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Math Help - Divergence ~ ln(ln(ln x))

  1. #1
    Super Member redsoxfan325's Avatar
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    Divergence ~ ln(ln(ln x))

    It is well-known in high-level mathematics that:

    \lim_{n\to\infty}\left[\left(\sum_{k=1}^n\frac{1}{k}\right)-\ln(n)\right]=\gamma (Euler-Mascheroni Constant)

    \lim_{n\to\infty}\left[\left(\sum_{p~prime}^n\frac{1}{p}\right)-\ln(\ln(n))\right]=M (Meissel-Mertens Constant)

    Is there a divergent series \sum a_k such that:

    \lim_{n\to\infty}\left[\left(\sum_{k=1}^n a_k\right)-\ln(\ln(\ln(n)))\right]=constant

    If so, can it be generalized to \ln(\ln(...(\ln(n))))?
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  2. #2
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    The approximation \sum_{k=1}^n\frac{1}{k}\approx\ln(n) can be justified by applying the procedure of the integral test for convergence to get \sum_{k=1}^n\frac{1}{k}\approx\int_1^n\frac1x\,dx = \ln(n).

    A similar procedure will give you \sum_{k=1}^n\frac{1}{k\ln k}\approx\ln(\ln(n)), \sum_{k=1}^n\frac{1}{k\ln(\ln k)}\approx\ln(\ln(\ln n))), \sum_{k=1}^n\frac{1}{k\ln(\ln(\ln k))}\approx\ln(\ln(\ln(\ln n)))), and so on.

    Edit. See corrections below.
    Last edited by Opalg; October 19th 2009 at 03:23 AM.
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  3. #3
    MHF Contributor chisigma's Avatar
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    Is...

    \frac{d}{dx} \ln x = \frac{1}{x} \rightarrow \int \frac{dx}{x} = \ln x + c (1)

    ... all right!...

    Is...

    \frac{d}{dx} \ln (\ln x) = \frac{1}{x\cdot \ln x} \rightarrow \int \frac{dx}{x\cdot \ln x} = \ln (\ln x) + c (2)

    ... all right!...

    But is...

    \frac{d}{dx} \ln (\ln (\ln x))= \frac{1}{x\cdot \ln x \cdot \ln (\ln x)} \rightarrow \int\frac {dx}{x\cdot \ln(\ln x)} \ne \ln (\ln (\ln x))+c (3)

    ... something wrong from me? ...

    Kind regards

    \chi \sigma
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  4. #4
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by chisigma View Post
    But is...

    \frac{d}{dx} \ln (\ln (\ln x))= \frac{1}{x\cdot \ln x \cdot \ln (\ln x)} \rightarrow \int\frac {dx}{x\cdot \ln(\ln x)} \ne \ln (\ln (\ln x))+c (3)

    ... something wrong from me? ...
    You're right, of course. The series should be \sum_{k=2}^\infty\frac1{k\ln k\ln(\ln k)}. (While we're about it, the series has to start at k=2, not k=1, for obvious reasons.)

    Similarly, the series that converges at the rate of ln(ln(ln(ln n))) should have been \sum_{k=2}^\infty\frac1{k\ln k\ln(\ln k)\ln(\ln(\ln k))}, and so on.
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  5. #5
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Opalg View Post
    You're right, of course. The series should be \sum_{k=2}^\infty\frac1{k\ln k\ln(\ln k)}. (While we're about it, the series has to start at k=2, not k=1, for obvious reasons.)

    Similarly, the series that converges at the rate of ln(ln(ln(ln n))) should have been \sum_{k=2}^\infty\frac1{k\ln k\ln(\ln k)\ln(\ln(\ln k))}, and so on.
    Is there a meaningful infinite set A\subset\mathbb{N}, such that

    \sum_{k\in A}\frac{1}{k} diverges ~ \ln(\ln(\ln n)))

    Like maybe let ~prime,~p\equiv1\mod4\}" alt="A=\{p\in\mathbb{N}~prime,~p\equiv1\mod4\}" />

    But, for all I know, that could converge. (I know that if ~prime,~p+2~also~prime\}" alt="A=\{p~prime,~p+2~also~prime\}" />, then the series does converge.)

    Actually, now that I think about it, that sum must converge, because

    \sum_{p\in A}\frac{1}{p}=\sum_{m,n\in I\subset\mathbb{N}}\frac{1}{m^2+n^2}<\sum_{k=1}^{\  infty}\frac{1}{k^2}=\frac{\pi^2}{6}

    EDIT: Although, the above calculation seems to imply that if we let ~prime,~p\equiv3\mod4\}" alt="B=\{p~prime,~p\equiv3\mod4\}" /> then \sum_{p\in B}\frac{1}{p} diverges.
    Last edited by redsoxfan325; October 19th 2009 at 11:38 AM.
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  6. #6
    MHF Contributor chiph588@'s Avatar
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     \sum_{p \equiv 1 \; \text{mod} \; 4} \frac{1}{p} diverges because one can show that given  (a,k)=1 and  A = \{ prime  p \; | \; p \equiv a \; \text{mod} \; k \},  \lim_{s \to 1^{+}} \frac{\sum_{p \in A} \frac{1}{p^s}}{\sum_{p} \frac{1}{p^s}} = \frac{1}{\phi(k)} .
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  7. #7
    Super Member redsoxfan325's Avatar
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    What's \phi(k)? (And also, where does my proof of convergence go wrong?)
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  8. #8
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by redsoxfan325 View Post
    What's \phi(k)? (And also, where does my proof of convergence go wrong?)
     \phi(k) is Euler's totient function. In your proof, What is  I ?
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  9. #9
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by chiph588@ View Post
     \phi(k) is Euler's totient function. In your proof, What is  I ?
    If p\equiv1\mod4, then p=m^2+n^2 for some m,n\in\mathbb{N}. I is the subset of \mathbb{N} that contains all of those m and n. It might be all of \mathbb{N}, but I wasn't sure, and I could think of a good way to write it, so I wrote I.
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  10. #10
    MHF Contributor chiph588@'s Avatar
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    I'd say this equality is incorrect:
    \sum_{p \in A} \frac{1}{p} = \sum_{m,n\in I\subset\mathbb{N}}\frac{1}{m^2+n^2}. This is because for a fixed  m (or  n ), the latter sum doesn't take into account how many times the  m (or  n ) appear in forming an arbitrary  p .
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