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Math Help - Solving an equation (2)

  1. #1
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    Solving an equation (2)

    Find all real numbers x \geq 0 such that \prod_{n=2}^{\infty} \left(1 - \frac{1}{n(\ln n)^x} \right) = 0.
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  2. #2
    MHF Contributor chisigma's Avatar
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    The 'infinite product' ...

    \prod _{n} (1-a_{n}) (1)

    ... diverges if the corresponding 'infinite sum'...


    \sum _{n} a_{n} (2)

    ... diverges. In this case the 'infinite sum' is...

    \sum _{n=2}^{\infty} \frac{1}{n\cdot \ln^{x} n} (3)

    ... and we can use the 'integral test' to verify its convergence. If x=1 we have...

    \int\frac{1}{t\cdot ln t} \cdot dt = \ln \ln |t| + c (4)

    ... and the integral from 2 to \infty diverges. If x>1 we have...

    \int\frac{1}{t\cdot ln^{x} t} \cdot dt = \frac{\ln^{1-x} t}{1-x} + c (5)

    ... and the integral from 2 to \infty converges. The conclusion is that the 'infinite product'...

    \prod _{n=2}^{\infty} (1-\frac{1}{n\cdot \ln^{x} n}) (6)

    ... converges for x>1 and diverges for x\le 1...



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  3. #3
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    Quote Originally Posted by chisigma View Post
    My new avatar will remain until the sentence of the European Court that removes the crucifix from the italian schools will be revised...
    I thought there was a rule here preventing "religious views". No?
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  4. #4
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by chisigma View Post
    The 'infinite product' ...

    \prod _{n} (1-a_{n}) (1)

    ... diverges if the corresponding 'infinite sum'...


    \sum _{n} a_{n} (2)

    ... diverges. In this case the 'infinite sum' is...

    \sum _{n=2}^{\infty} \frac{1}{n\cdot \ln^{x} n} (3)

    ... and we can use the 'integral test' to verify its convergence. If x=1 we have...

    \int\frac{1}{t\cdot ln t} \cdot dt = \ln \ln |t| + c (4)

    ... and the integral from 2 to \infty diverges. If x>1 we have...

    \int\frac{1}{t\cdot ln^{x} t} \cdot dt = \frac{\ln^{1-x} t}{1-x} + c (5)

    ... and the integral from 2 to \infty converges. The conclusion is that the 'infinite product'...

    \prod _{n=2}^{\infty} (1-\frac{1}{n\cdot \ln^{x} n}) (6)

    ... converges for x>1 and diverges for x\le 1
    I think I'm not understanding something. How do you know it converges to 0?
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  5. #5
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by redsoxfan325 View Post
    I think I'm not understanding something. How do you know it converges to 0?...
    A precise enough definition of 'infinite product' is given here...

    Infinite product - Wikipedia, the free encyclopedia

    If You have an infinite product in the form...

    \prod_{n=1}^{\infty} a_{n} = \lim_{n \rightarrow \infty} \prod_{k=1}^{n} a_{k} (1)

    ... it is said to be convergent if the limit exists and it is not zero. In all other cases it is said to be divergent. If \forall n , a_{n}>0 we can write...

    \ln \prod_{n=1}^{\infty} a_{n} = \sum_{n=1}^{\infty} \ln a_{n} (2)

    ... so that the convergence of the infinite product is equivalent of the convergence of the 'infinite sum' of logarithms. If a product of infinite positive terms diverges the limit can be 0 or + \infty. An useful criterion of convergence of an infinite product is allowable when the general term can be written as a_{n}= 1 + p_{n}. In this case the (1) converges if converges the 'infinite sum'...

    \sum_{n=1}^{\infty} p_{n} (3)



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  6. #6
    Super Member redsoxfan325's Avatar
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    Oh, okay. I didn't realize that a product of zero was considered divergent. (You can imagine my confusion when I read your post not knowing that!)
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