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Finding if a series is convergent.

  1. Apr 13, 2010 #1
    Finding if a series is convergent-Answered

    1. The problem statement, all variables and given/known data
    Find for which values of K is the fallowing series convergent.


    N is the variable.
    K is a constant or a list of constant (eg. "(2,91]")

    2. Relevant equations

    I believe the ratio test, which states that if (f(n+1)/(f(n) as n approaches infinity is less then 1, it converges.

    3. The attempt at a solution

    I believe the obvious way to go about this would be the ratio test which is as fallows"

    [PLAIN]http://img688.imageshack.us/img688/2140/equation1.png [Broken]
    [PLAIN]http://img146.imageshack.us/img146/2783/equation2.png [Broken]
    [PLAIN]http://img232.imageshack.us/img232/1554/equation3.png [Broken]

    1>((n+1)!*(n+1)!/(k(n+1))!*((Kn)!/(n!*n!) as n[tex]\rightarrow[/tex] [tex]\infty[/tex]

    1>(n+1)(n+1)/(k(n+1)) as n[tex]\rightarrow[/tex] [tex]\infty[/tex]

    1> (n+1)/k

    which is not true, therefore this series must diverge for any possible K.

    my question: am i doing anything wrong or did the teacher give a trick question?
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Apr 13, 2010 #2


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    actually sorry, I take that back, i think there is a mistake
    [tex] \frac{ (kn)! }{ (k(n+1))! } = \frac{ (kn)! }{ (kn+k)! } \neq \frac{1}{k(n+1)} [/tex]
    Last edited: Apr 13, 2010
  4. Apr 13, 2010 #3


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    as a guide, try k = 1,2,3 and see whether you get a converging series

    another way do this covering all integer k>1 would be to use Stirlings's Approximation for n! as n gets large
  5. Apr 14, 2010 #4
    Oh, Thanks for your help lanedance. i found it pretty intuitive that 2+ was the answer but i wouldn't have thought to be able to solve it mathematically i would need to distribute the K then simply insert a value. you've been very helpful.
  6. Apr 14, 2010 #5


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    to be rigorous, you need to show its true for all k>2, not just one larger than 2 (3)

    so options are
    - have a look at Stirlings Approximation [itex] n! \approx n^{n}e{-n}\sqrt{2\pi n }[/itex]
    - try mathematical induction

    i haven't tried either fully but think both should work
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