We know that
$$
\sum_{n \le x}\frac{1}{n\ln n} = \ln\ln x + c_1 + O(1/x)
$$
where $c_1$ is a constant. Again Mertens' theorem says that the primes $p$ satisfy
$$
\sum_{p \le x}\frac{1}{p} = \ln\ln x + M + O(1/\ln x).
$$
Thus both these divergent series grow at the same rate. Mertens' theorem was proved without using the prime number theorem, some 25 years before PNT was proved. However from these two examples, we cannot conclude that
$$
\lim_{n \to \infty} \frac{p_n}{n\ln n} = 1
$$
otherwise Mertens' would have been the first to prove PNT. My question is – based on the above two series, what are the technical difficulties that prevent us from reaching the conclusion that $p_n/n\ln n = 1$. There may be counter examples with other series, so such conclusions may not be true in general. However I am not interested in the general case. Instead I am asking only in case of the sequence $1/n\ln n$ and $1/p_n$.
Best Answer
Here is a heuristic that I plan to make into a blog post some day. Suppose that there were only finitely many primes with first digit $9$. Is your estimate good enough to see that?
To be more precise, suppose that there were no primes between $9 \times 10^k$ and $10^{k+1}$ for all sufficiently large $k$. And suppose that the number of primes between $a$ and $b$, for $10^k \leq a < b \leq 9 \times 10^k$ is $\approx \frac{\log 10}{\log 9} \int_{a}^b dt/\log t$ (when $a$ is not too close to $b$). We'll see later where the fraction $\log 10/\log 9$ comes from.
The first thing to note is that this would violate the prime number theorem. In this scenario, we would have $\pi(9 \times 10^k) = \pi(10 \times 10^k)$ for $k$ large. But the prime number theorem says that $$\pi(10\times 10^k) - \pi(9 \times 10^k) \sim \frac{10 \times 10^k}{k \log 10 + \log {10}} - \frac{9 \times 10^k}{k \log 10 + \log {9}} \sim \frac{10^k}{k \log 10} \to \infty.$$ So proving the prime number theorem involves disproving this ridiculous scenario.
Now, let's see that the scenario is consistent with $\sum_{p \leq N} 1/p = \log \log N + M + O(1/\log N)$. The sum over the primes between $10^k$ and $10^{k+1}$ would be roughly $$\frac{\log 10}{\log 9} \int_{10^k}^{9 \times 10^k} \frac{dt}{t \log t} = \frac{\log 10}{\log 9} \left( \log \log (9 \times 10^k) - \log \log 10^k \right)$$ $$=\frac{\log 10}{\log 9} \left( \log( k \log 10+\log 9) - \log (k \log 10) \right) = \frac{\log 10}{\log 9} \left( \log \left( 1+\frac{\log 9}{k \log 10} \right) \right) $$ $$ = \frac{\log 10}{\log 9} \frac{\log 9}{k \log 10} + O(1/k^2)= \frac{1}{k} + O(1/k^2)$$
So $$\sum_{p \leq 10^{n+1}} \frac{1}{p} = \sum_{j=1}^n \left( \frac{1}{j} + O(1/j^2) \right)=$$ $$\log n + B + O(1/n) = \log \log 10^{n+1} + C + O(1/\log 10^{n+1}).$$ Very important exercise left for you: If you redo this computation for $\sum_{p \leq 9 \times 10^k} 1/p$, you get $\log \log (9 \times 10^k) + C + O(1/\log(9 \times 10^k))$ for the same constant $C$. The point is that $\log \log 10^{k+1} - \log \log (9 \times 10^k) = O(1/\log 10^k)$, so this estitmate is consistent with $\sum_{p \leq N} 1/p$ not growing at all between $9 \times 10^k$ and $10 \times 10^k$.
This trick is useful for refuting other simple approaches to the PNT. For example, the "primes hate to start with $9$ scenario" is also consistent with $\sum \log p/p^s = 1/(s-1) + O(1)$ as $s \to 1^{+}$, so that is also not enough to prove PNT.