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Thread: 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 $\displaystyle p_n / {\log{p_n}} \sim n$, where $\displaystyle p_n$ denotes the nth prime. I can also show that $\displaystyle x/{\log{x}}$ is an increasing function (for $\displaystyle x>e$). I'm having problems deducing that there are only finitely many $\displaystyle n$ such that $\displaystyle p_n > n (\log{n})^2$. I've tried showing that the limit (if it exists) of $\displaystyle p_n / {n (\log{n})^2}$ 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 : $\displaystyle \displaystyle\frac{p_n}{\log p_n} = n\cdot e^{o(1)}$

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

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

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

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

    Notation : $\displaystyle f(n)=o(1)$ as $\displaystyle n\to +\infty$ if and only if $\displaystyle \displaystyle\lim_{n\to + \infty}f(n)=0$
    Last edited by PaulRS; Oct 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 $\displaystyle \log{p_n}\leq \log{n} + 2 \log \log{n}$ 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 $\displaystyle {p_n}/{n \log{p_n}} \rightarrow 1$ as $\displaystyle n \rightarrow \infty$ where $\displaystyle p_n$ 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 $\displaystyle \log p_n$ so that we can get rid of it in the expression $\displaystyle \displaystyle\frac{p_n}{\log p_n}$

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

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