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I want to show that for large $n$ the $n$-th prime grows like $n\ln (n)$. Is this correct?

By PNT $\mathop {\lim }\limits_{x \to \infty } \frac{{\pi (x)\ln (x)}}{x} = 1.$ Let $x = {p_n}$, so that $\pi (x) = n$, where ${p_n}$ is the $n$-th prime, then $\mathop {\lim }\limits_{n \to \infty } \frac{{n\ln (n)}}{{{p_n}}} = 1$ or ${p_n} \sim n\ln (n).$

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    I think taking the logarithm of ${\lim _{x \to \infty }}\pi (x)\ln (x)/x$, simplifying and multiplying by ${\lim _{x \to \infty }}\pi (x)\ln (x)/x$ yields ${\lim _{n \to \infty }}n\ln (n)/{p_n}$.2012-10-02

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To complete your argument you have to show that $\ln(p_n)/\ln(n) = 1.$

Now, if you have that $p_n \sim n\ln (n),$ then $\ln(p_n) = \ln(n) + \ln\ln(n) + o(1)$, and so indeed $\ln(p_n)/\ln(n) \to 1$ as $n \to \infty$. So at least this is consistent with what you are trying to prove.

On the other other hand, what you know is that $n\ln(p_n)/p_n \to 1$ as $n \to \infty$. Taking $\ln$ gives $\ln(n) + \ln\ln(p_n) = \ln(p_n) + o(1),$ and so $\ln(n)/\ln(p_n) + \ln\ln(p_n)/\ln(p_n) = 1 + o(1).$ Can you finish from there?

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    $T$han$k$s. I had a similar argument in mind.2012-10-02
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There is a problem with this approach. You took $x = p_n$, but that means $\ln(x) = \ln(p_n)$ and not $\ln(n)$.

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    Any ideas on how to prove ${p_n} \sim n\ln (n)$ using PNT?2012-10-02
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Assume $(1):\hspace{3mm}\frac{\pi(x) \log x}{x}\sim 1 $, take logs of both sides:

$\lim_{x \to \infty}(\log \pi(x)+ \log\log x - \log x) = 0$

or $ \lim_{n \to \infty} [ \log x(\frac{\log \pi(x)}{\log x} + \frac{\log\log x}{\log x} -1)] = 0$

Because $\log x \to \infty$ as $ x \to \infty,$

$\lim (\frac{\log \pi(x)}{\log x} + \frac{\log\log x}{\log x} -1) = 0 $

So $(2) \hspace{3mm}\lim_{x\to\infty} \frac{\log \pi(x)}{\log x} = 1$ and from (1) we have

$(3)\hspace{3mm}\lim \frac{\pi(x) \log \pi(x)}{x} = 1 $ and if $x = p_n, \pi(x) = n, $ so $\pi(x) \log \pi(x) = n \log n$, so (3) implies that

$\lim_{n\to \infty }\frac{n \log n}{p_n } = 1. $

This is basically Apostol's proof.