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Math Help - Sum of Reciprocal of primes

  1. #1
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    Sum of Reciprocal of primes

    I'm trying to prove that if s is real and s>1 then



    I have managed to use the euler product for the zeta function and the taylor expansion log function on the LHS to give

    |sigma {p} sigma {m=2 to infinity} (1/(mp^ms))| (sorry my latex is bad)

    Does anyone know where I go from here
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  2. #2
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by klw289 View Post
    I'm trying to prove that if s is real and s>1 then



    I have managed to use the euler product for the zeta function and the taylor expansion log function on the LHS to give

    |sigma {p} sigma {m=2 to infinity} (1/(mp^ms))| (sorry my latex is bad)

    Does anyone know where I go from here
    From the Euler's product...

    \zeta(s)= \prod_{p} \frac{1}{1-p^{-s}} (1)

    ... You derive with simple steps...

    \ln \zeta(s)= \sum_{p} p^{-s} + \frac{1}{2} \sum_{p} p^{-2s} + \frac{1}{3} \sum_{p} p^{-3s} + ... (2)

    ... so that is...

    \ln \zeta(s) - \sum_{p} p^{-s} = \frac{1}{2} \sum_{p} p^{-2s} + \frac{1}{3} \sum_{p} p^{-3s} + ... (3)

    Now we can proceed expanding in series the ratio \frac{\zeta(2s)}{1-2^{-s}} ... and that is left to You...

    Kind regards

    \chi \sigma
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  3. #3
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    Quote Originally Posted by chisigma View Post
    From the Euler's product...

    \zeta(s)= \prod_{p} \frac{1}{1-p^{-s}} (1)

    ... You derive with simple steps...

    \ln \zeta(s)= \sum_{p} p^{-s} + \frac{1}{2} \sum_{p} p^{-2s} + \frac{1}{3} \sum_{p} p^{-3s} + ... (2)

    ... so that is...

    \ln \zeta(s) - \sum_{p} p^{-s} = \frac{1}{2} \sum_{p} p^{-2s} + \frac{1}{3} \sum_{p} p^{-3s} + ... (3)

    Now we can proceed expanding in series the ratio \frac{\zeta(2s)}{1-2^{-s}} ... and that is left to You...

    Kind regards

    \chi \sigma
    Unfortunatly as I wrote in my first post I had managed already to reach

    \ln \zeta(s) - \sum_{p} p^{-s} = \sum_{p} \sum_{m=2}\frac{1}{m} \ p^{-ms}

    That is where I am now confused I had tried expanding the series but I really couldn't see the link
    Last edited by klw289; June 3rd 2011 at 05:00 AM. Reason: bad english
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  4. #4
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by chisigma View Post
    From the Euler's product...

    \zeta(s)= \prod_{p} \frac{1}{1-p^{-s}} (1)

    ... You derive with simple steps...

    \ln \zeta(s)= \sum_{p} p^{-s} + \frac{1}{2} \sum_{p} p^{-2s} + \frac{1}{3} \sum_{p} p^{-3s} + ... (2)

    ... so that is...

    \ln \zeta(s) - \sum_{p} p^{-s} = \frac{1}{2} \sum_{p} p^{-2s} + \frac{1}{3} \sum_{p} p^{-3s} + ... (3)

    Now we can proceed expanding in series the ratio \frac{\zeta(2s)}{1-2^{-s}} ... and that is left to You...

    Kind regards

    \chi \sigma
    Is...

    \frac{1}{2}\ \frac{\zeta(2s)}{1-2^{-s}} = \frac{1}{2} \sum_{n} n^{-2s} + 2^{-(1+s)}\  \sum_{n} n^{-2s} + 2^{-(1+2s)}\ \sum_{n} n^{-2s} +... >

    > \frac{1}{2} \sum_{n} n^{-2s} + \frac{1}{2}\ \sum_{n} n^{-3s} + ... (1)

    ... and if you compare the series (1) with the series...

    \ln \zeta(s) - \sum_{p} p^{-s} = \frac{1}{2} \sum_{p} p^{-2s} + \frac{1}{3} \sum_{p} p^{-3s} + ... (2)

    ... it is clear which is greater...

    Kind regards

    \chi \sigma
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