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Thread: 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:

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

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

    Is there a divergent series $\displaystyle \sum a_k$ such that:

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

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

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

    Edit. See corrections below.
    Last edited by Opalg; Oct 19th 2009 at 02:23 AM.
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  3. #3
    MHF Contributor chisigma's Avatar
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    Is...

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

    ... all right!...

    Is...

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

    ... all right!...

    But is...

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

    ... something wrong from me? ...

    Kind regards

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

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

    ... something wrong from me? ...
    You're right, of course. The series should be $\displaystyle \sum_{k=2}^\infty\frac1{k\ln k\ln(\ln k)}$. (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 $\displaystyle \sum_{k=2}^\infty\frac1{k\ln k\ln(\ln k)\ln(\ln(\ln k))}$, 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 $\displaystyle \sum_{k=2}^\infty\frac1{k\ln k\ln(\ln k)}$. (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 $\displaystyle \sum_{k=2}^\infty\frac1{k\ln k\ln(\ln k)\ln(\ln(\ln k))}$, and so on.
    Is there a meaningful infinite set $\displaystyle A\subset\mathbb{N}$, such that

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

    Like maybe let $\displaystyle A=\{p\in\mathbb{N}~prime,~p\equiv1\mod4\}$

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

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

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

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