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Ratio Test, SUPER , help?

  1. Apr 25, 2012 #1
    Ratio Test, SUPER URGENT, help?

    Consider the series
    ∑ n=1 to infinity of asubn, where asubn = [8^(n+4)] / [(8n^2 +7)(5^n)]
    use the ratio test to decide whether the series converges. state what the limit is.

    From the ratio test I got the limit n-> infinity of
    [8^(n+1+4)] / [(8(n+1)^2 +7)(5^n+1)] / [8^(n+4)] / [(8n^2 +7)(5^n)]
    = [8^n+5(8n^2 + 7)5^n] / (8(n+1)^2 +7)5^(n+1) (8^n+4)
    =[8^n(8^5)(8n^2 + 7) 5^n] / (8n^2 + 16n + 15)(5^n)(5)(8^n)(8^4)

    the lim n-> infinity of [262144n^2 +229376] / [163840n^2 + 327680n + 307200]
    = 8/5
    which is divergence. Is this correct?

    Help?
     
    Last edited: Apr 25, 2012
  2. jcsd
  3. Apr 25, 2012 #2

    SammyS

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    Re: Ratio Test, SUPER URGENT, help?

    Is this the series you're working with ?

    [itex]\displaystyle \sum_{n=1}^{\infty} \frac{8^{n+4}}{(8n^2 +7)(5^n)}[/itex]
    Are you saying your result is wrong because the series doesn't converge absolutely?
     
  4. Apr 25, 2012 #3
    Re: Ratio Test, SUPER URGENT, help?

    yes, that is the series I'm working with.

    And actually, I've redone the problem and edited my work above.
    Now I got a limit of 8/5 = 1.6 so the series diverges.
    Is this correct (did I calculate the limit correct)?
     
  5. Apr 25, 2012 #4

    Dick

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    Re: Ratio Test, SUPER URGENT, help?

    Yes, that would be correct.
     
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