# Thread: Deduce this result from the Prime Number Theorem

1. ## 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.

2. 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 $\displaystyle\lim_{n\to + \infty}f(n)=0$

3. 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.

4. 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 $n$.