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Homework Help: Cauchy condensation test

  1. Jun 22, 2010 #1
    1. The problem statement, all variables and given/known data

    I need to determine, using the Cauchy Condensation Test, whether or not
    the series 1/(n * Log(n)) converges.


    2. Relevant equations



    3. The attempt at a solution

    I believe that this series converges iff 2^n(1/(2^n*Log(2^n)) converges (Cauchy Condensation Test). I believe the series actually diverges, but I am not sure how to work through it. Thanks for your help.
     
  2. jcsd
  3. Jun 22, 2010 #2

    Office_Shredder

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    Try simplifying those new terms that you got.
     
  4. Jun 22, 2010 #3
    Basically not hard at all, if you know that [tex] 1/n [/tex] diverges.
    use log rules. and eliminate doubles in nominator and denominator.

    Also can be checked that it diverges by the Integral Test.

    Good luck.
     
  5. Jun 22, 2010 #4
    Thank you so much!
     
  6. Jun 22, 2010 #5
    I have another one for you tarheelborn ! :)

    "determine, using the Cauchy Condensation Test, whether or not
    the series 1/[Log(n)^Log(n)] converges."

    If you like you can try any other test !

    *In Israel we never use Log(n) when we intend to say Ln(n) (as in Log in the natural base e), and in American literature they always get me confused with Log/Ln.
    just a curiosity :)
     
  7. Jun 22, 2010 #6
    My professor used Log although he said that here is means Ln; I don't know quite why. I was accustomed to using Ln, too, in previous classes.

    Meanwhile, to your problem... So 1/[Log(n)^Log(n)] converges iff 2^n*(1/[Log(2^n)^Log(2^n)] converges.

    2^n*(1/[Log(2^n)^Log(2^n)] = 2^n/[Log(2^n)*Log(2^n)]
    = 2^n/[n log(2) * n log(2)]
    = 2 log(2) * Sum [2^n/n^2]
    = 2 log(2) * Sum [1/n^2] which converges so
    1/[Log(n)^Log(n)] converges.

    Thanks for the challenge!
     
  8. Jun 22, 2010 #7
    Very good try !

    but there is a little problem :)

    "2^n*(1/[Log(2^n)^Log(2^n)] = 2^n/[Log(2^n)*Log(2^n)]"

    you changed "Log(2^n)^Log(2^n)" to "Log(2^n)*Log(2^n)" so you basically did
    ln(x)^a = a*ln(x), which isn't true..
    ln(x^a) = a*ln(x) : this one is OK :)
    have another try!
     
  9. Jun 23, 2010 #8
    Ok, here is my next attempt:

    2^n*(1/[Log(2^n)^Log(2^n)] = 2^n/[n*Log(2)^(n*Log(2))]. Now 2^n goes to 1.
    Log(2) is constant, so we are basically left with 1/n^n which is a quickly increasing version of 1/n which does not converge.

    I am not sure if I can make those assumptions, but I don't know how else to deal with the Log. Again, thanks for the challenge; it is fun.
     
  10. Jun 23, 2010 #9
    You have to use the cauchy's test?
    If not: For an easier life, use the integral test.
     
  11. Jun 23, 2010 #10
    Another good try :)

    Basically right, but 2^n does not go to 1, but you right that it grows much slower then n^n, so n^n is the dominate expression.

    When you simplify the expression you get:

    2^n / [n^(n*ln(2))*ln(2)^(n*ln(2))]

    Which you can test by Cauchy's n'th Root Test to be convergent:

    2 / [n^(ln(2))*ln(2)^(ln(2))] < 1

    Another way altogether is to say that after some n: (e^2) is smaller then (ln(n)) :

    1/[ln(n)^ln(n)] < 1/[(e^2)^ln(n)] = 1/[(e^(2*ln(n))] = 1/[(e^(ln(n^2))] = 1/[n^2] !
    1/[n^2] is convergent. so the original expression is also convergent thorough direct comparison test :)

    *Know your logarithmic rules ! :)

    Nevermind, Good Luck to you !
     
  12. Jun 23, 2010 #11
    I do need to know the logarithmic rules more thoroughly! Sometimes I wonder if I will ever know all the rules and be able to remember them as needed. Practice, practice!

    Thank you for this challenge.
     
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