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Math Help - Infinite product

  1. #1
    Senior Member Dinkydoe's Avatar
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    Infinite product

    A part of my course complex-analysis we deal with infinite products. And as a homework assignment I got the following exercise.


    If b_n>1 for all n, then prove that \prod_n b_n converges iff \sum_{n}\log(b_n) < \infty
    I'm used to not so easy homework-assignments, so that's why I feel I'm missing something here.

    My answer would be:

    \Rightarrow

    if \prod_n b_n=M we have M\geq 1 since b_n>1 for all n. Hence \log(M)<\infty and thus we obtain \log(M)=\log(\prod_n b_n)= \sum_n \log(b_n) <\infty

    \Leftarrow
    If \sum_n \log(b_n) = \log(\prod_n b_n) = M we have e^M= \prod_n b_n<\infty

    So am I missing something?
    Last edited by Dinkydoe; May 8th 2010 at 09:44 AM.
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  2. #2
    MHF Contributor chisigma's Avatar
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    A more general criterion of convergence is that b_{n} >0 , \forall n and \sum_{n} \ln (b_{n}) converges...

    Kind regards

    \chi \sigma
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  3. #3
    Senior Member Dinkydoe's Avatar
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    A more general criterion of convergence is that and converges...
    I made 2 typo's i saw...i wrote n instead of b_n

    However, what exactly am I supposed to show here? And what did I miss?
    I feel my argument is incomplete, so what is left to show?
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  4. #4
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by Dinkydoe View Post
    A part of my course complex-analysis we deal with infinite products. And as a homework assignment I got the following exercise.


    If b_n>1 for all n, then prove that \prod_n b_n converges iff \sum_{n}\log(b_n) < \infty
    I'm used to not so easy homework-assignments, so that's why I feel I'm missing something here.

    My answer would be:

    \Rightarrow

    if \prod_n b_n=M we have M\geq 1 since b_n>1 for all n. Hence \log(M)<\infty and thus we obtain \log(M)=\log(\prod_n b_n)= \sum_n \log(b_n) <\infty

    \Leftarrow
    If \sum_n \log(b_n) = \log(\prod_n b_n) = M we have e^M= \prod_n b_n<\infty

    So am I missing something?
    Go back to the definitions. To say that \prod_n b_n converges means that \lim_{N\to\infty}\prod_{n=1}^N b_n exists (and is nonzero). To say that \sum_{n}\log(b_n) converges means that \lim_{N\to\infty}\sum_{n=1}^N \log(b_n) exists. Also, if b_n>1 for all n then \sum_n\log(b_n) is a series of positive terms, so it converges iff its sum is finite.

    Now use the fact that the logarithm function and its inverse are continuous to argue that \sum_{n=1}^N \log(b_n) = \log\Bigl(\prod_{n=1}^N b_n\Bigr) converges iff \prod_{n=1}^N b_n converges, as N\to\infty.
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