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Proving convergence of 1/log(n)

  1. Apr 2, 2012 #1
    Has anybody got any idea as to how to prove that Ʃ 1/(n(log(n))^p) converges? (where p>1)
  2. jcsd
  3. Apr 3, 2012 #2
    0 < log(n) < n for n>3
  4. Apr 3, 2012 #3

    First way, the integral test: [itex]\int_2^\inf \frac{1}{x\log^px} dx=\frac{\log^{1-p}(x)}{1-p}|_2^\inf \rightarrow \frac{log^{1-p}(2)}{p-1}[/itex] .

    Second way, Cauchy's Condensation test: taking [itex]n=2^k[/itex] , the series's general term is [itex]\frac{1}{2^kk^p\log^p(2)}[/itex] , so multiplying this by [itex]2^k[/itex] we get [itex]\frac{1}{k^p\log^p(2)}[/itex] , which is a multiple of the series of [itex]\frac{1}{k^p}[/itex] , which we know converges for [itex]p>1[/itex] .

  5. Apr 3, 2012 #4
    Thing is I've got to prove determine it's convergence from 1 to infinity.
  6. Apr 3, 2012 #5
    Just to clarify, the previous message was supposed to be:
    Thing is I've got to determine it's convergence from 1 to infinity.
  7. Apr 3, 2012 #6


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    If the sum starts at 1, you have a problem because log(1)=0
  8. Apr 3, 2012 #7
    The sum from 1 to any fixed number will always converge (unless you divide by zero).
    Therefore, if the sum from any fixed number to infinity converges, then the sum from 1 to infinity converges.

    You can choose any fixed number you like. I chose 3 because 0<log(n) < n for all n>=3.

    You can choose 11 if your log is base 10 rather than natural, or because, you know, usual amps only go up to 10, but yours goes all the way up to 11.

    The sum from 3 to infinity of your series is smaller than the sum from 3 to infinity over 1/n^(p+1), which is smaller than 1/n^2 which we know converges.
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