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Math Help - Determining convergence/divergence of a series.

  1. #1
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    Determining convergence/divergence of a series.

    \sum_{n=0}^{\infty}\frac{e^4^n}{e^3^n}

    This is a practice problem for a test. The test doesn't cover the comparison test, ratio test, root test or limit comparison test. I should be able to do this using the nth-term test, p-series, integral test or other basic methods of evaluating sequences and series. With that said, I don't think the integral test can be used here as the function is not decreasing. If we take the limit of the function we get:

    \lim_{x \to \infty}\frac{e^4^n}{e^3^n}=\frac{\infty}{\infty}

    Which is an indeterminate form. However, if I use L'Hopital's rule I don't think I will get anywhere:

    \lim_{x \to \infty}\frac{e^4^n}{e^3^n}=\frac{4e^4^n}{3e^3^n}= \frac{16e^4^n}{9e^3^n} ...

    And so on. No am I starting to get desperate. I think maybe I need to rewrite this. Or I have overlooked a simpler answer. Or my answer thus far is wrong.

    Can I rewrite this as:
     \left (\frac{e^4}{e^3} \right )^n

    That would mean that:
    \left | r \right |\geq 1

    And so the series diverges.
    Last edited by jamesrb; March 31st 2013 at 07:21 PM.
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  2. #2
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    Re: Determining convergence/divergence of a series.

    Quote Originally Posted by jamesrb View Post
    \sum_{n=0}^{\infty}\frac{e^4^n}{e^3^n}
    Can I rewrite this as:
     \left (\frac{e^4}{e^3} \right )^n

    That would mean that:
    \left | r \right |\geq 1

    And so the series diverges.
    Yes, you can do that.
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  3. #3
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    Re: Determining convergence/divergence of a series.

    You should immediately see that this series doesn't even go to 0 as n goes to infinity. It fails that test so it has to diverge.
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  4. #4
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    Re: Determining convergence/divergence of a series.

    Yes, SworD is correct - the individual terms \frac{e^{4n}}{e^{3n}}=e^n do not go to zero as n goes to infinity, so the series must diverge. That was my first thought.

    But your way works, too.

    - Hollywood
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