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Thread: 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 $\displaystyle b_n>1$ for all $\displaystyle n$, then prove that $\displaystyle \prod_n b_n$ converges iff $\displaystyle \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:

    $\displaystyle \Rightarrow$

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

    $\displaystyle \Leftarrow $
    If $\displaystyle \sum_n \log(b_n) = \log(\prod_n b_n) = M$ we have $\displaystyle 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 $\displaystyle b_{n} >0 , \forall n$ and $\displaystyle \sum_{n} \ln (b_{n})$ converges...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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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 $\displaystyle b_n$

    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 $\displaystyle b_n>1$ for all $\displaystyle n$, then prove that $\displaystyle \prod_n b_n$ converges iff $\displaystyle \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:

    $\displaystyle \Rightarrow$

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

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

    So am I missing something?
    Go back to the definitions. To say that $\displaystyle \prod_n b_n$ converges means that $\displaystyle \lim_{N\to\infty}\prod_{n=1}^N b_n$ exists (and is nonzero). To say that $\displaystyle \sum_{n}\log(b_n)$ converges means that $\displaystyle \lim_{N\to\infty}\sum_{n=1}^N \log(b_n)$ exists. Also, if $\displaystyle b_n>1$ for all n then $\displaystyle \sum_n\log(b_n)$ 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 $\displaystyle \sum_{n=1}^N \log(b_n) = \log\Bigl(\prod_{n=1}^N b_n\Bigr)$ converges iff $\displaystyle \prod_{n=1}^N b_n$ converges, as $\displaystyle N\to\infty$.
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