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Convergence/divergence test

  1. Nov 30, 2014 #1
    1. The problem statement, all variables and given/known data
    Show if this sequence (with n=1 to infinity) diverge or converge

    2. Relevant equations

    CodeCogsEqn-5.gif

    3. The attempt at a solution

    If I use the Limit Comparison Test:

    compare with CodeCogsEqn-6.gif so you get CodeCogsEqn-7.gif that equals CodeCogsEqn-8.gif


    lim n -> inf CodeCogsEqn-8.gif => inf.

    Can I use the Test like this? What does this tell me? (If anything at all...)
     
  2. jcsd
  3. Nov 30, 2014 #2

    Orodruin

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    I suggest doing limit comparison with ##(2/3)^n## instead.
     
  4. Nov 30, 2014 #3
    Ah the limit is now 1 .
    So:
    Limit < infinity.
    And ##(2/3)^n## converges.
    So the original function must converge as well.
     
  5. Nov 30, 2014 #4

    Ray Vickson

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    Right, and a direct comparison is easier. First, for ##a > 1## the function ##a^n## always swamps ##n^k## for any fixed power ##k##, meaning that for large enough ##n## we always have ##a^n > n^k##. So, if
    [tex] t_n = \frac{2^n + n^2}{3^n + n^3}[/tex]
    we have that the numerator is ##2^n + n^2 < 2^n + 2^n = 2 \cdot 2^n## for all ##n > n_0## (where you can even find ##n_0## if you want). The denominator is ##3^n + n^3 > 3^n##, so ##0 < t_n < \frac{2 \cdot 2^n}{3^n} = 2(2/3)^n##, and now you are almost done.
     
    Last edited: Nov 30, 2014
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