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Math Help - Sum of reciprocals of Prime Numbers

  1. #1
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    Sum of reciprocals of Prime Numbers

    Hello,

    Could anyone please give me a proof using basic/elementary number theory and or calculus of the following:

    The sum of the reciprocals of primes is approximately equal to log(log(x))

    e.g 1/2 + 1/3 + 1/5 + 1/7 + 1/11 + ... ~ log(log(x))

    Thanks in advance
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  2. #2
    MHF Contributor chisigma's Avatar
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    Simple question: what is 'x' in your formula?...

    Kind regards

    \chi \sigma
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  3. #3
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    the sum of all reciprocal P's that are smaller or equal to x.

    That is the x

    Thanks for pointing it out.
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  4. #4
    MHF Contributor chisigma's Avatar
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    All right!... the procedure that follows is due to Leohard Euler and starts with the 'infinite product' that brings his name...

    \sum_{k=1}^{\infty} \frac{1}{k^{n}}= \prod_{p} \frac{1}{1-p^{-n}} (1)

    ... where with 'p' we indicate the sequence of prime numbers. For n=1 the (1) becomes...

    \sum_{k=1}^{\infty} \frac{1}{k}= \prod_{p} \frac{1}{1-p^{-1}} (2)

    Taking logarithm on both sides of (2)...

    \ln\sum_{k=1}^{\infty} \frac{1}{k}= \ln\prod_{p} \frac{1}{1-p^{-1}}= - \sum_{p} \ln (1-p^{-1}) = \sum_{p} (\frac{1}{p} + \frac{1}{2p^{2}} + \frac{1}{3p^{3}} + \dots) =

    = \sum_{p} \frac{1}{p} + \sum_{p} \frac{1}{p^{2}}\cdot (\frac{1}{2} + \frac{1}{3p} + \frac{1}{4p^{2}}+ \dots) < \sum_{p} \frac{1}{p} + \sum_{p} \frac{1}{p^{2}}\cdot ( 1 + \frac{1}{p} + \frac{1}{p^{2}}+ \dots)=

    = \sum_{p} \frac{1}{p} + \sum_{p} \frac{1}{p(p-1)} = \sum_{p} \frac{1}{p} + \chi = \sum_{k=1}^{\infty} \pi(k) + \chi (3)

    ... where \chi is a positive constant and is  \pi(k)= \frac{1}{k} for k prime and \pi(k)=0 elsewhere. Euler had demonstrated before that...

    \lim_{n\rightarrow \infty} (\sum_{k=1}^{n}\frac{1}{k}- \ln n) = \gamma (4)

    ... where \gamma is another positive constant and is 0<\gamma<1, so that for n 'large enough' is ...

    \sum_{k=1}^{n}\frac{1}{k} \approx \ln n + \gamma (5)

    ... i.e. \sum_{k=1}^{n}\frac{1}{k} is asymptotic to \ln n. Now observing (3) we conclude that \sum_{p} \frac{1}{p} is asymptotic to \ln(\ln n) i.e. for n 'large enough' is...

    \ln \sum_{k=1}^{n}\frac{1}{k} \approx \sum_{k=1}^{n}\pi(k) \approx \ln(\ln n) (6)

    ... or more precisely is...

    \lim_{n\rightarrow \infty} \frac{\sum_{k=1}^{n} \pi(k)}{\ln(\ln n)}=1 (7)

    Kind regards

    \chi \sigma
    Last edited by chisigma; April 21st 2009 at 08:22 PM.
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  5. #5
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    Thank you
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  6. #6
    MHF Contributor chisigma's Avatar
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    In order to well undestand please observe this figure...



    ... where are reported...

    a) in blue the function...

     \sum_{k=1}^{n} \pi(k)

    ... where...

     \pi(k)= \frac{1}{k}, k prime, 0 elsewhere...

    b) in red the function...

    \ln\sum_{k=1}^{n} \frac{1}{k}

    c) in grey the function...

    \ln (\ln n)

    What Euler has demonstrated two and half centuries ago [and that i replied now...] is that...

    \lim_{n\rightarrow \infty} \frac{\sum_{k=1}^{\infty} \pi(k)}{\ln (\ln n)}=1

    ... and nothing else...

    Kind regards

    \chi \sigma
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