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Sequence math problem

  1. Jan 12, 2012 #1
    1. The problem statement, all variables and given/known data

    Prove that if f(x) is continuous for 0<f(x)<1, then [tex] lim_{n->\infty}\frac{1}{n}[(n+1)(n+2)(n+3)...(2n)]^{\frac{1}{n}}=\frac{4}{e}[/tex].

    2. Relevant equations


    f(x)=log(1+x)
    3. The attempt at a solution
    We know that [tex]lim_{n->\infty}\frac{1}{n}[f(\frac{1}{n})+f(\frac{2}{n})+f(\frac{3}{n})+....f(\frac{n}{n})]=\int_0^1 f(x)dx[/tex]

    So that [tex]\int_0^1 log(1+x)(=f(x))dx [/tex] equals the sequence in the limit.
    But evaluation of the integral shows that it is a divergent one.
     
  2. jcsd
  3. Jan 12, 2012 #2

    Char. Limit

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    Re: sequence

    There's no way that integral can be divergent. In the interval between 0 and 1, log(1+x) is bounded by 0 and log(2), respectively. So the integral of log(1+x) needs to be between 0 and log(2). It should be relatively easy to prove that the integral is bounded above and below by those two numbers.
     
  4. Jan 13, 2012 #3
    Re: sequence

    Right, I should've fixed the limits of integration from 0to 1 to 1 to 2 when i did it. Thanks for point it out. Although I think the 1/n power indicated in the problem is not necessary.
     
    Last edited: Jan 13, 2012
  5. Jan 13, 2012 #4

    Char. Limit

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    Re: sequence

    So after you've found the value of the integral, what's your next step? Do you know?
     
  6. Jan 13, 2012 #5
    Re: sequence

    It turned out to be [tex] \lim_{n->\infty}\frac{1}{n}(n+1)(n+2)(n+3)...(2n)=\frac{e}{4}[/tex]
     
  7. Jan 13, 2012 #6

    Char. Limit

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    Re: sequence

    Hmm, but did you remember the 1/n power? First thing I'd try doing is taking the logarithm of the left and right sides.
     
  8. Jan 13, 2012 #7
    Re: sequence

    oops.
     
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