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ieatpi

Really stumped on this one, anyone good at divergence of series?

  • one year ago
  • one year ago

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  1. ieatpi
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    • one year ago
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  2. godorovg
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    hey there is a of getting a bigger picture it is a little a hard to read for me at least just asking..

    • one year ago
  3. ieatpi
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    \[\sum_{3}^{∞} \frac{ 1 }{ n \ln n \sqrt{(\ln n)^2-1} }\]

    • one year ago
  4. abb0t
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    It is kinda small, but it looks like you'd want to use comparison test. I think you might be able to use integral test also since you are workingn with log, and you know the values must be positive.

    • one year ago
  5. ieatpi
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    any suggestions for what to compare it with? I redid the equation for readability

    • one year ago
  6. godorovg
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    one sec

    • one year ago
  7. godorovg
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    try this website http://tutorial.math.lamar.edu/Classes/CalcIII/DivergenceTheorem.aspx

    • one year ago
  8. abb0t
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    Use integral test.

    • one year ago
  9. abb0t
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    @godorovg this is for series, not functions in vector fields.

    • one year ago
  10. godorovg
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    sorry it has been a while for me. I am really sorry man..

    • one year ago
  11. abb0t
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    You can use to evaluate the series just as you would an improper integral. Try it to see what you get. you are going to be using u-substitution.

    • one year ago
  12. godorovg
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    I am trying to find the right way to help here http://tutorial.math.lamar.edu/Classes/CalcII/ConvergenceOfSeries.aspx

    • one year ago
  13. godorovg
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    is this what I need to look at? I am making sure I do this right is all..

    • one year ago
  14. abb0t
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    If you are lookng at that site: http://tutorial.math.lamar.edu/Classes/CalcII/IntegralTest.aspx Example one would be a similar example to the problem you are working on now.

    • one year ago
  15. godorovg
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    okay let me review this

    • one year ago
  16. ieatpi
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    The thing that bothers me, is for the integral test to apply, the derivative must be less than 0, which it doesn't appear to be

    • one year ago
  17. godorovg
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    that is what I was think also it must be + not - for this test to work as well

    • one year ago
  18. godorovg
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    according to what I have read on the website at least.

    • one year ago
  19. ieatpi
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    yeah, back to square one I guess

    • one year ago
  20. godorovg
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    no it does work here because when you look the rule of squares and taking LN that is squared would be at 1 at least not zero.

    • one year ago
  21. godorovg
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    also given that when infinty goes to 3 zero would not be included right?

    • one year ago
  22. godorovg
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    so I am thinking the series it is itself would 1,2,3 and so on leaving out and the end result than would be postive so therefore test works

    • one year ago
  23. godorovg
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    Am I making sense to you?

    • one year ago
  24. godorovg
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    because the limit would be 3 in this case

    • one year ago
  25. godorovg
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    you still with me??

    • one year ago
  26. ieatpi
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    so let me make sure I do this right, the integral is lim n->inf sec^-1(lnx) evaluated from 3 to t right?

    • one year ago
  27. godorovg
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    yes

    • one year ago
  28. godorovg
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    if you come up with zero at any time test fails..

    • one year ago
  29. godorovg
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    can you check the answer after you are finished?

    • one year ago
  30. ieatpi
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    i got pi/2 -arcsec(ln3), implying that the sum converges

    • one year ago
  31. godorovg
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    that looks right

    • one year ago
  32. terenzreignz
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    I realise that this might be moot, but how about checking the convergence of \[\frac{1}{n(\ln n)^2}\] instead? Then it can be shown to be convergent, and by the limit comparison test, you'll be able to show that your original series was also convergent.

    • one year ago
  33. godorovg
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    did you check the answer or can even do that depending if this an odd or even in the book you are working out of? So, how are you doing so far??

    • one year ago
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