I am trying to learn some basic number theory now, and I apologize for this trivial question, but I just can't see how to get $p_n\sim n\log(n)$ from $\pi(x)\sim\frac x{\log x}$?
The prime number theorem is equivalent to the statement that the $n$th prime number $p_n$ satisfies $\pi(x)\sim\frac x{\log x}$?
I tried this: If $x=p_n$, then $\pi(p_n)\sim\frac {p_n}{\log p_n}$. Since $\pi(p_n)$ represents number of primes $\leq p_n$ (which is essentially $n$ since $p_n$ is $n$th prime), then $n\sim \frac {p_n}{\log p_n}$ and after multiplying both sides by ($\log p_n$), I get $n \log p_n \sim p_n$. How do I go from here to show that $n \log p_n \sim p_n$ = $p_n\sim n\log(n)$. I also think I went somewhere wrong because $\sim$ is not always symmetric (I think so, please correct me if I am wrong)?
Can someone please help me here?
