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In this Math SE post, the accepted answer begins with:

"Because of the Prime Number Theorem, ultimately we have: $$.9\,n\,\log n \leq p_n \leq 1.1 \,n\,\log n.$$ where $p_n$ is the $n$th prime".

Where do the bounds of $.9$ and $1.1$ come from? Wikipedia states that:

$$ 0.921<\frac{\pi(n)}{n\log(n)}<1.018, $$

for $10<n<10^{25}$, but other than experimental evidence, how can one be sure that these are really the correct bounds?

Klangen
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  • @DietrichBurde Thank you for the link. However, my question refers to the explicit bounds of $.9$ and $1.1$, those are not discussed in that other post. – Klangen Oct 16 '17 at 12:27
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    These are arbitrary choices. From the prime number theorem, it follows that $p_n \sim n\log n$, so for every $\varepsilon > 0$ there is an $N(\varepsilon)$ such that $(1-\varepsilon)n\log n < p_n < (1+\varepsilon) n\log n$ for all $n \geqslant N(\varepsilon)$. – Daniel Fischer Oct 16 '17 at 12:33

2 Answers2

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The prime number theorem gives that $\pi(x) \sim \frac{x}{\log x}$, in the sense that $$ \lim_{x \to \infty} \pi(x) / \frac{x}{\log x} = 1.$$ This is equivalent to the fact that $p_n \sim n \log n$, also in the sense that $$ \lim_{n \to \infty} \frac{p_n}{n \log n} = 1.$$ In particular, this means that there exists some $N$ such that for all $n > N$, we have $$ 0.9 < \frac{p_n}{n \log n} < 1.1.$$ There is nothing special about $0.9$ and $1.1$ in this, except that $0.9 < 1$ and $1 < 1.1$. We could replace those two numbers with any other such numbers. We should note that it is (currently) unclear how large $N$ should actually be, and efforts to understand the explicit rate of convergence are wide and varied within the literature (and a speciality of people like Dusart).

davidlowryduda
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Those coefficients have nothing special.

We know that $$\lim_{n\to\infty}\frac{p_n}{n\log n}=1$$

That is, for any $\epsilon>0$, there exists some $N\in\Bbb N$ such that if $n\ge N$ we have $$1-\epsilon<\frac{p_n}{n\log n}<1+\epsilon$$ Now take $\epsilon=0.1$.

ajotatxe
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