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Divergent Series - How to test it

  1. Aug 23, 2012 #1
    1. The problem statement, all variables and given/known data

    Ʃ (1+n+n2)/(√1+n2+n6)) n=1 to infinity
    Answer is divergent


    2. Relevant equations
    Comparison Test / P Series
    0≤an≤bn

    3. The attempt at a solution
    Hello, I simplified the problem to

    Ʃ (1+n+n2)/(1+n2+n6)1/2

    Is it incorrect of me to say immediately right here that because the power of the denominator p<1 it is divergent?

    Or if I take a different approach use comparison test and say 0≤an≤1/(1+n2+n6)1/2 can I say it is divergent because bn is divergent due to p-series?

    WA says it is divergent due to comparison test but how would I get something like say 1/n^6 out of that square root to use as bn?
     
  2. jcsd
  3. Aug 23, 2012 #2

    Simon Bridge

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    I would have said it was divergent because it behaves as 1/n for large n... and we know the sum of 1/n diverges from a previous proof.
     
  4. Aug 23, 2012 #3

    Dick

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    True, but 1/n is on the boundary between divergence and convergence for power series. You need an actual comparison to make. Being asymptotically 'like' 1/n is not good enough. Choose a series with a smaller denominator and a larger denominator and show it diverges. I'm a little worried that what Saterial has said so far makes little sense.
     
  5. Aug 23, 2012 #4

    LCKurtz

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    Yes, that would be incorrect.

    No. It isn't a p series

    Look up the "general comparison test" or it may be called the "limit comparison test" in your text.
     
  6. Aug 23, 2012 #5

    LCKurtz

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    That would be a correct intuitive argument, but you need the limit comparison test to prove it.
     
  7. Aug 24, 2012 #6

    Dick

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    A direct comparison test will do it. And it's simpler.
     
  8. Aug 24, 2012 #7

    LCKurtz

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    Probably so and point taken. I haven't actually worked on it. But I do think it is good for students to recognize situations that are a natural setting for the limit comparison test.
     
  9. Aug 24, 2012 #8

    Dick

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    I'll repeat what I said before. Make the denominator smaller and the numerator larger in a way that the resulting series still diverges and you've got it. Limit comparison test may not be in Saterial's repetoire. And it's not needed. It's really simple once you've seen it.
     
    Last edited: Aug 24, 2012
  10. Aug 24, 2012 #9

    Curious3141

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    Making the denominator smaller and the numerator larger (and hence every term larger) would not help you prove divergence.

    Maybe you meant make the denominator larger and the numerator smaller, and prove that the resulting series still diverges?
     
  11. Aug 24, 2012 #10
    Yes, the divergence follows rather trivially when one does that because the series collapses into the harmonic series.
     
  12. Aug 24, 2012 #11
    (1+n+n2)/(√1+n2+n6)) > (1)/(√1+n2+n6)) > (1)/(√n2+n2+n6)) = (1)/(√2n2+n6)) = (1)/(√n2(2+n4))) = (1)/(n√(2+n4))) > 1/n
     
  13. Aug 24, 2012 #12

    Dick

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    Oh yeah. That's what I meant. Denominator is the thing on the BOTTOM. Duh. Thanks!
     
  14. Aug 24, 2012 #13

    Dick

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    Sorry. (1)/(n√(2+n4))) is less than 1/n. Not greater.
     
  15. Aug 24, 2012 #14

    Dick

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    How about just noting that the original series is greater than n^2/sqrt(n^6+n^6+n^6)?
     
  16. Aug 24, 2012 #15

    Curious3141

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    Yes, that would work. Much simpler, too. :approve:

    I just showed the first thing I thought of.
     
  17. Aug 24, 2012 #16

    uart

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    Or as another alternative noting that it's greater than [itex]\frac{n^2}{n^3 + 1}[/itex], which in turn is greater than [itex]\frac{1}{n+1}[/itex]. We can then compare it directly to the harmonic series starting at n=2.
     
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