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Math Help - Sum of the reciprocal of the primes diverges

  1. #1
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    Sum of the reciprocal of the primes diverges

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

    |\log\zeta\left(s\right)-\underset{p}{\sum}p^{-s}|<\frac{1}{2}\left(1-2^{-s}\right)^{-1}\zeta\left(2s\right)

    By letting s\rightarrow1 I can then show that \underset{p}{\sum}p^{-1} diverges (I hope).

    I've tried all sorts of expressions for the left hand side of the inequality and don't seem to be able to get anywhere.

    Any ideas?
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  2. #2
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Cairo View Post
    I'm trying to prove that if s is real and s>1 then

    |\log\zeta\left(s\right)-\underset{p}{\sum}p^{-s}|<\frac{1}{2}\left(1-2^{-s}\right)^{-1}\zeta\left(2s\right)

    By letting s\rightarrow1 I can then show that \underset{p}{\sum}p^{-1} diverges (I hope).

    I've tried all sorts of expressions for the left hand side of the inequality and don't seem to be able to get anywhere.

    Any ideas?
    The fact that \sum_{p} \frac{1}{p} diverges has been demonstrated in more elementar way in...

    http://www.mathhelpforum.com/math-he...ers-84832.html

    Kind regards

    \chi \sigma
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  3. #3
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    I totally agree, but I am independently working through a textbook that asks to prove the divergence of the sum of the reciprocal of the primes this way, via an exercise.
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  4. #4
    Senior Member Shanks's Avatar
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    The Euler Product of zeta function and the taylor expansion of log function would help you in this problem.
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  5. #5
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    Thanks for this Shanks.

    I've now managed to reach the required inequality, but have no idea how I can show the divergence of the sum as s tends to 1.

    Can anyone offer a rigorous argument here?
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  6. #6
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by Cairo View Post
    I'm trying to prove that if s is real and s>1 then

    |\log\zeta\left(s\right)-\underset{p}{\sum}p^{-s}|<\frac{1}{2}\left(1-2^{-s}\right)^{-1}\zeta\left(2s\right)

    By letting s\rightarrow1 I can then show that \underset{p}{\sum}p^{-1} diverges (I hope).

    I've tried all sorts of expressions for the left hand side of the inequality and don't seem to be able to get anywhere.

    Any ideas?
    As s\to 1, the right side is bounded. However, \log(\zeta(s)) is unbounded as s\to 1, so...
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  7. #7
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    Ok.

    So the right side is bounded (presumably by pi^2/6) as s tends to 1. Also, zeta(1) is just the harmonic series (which diverges) and so does its logarithm. I'm fine with that.

    So how does this show that the sum also diverges? Is there a theorem about the sum of two divergent series or something?
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  8. #8
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by Cairo View Post
    Ok.

    So the right side is bounded (presumably by pi^2/6) as s tends to 1. Also, zeta(1) is just the harmonic series (which diverges) and so does its logarithm. I'm fine with that.

    So how does this show that the sum also diverges? Is there a theorem about the sum of two divergent series or something?
    Well, if a_n, b_n and c_n are sequences such that a_n is bounded and b_n diverges to infinity, and |b_n-c_n|<a_n, then necessarily c_n diverges also...
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