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Ratio Test Radius of Convergence

  1. Nov 10, 2014 #1
    1. The problem statement, all variables and given/known data
    ∑ x2n / n!

    The limits of the sum go from n = 0 to n = infinity

    2. Relevant equations


    3. The attempt at a solution

    So I take the limit as n approaches infinity of aa+1 / an. So that gives me:

    ((x2n+2) * (n!)) / ((x2n) * (n + 1)!)

    Canceling everything out gives me x2 / (n + 1)

    The limit as n approaches infinity is x2 1 / (infinity + 1), which is x^2 * 0

    So now where do I go from here? x^2 * 0 is 0. So does that mean my interval of convergence is just the point x = 0?
     
  2. jcsd
  3. Nov 10, 2014 #2
    You messed up, it's definitely worth fixing and understanding.

    You should get ##\lim n \to \infty \frac{x}{n + 1} = 0##. Convince yourself:)

    Here ##L = 0##. The ratio test says the series converges absolutely if ##L < 1##. For what values of ##x## is this true?
     
  4. Nov 10, 2014 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Just the opposite. What you have shown is that the ratio is always less than 1 so it always converges. In general if the limit in the ratio test is 'A' then the radius of convergence is 1/A. If ratio is never less than 1, the radius of convergence is 0 (so the interval of convergence is just x= 0) and if the ratio is always less than 1, the radius of convergence is infinity.
     
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