Logarithm of the n'th prime.

Question

Let $P_n$ denote the n'th prime number. How could we conclude the following from the prime number theorem? $$ \log(P_n)=\log n + \log\log n + o(1) $$

Maybe by showing that $P_n=An\log n $ for a suitable $A$?

Answer 1

For some constants $c,C$ we have $$ c \frac{n}{\log n} \leq \pi(n) \leq C \frac{n}{\log n}. $$ Since $\pi(P_n) = n$, we get $$ c \frac{P_n}{\log P_n} \leq n \leq C \frac{P_n}{\log P_n}. $$ This can be manipulated to show that $$ P_n = \log n + \log\log n + O(1). $$ If you want to reduce $O(1)$ to $o(1)$, use a better version of the prime number theorem, namely one that gives $\pi(n) = \frac{n}{\log n} + o(\frac{n}{\log n})$.