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Math Help - prime infinite series

  1. #1
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    prime infinite series

    All i really want to know is does the series 1+(1/2)+(1/3)+(1/5)+ ... +(1/n). When n is a prime , converge. and if it does can anyone explain why.

    Thanks for your help
    nag
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  2. #2
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    Quote Originally Posted by N.A.G View Post
    All i really want to know is does the series 1+(1/2)+(1/3)+(1/5)+ ... +(1/n). When n is a prime , converge. and if it does can anyone explain why.

    Thanks for your help
    nag
    The series diverges. As Euler showed.

    Define \pi (n) to be the primes \leq n. For a given n (sufficiently large) let p_1,p_2,...,p_{\pi (n)} be the primes \leq n.
    Define \Pi (n) = \prod_{k=1}^{\pi (n)} \left( 1 - \frac{1}{p_k} \right)^{-1}.

    Notice that \left( 1 - \frac{1}{p_k} \right)^{-1} = \sum_{a_k =0}^{\infty} \frac{1}{p^{a_k}}.

    Therefore, we see that \Pi (n) = \sum_{a_k \geq 0} (p_1^{a_1}p_2^{a_2}...p_{\pi(n)} ^{a_{\pi(n)}} )^{-1}.

    By the fundamental theorem of arithmetic any number 1\leq m\leq n be be written as m=p_1^{b_1}...p_{\pi (n)}^{b_{\pi(n)}} where b_k\geq 0.
    It follows that,
    \sum_{k=1}^n \frac{1}{k} \leq \sum_{a_k \geq 0} (p_1^{a_1}p_2^{a_2}...p_{\pi(n)} ^{a_{\pi(n)}} )^{-1} = \Pi (n).

    Since the harmonic series diverges it means \lim ~ \Pi (n) = \infty \implies  \lim ~ \log \Pi (n) = \infty.

    However,
    \log \Pi (n) = \log \prod_{k=1}^{\pi (n)} \left( 1 - \frac{1}{p_k} \right)^{-1} = - \sum_{k=1}^{\pi(n)}  \log \left( 1 - \frac{1}{p_k} \right) = \sum_{k=1}^{\pi (n)} \sum_{j=1}^{\infty}\frac{1}{jp_k^j}

    Thus,
    \log \Pi (n) = \left( \frac{1}{p_1}+...+\frac{1}{p_{\pi(n)}} \right)  + \sum_{k=1}^{\pi (n)} \sum_{j=2}^{\infty}\frac{1}{jp_k^j}

    Notice that,
    \sum_{j=2}^{\infty} \frac{1}{jp_k^j} \leq \sum_{j=2}^{\infty} \frac{1}{p_k^j} = \frac{1}{p_k^2(1-\tfrac{1}{p_k})} \leq \frac{2}{p_k^2}

    We have shown,
    \log \Pi (n) \leq \left( \frac{1}{p_1}+...+\frac{1}{p_{\pi(n)}} \right) + 2 \left( \frac{1}{p_1^2}+...+\frac{1}{p_{\pi(n)}^2} \right) \leq \sum_{k=1}^{\pi(n)} \frac{1}{p_k} + 2 \sum_{j=1}^{n} \frac{1}{j^2}

    Remember that \sum_{j=1}^{\infty} \frac{1}{j^2} < \infty.

    What this means is that if \sum_{k=1}^{\infty} \frac{1}{p_k} < \infty then \log \Pi (n) would be bounded.
    This is impossible, it is not bounded.

    Thus, the sum of prime reciprocals must diverge.
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  3. #3
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    a very clever proof of this result is due to Paul Erdos. see "Second Proof" in here. note that in the "Lower estimate" part of the proof the inclusion \subset is actually an equality.

    although this doesn't change anything in the proof, but still the wikipedia author should have been more careful!
    Last edited by NonCommAlg; October 24th 2008 at 10:18 PM.
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  4. #4
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    Quote Originally Posted by NonCommAlg View Post
    a very clever proof of this result is due to Paul Erdos. see "Second Proof" in here. note that in the "Lower estimate" part of the proof the inclusion \subset is actually an equality.

    although this doesn't change anything in the proof, but still the wikipedia author should have been more careful!
    It's Wikipedia, if you don't like the article or its wrong it's your job to change it!

    CB
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