Estimating the $n^{th}$ prime $p_n$ in terms of $n$ for $n$ large

Question

From the prime number theorem we know that for $n$ large ,

$n=\pi(p_n)\sim\frac{p_n}{\log p_n}$

$\implies \log n \sim \log(p_n)-\log\log p_n \ \ -(i)$

Now $p_n \sim n\log p_n$. Some calculations with primes between $10000$ and $100000$ makes me feel $p_n \sim n\log n$ holds. To prove this I'd have to show $\lim_{n\rightarrow \infty}\frac{\log \log p_n}{log n}=0$ .So that by $(i)$ I'd be done.

Can someone tell me if my intuition is correct and if so how to prove the step I'm stuck at.Thanks in advance.

Answer 1

$$\pi(x) \sim \frac{x}{\log x}$$ gives $$\pi(x)\log \pi(x) \sim \frac{x}{\log x}\log (\frac{x}{\log x}) \sim x$$

letting $x=p_n$ gives your result $$n \log n \sim p_n$$