Assuming PNT $$\pi(x)\sim \frac{x}{\log{x}}$$ How can we show that given any $\epsilon>0$ $$p_n^{1-\epsilon}< n,$$ for all sufficiently large $n$ ($p_n$ denotes the $n^{th}$ prime.)

My work: Setting $x=p_n$ we get $\lim\limits_{n\to\infty}\dfrac{n\log{p_n}}{p_n}=1\Rightarrow 1-\epsilon<\dfrac{n\log{p_n}}{p_n}$, for sufficiently large $n$. Now if I can show that $$\dfrac{n\log{p_n}}{p_n}\le\dfrac{\log n}{\log{p_n}}$$ then the result follows. But I am unable to show the last inequality.

Can someone please help with this? Other approaches are welcome as well. [NOTE: Using this I want to show that $\lim\limits_{n\to\infty}\frac{\log{n}}{\log{p_n}}=1$ so please don't use that, although it can be derived independently]

Thank you


$\color{red}{\text{1. One way}}$

I will use 2 results

  • PNT $$\lim\limits_{n\rightarrow\infty}\frac{\pi(n)\ln{(n)}}{n}=1 \Rightarrow \lim\limits_{n\rightarrow\infty}\frac{n}{\pi(n)\ln{(n)}}=1 \tag{1}$$
  • and $$\lim\limits_{n\rightarrow\infty}\frac{p_n}{n\ln{(n)}}=1 \tag{2}$$

Proposition 1.1 $$\lim\limits_{n\rightarrow\infty} \frac{\ln{(n)}}{\ln{(p_n)}}=1$$

$\{p_n\}$ is a subsequence of $\{n\}$, thus, from $(1)$, $$\lim\limits_{p_n\rightarrow\infty}\frac{\pi(p_n)\ln{(p_n)}}{p_n}=1 \Rightarrow \lim\limits_{n\rightarrow\infty}\frac{\pi(p_n)\ln{(p_n)}}{p_n}=1 \Rightarrow ...$$ because $\pi(p_n)=n$ $$...\lim\limits_{n\rightarrow\infty}\frac{n\ln{(p_n)}}{p_n}=1 \tag{3}$$ Now $$\lim\limits_{n\rightarrow\infty} \frac{\ln{(n)}}{\ln{(p_n)}}= \lim\limits_{n\rightarrow\infty} \left(\frac{n\ln{(n)}}{p_n}\cdot\frac{p_n}{n\ln{(p_n)}}\right)=\\ \lim\limits_{n\rightarrow\infty} \left(\frac{n\ln{(n)}}{p_n}\right)\cdot \lim\limits_{n\rightarrow\infty} \left(\frac{p_n}{n\ln{(p_n)}}\right)\overset{(2)(3)}{=}1$$

Proposition 1.2 For large enough $n$ $$p_n^{1-\varepsilon}<n$$

From $$\lim\limits_{n\rightarrow\infty} \frac{\ln{(n)}}{\ln{(p_n)}}=1$$ using the definition of limit, $\forall\varepsilon >0, \exists N(\varepsilon)\in\mathbb{N}$ s.t. $\forall n> N(\varepsilon)$ $$\left|\frac{\ln{(n)}}{\ln{(p_n)}}-1\right|<\varepsilon \Rightarrow 1-\varepsilon <\frac{\ln{(n)}}{\ln{(p_n)}}< 1+\varepsilon \Rightarrow \\ (1-\varepsilon)\ln{(p_n)} <\ln{(n)}< (1+\varepsilon)\ln{(p_n)} \Rightarrow \\ \ln{(p_n)^{(1-\varepsilon)}} <\ln{(n)}< \ln{(p_n)^{(1+\varepsilon)}} \Rightarrow ...$$ $e^x$ is ascendng, thus $$... p_n^{1-\varepsilon} <n< p_n^{1+\varepsilon} $$

$\color{red}{\text{2. Another way}}$

Using Vallée-Poussin, for large enough $x$ $$\pi(x)>\frac{x}{\ln(x)-(1-\varepsilon)}>\frac{x}{\ln(x)}$$ Let's show that for large enough $x$ we also have $$\frac{x}{\ln(x)}>x^{1-\varepsilon}$$ which is the same as showing $$\frac{x^{\varepsilon}}{\ln{x}}>1$$ for large $x>0$.

Propositions 2.1 Function $f(x)=\frac{x^{\varepsilon}}{\ln{x}}$ is ascending for large $x>0$.

Because $$f'(x)=\frac{x^{\varepsilon-1} (\varepsilon \ln{x}-1)}{\ln^2{x}}>0 \iff \varepsilon \ln{x}-1>0 \Rightarrow x> e^{\frac{1}{\varepsilon}}$$

Propositions 2.2 $\lim\limits_{x\rightarrow \infty}f(x) \rightarrow \infty$

If we assume it is bounded by a large $\alpha>0, \forall x>1$ and we know $\ln{x}$ is ascending $$\frac{x^{\varepsilon}}{\ln{x}} < \alpha \iff 1<x^{\varepsilon}< \alpha \ln{x} \iff \color{red}{0}<\varepsilon<\frac{\ln{\alpha}}{\ln{x}}+\frac{\ln{\ln{x}}}{\ln{x}}\rightarrow \color{red}{0}, x\rightarrow\infty$$ which is a contradiction.

So, for large $x$ we have $$\pi(x)>x^{1-\varepsilon}$$ which means, for large $n$ we have $$n=\pi(p_n)>p_n^{1-\varepsilon}$$

  • $\begingroup$ Well I already mentioned in the note that I am trying to derive $\log n\sim \log{p_n}$ using my question (and later on to show that $p_n\sim n\log n$). $\endgroup$ – gustaffIR Jun 24 '18 at 11:39
  • $\begingroup$ You also mentioned "Other approaches are welcome as well.". You can also use Vallée Poussin en.wikipedia.org/wiki/… $\endgroup$ – rtybase Jun 24 '18 at 11:57
  • $\begingroup$ Well of course the effort is appreciated, thanks for help. $\endgroup$ – gustaffIR Jun 24 '18 at 12:06
  • 1
    $\begingroup$ @Vulthuryol is it better now? $\endgroup$ – rtybase Jun 24 '18 at 13:10
  • $\begingroup$ how does the PNT implies vallee-poussin's inequality? Can you link a proof? $\endgroup$ – gustaffIR Jun 25 '18 at 3:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.