For the primes it's true that $$ \sum_{p \le x}\frac{1}{p} = \ln\ln x + M + O(1/x) $$ where, $M$ is suitable constant, and, moreover, the prime number theorem gives that $$ \lim_{x\to\infty}\frac{\pi(x)}{x/\ln x}=1 $$ with $\pi(x)$ is the prime counting function. David Speyer gives a nice heuristic here in order to explain why Mertens formulas aren't enough for pnt. However, I'm concerned with finding a series of integer numbers $a$ such that $$ \sum_{a \le x}\frac{1}{a} = \ln\ln x + C + O(1/x) $$
with $C$ suitable constant, but it isn't true that $$ \lim_{x\to\infty}\frac{f(x)}{x/\ln x}=1 $$ where $f$ is the counting function of numbers $a$. This would give a a concrete counterexample for $$ \text{Mertens}\rightarrow\text{pnt}. $$

For the primes it's true that $$ \sum_{p \le x}\frac{1}{p} = \ln\ln x + M + O(1/\ln x) $$ where, $M$ is suitable constant, and, moreover, the prime number theorem gives that $$ \lim_{x\to\infty}\frac{\pi(x)}{x/\ln x}=1 $$ with $\pi(x)$ is the prime counting function. David Speyer gives a nice heuristic here in order to explain why Mertens formulas aren't enough for pnt. However, I'm concerned with finding a series of integer numbers $a$ such that $$ \sum_{a \le x}\frac{1}{a} = \ln\ln x + C + O(1/\ln x) $$
with $C$ suitable constant, but it isn't true that $$ \lim_{x\to\infty}\frac{f(x)}{x/\ln x}=1 $$ where $f$ is the counting function of numbers $a$. This would give a a concrete counterexample for $$ \text{Mertens}\rightarrow\text{pnt}. $$