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Math Help - Deduce this result from the Prime Number Theorem

  1. #1
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    Deduce this result from the Prime Number Theorem

    Using the prime number theorem I can show that p_n / {\log{p_n}} \sim n, where p_n denotes the nth prime. I can also show that x/{\log{x}} is an increasing function (for x>e). I'm having problems deducing that there are only finitely many n such that p_n > n (\log{n})^2. I've tried showing that the limit (if it exists) of p_n / {n (\log{n})^2} is strictly less than 1, but I can't get anywhere.
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  2. #2
    Super Member PaulRS's Avatar
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    You have : \displaystyle\frac{p_n}{\log p_n} = n\cdot e^{o(1)}

    Taking logs we find: \log p_n = \log( n) + \log\log (p_n) + o(1)

    2^{\pi(N)} \geq \sqrt{N} see here (*), thus 2^{n} \geq \sqrt{p_n} and \log( n) \geq \log\log(2^n)\geq \log\log(p_n) - \log(2)

    Thus \log p_n \leq 2\cdot \log( n) + \log( 2) + o(1) try to go on from here.

    (*) Every number in \left\{1,2,...,N\right\} can be written in the form n=a_n\cdot (b_n)^2 where a_n is square free. There are no more than 2^{\pi(N)} possible values for a_n ( each one corresponds to a different set of primes <= N), and no more than \sqrt{N} possible values for b_n thus 2^{\pi(N)}\cdot \sqrt{N}\geq N

    Notation : f(n)=o(1) as n\to +\infty if and only if \displaystyle\lim_{n\to + \infty}f(n)=0
    Last edited by PaulRS; October 7th 2010 at 06:06 AM.
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  3. #3
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    I'm not really sure how this helps. If we're taking logs then we want to show that \log{p_n}\leq \log{n} + 2 \log \log{n} for large n. I get the feeling this isn't meant to be a difficult question; I think it's supposed to follow naturally from the fact that {p_n}/{n \log{p_n}} \rightarrow 1 as n \rightarrow \infty where p_n is the nth prime.
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  4. #4
    Super Member PaulRS's Avatar
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    Not really, what I am doing is estimating \log p_n so that we can get rid of it in the expression \displaystyle\frac{p_n}{\log p_n}

    So we get p_n = (\log p_n)\cdot n\cdot e^{o(1)} \leq \left( 2\log(n) + \log(2)+o(1) \right)\cdot n \cdot e^{o(1)}

    Note that the right-hand-side is way smaller than n\cdot \left(\log( n)\right)^2 for sufficiently large n.
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