's not of the form ##1/k^p##.
     
  4. Oct 31, 2014 #3

    RJLiberator

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    Thank you for your help

    1) Ah, I cannot say it is a p-series because it is not in correct form. Great information.
    2) Limit comparison test seems like a great idea, I may not be able to use it since we will be learning that later in the course.
    3) I will need to use the direct comparison test that you pointed out to me. I am going to try this work now.

    Thanks.
     
  5. Oct 31, 2014 #4

    RJLiberator

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    Since the sum starts at a value of 0 I cannot use k^(-1/2) alone in the denominator.

    Can I compare it with c/(sqrt(k+1)) ?
     
  6. Oct 31, 2014 #5

    vela

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    It would be simpler to throw away the first term. In other words, if the sum from k=1 converges, then the sum from k=0 will converge as well because the two series only differ by a finite number of terms.

    The reason a series of the form ##c/\sqrt{k}## is nice to have is because it is a p-series, and you can therefore conclude it diverges.
     
  7. Oct 31, 2014 #6

    RJLiberator

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    Ah..... That makes sense. Being able to throw away initial 0 as it doesn't really add anything to the convergence/divergence of the sum. Thank you.
     
  8. Oct 31, 2014 #7

    Ray Vickson

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    You tell us. You need to have some confidence in your own knowledge/abilities.
     
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