# Can I use PNT in this way?

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 $${p_n} \sim n\ln (n).$$

-
How do you go from the expression in terms of $x$ to the expression in terms of $n$? $\ln(x)$ and $\ln(n)$ aren't interchangeable, for one. The result is correct (roughly), but your derivation isn't... –  Steven Stadnicki Oct 2 '12 at 0:56
I knew there was a mistake. –  glebovg Oct 2 '12 at 0:59
I think it's assumed that p_n < x < p_(n+1). This is exactly how Apostol begins his proof on p. 80-81. –  daniel Oct 2 '12 at 1:07
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 ${\lim _{n \to \infty }}n\ln (n)/{p_n}$. –  glebovg Oct 2 '12 at 1:11

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?

-
Thanks. I had a similar argument in mind. –  glebovg Oct 2 '12 at 1:20

There is a problem with this approach. You took $x = p_n$, but that means $\ln(x) = \ln(p_n)$ and not $\ln(n)$.

-
Any ideas on how to prove ${p_n} \sim n\ln (n)$ using PNT? –  glebovg Oct 2 '12 at 1:04

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.

-