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
$$
\text{Mertens}\rightarrow\text{pnt}.
$$
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$\begingroup$ Duplicate of mathoverflow.net/questions/95743/… . $\endgroup$– JoshuaZCommented Nov 7, 2020 at 12:07
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$\begingroup$ Does this answer your question? Why could Mertens not prove the prime number theorem? $\endgroup$– JoshuaZCommented Nov 7, 2020 at 12:07
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$\begingroup$ No, actually I linked that post because the questions are related, but there they only present reasoning about why Mertens isn't enough for pnt. What I need is an explicit subset of $\mathbb{N}$ of elements $a$ that gives a counterexample for Mertens>pnt. $\endgroup$– user627482Commented Nov 7, 2020 at 12:11
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3$\begingroup$ Does David's example using powers of ten not lead to such a set? $\endgroup$– JoshuaZCommented Nov 7, 2020 at 12:19
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1$\begingroup$ @user627482 Take the primes P. Then, for all the primes between $9*(10^k)$ and $10^{k+1}$, if the prime is less closer to 9*(10^k), replace it with the next available integer below $9*(10^k)$, and if the prime is closer to $10^{k+1}$ replace with the next available integer above $10^{k+1}$. This set will obey Mertens theorem by David's argument, but the ratio of $\Pi(x)/ (x/\log x)$ will not have a limit. Does this work for your purposes? (I haven't checked that the error term is precisely of the order you want.) $\endgroup$– JoshuaZCommented Nov 7, 2020 at 12:38
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1 Answer
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Take $a_n\in 0,1$ such that $$\sum_{n\le x} a_n=\frac{x}{\ln x}(1+\frac14 \cos(\ln x))+O(1)$$ then do a partial summation to find the asymptotic of $$\sum_{n\le x} \frac{a_n}{n}$$ And you meant $\sum_{p\le x} \frac1p= \ln\ln x + C + O(1/\ln x)$.
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$\begingroup$ In the first sum, did you mean $n\leq x$? $\endgroup$ Commented Nov 7, 2020 at 19:44
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$\begingroup$ Thanks for the answer anyway, you nailed it! But it's not clear to me why such a sequence exists. $\endgroup$ Commented Nov 8, 2020 at 14:25
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$\begingroup$ The derivative of $\frac{x}{\ln x}(1+\frac14 \cos(\ln x))$ is $\ge 0$ and $\to 0$. ie. $a_n = \lfloor \frac{n}{\ln n}(1+\frac14 \cos(\ln n))\rfloor - \lfloor \frac{n-1}{\ln( n-1)}(1+\frac14 \cos(\ln (n-1)))\rfloor$ $\endgroup$– reunsCommented Nov 8, 2020 at 14:49
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$\begingroup$ I've another question...do you have any idea about what kind of "$\Lambda$" function does this sequence give? I mean, for the primes we have the von Mangoldt function $\Lambda(n)=\log p$ for $n=p^{k}$, and the Mertens' formula $\sum_{n\geq1}\Lambda(n)/n=\log x+O(1)$, with the sequence you have found what do we get? $\endgroup$ Commented Nov 11, 2020 at 18:07