Prime number theorem $n^{\rm th}$ prime bounds 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?
 A: 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).
A: 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 $\epsilon=0.1$.
