Asymptotic Growth of Function of Prime Counting Function

Question

Consider $f(x)$ defined by $$f(x)=\sum_{k=1}^\infty \pi\Big{(}\frac{x}{k}\Big{)}$$ Does there exist a function $g(x)$ be such that $$\lim_{x\to\infty}\frac{f(x)}{g(x)}=1$$I have tried $g(x)=c x \log(\log(x))$, yet this seems to grow slightly too slowly.

Answer 1

Your sum is identical to $\sum_{p\le x} \lfloor \frac{x}{p}\rfloor$, so it is indeed asymptotic to $x \log \log x$ by Mertens's second theorem.

In fact we get the second-order approximation $x \log \log x + Mx$ where $M$ is the Meissel-Mertens constant. The error term is $O(x/\log x)$ which matches the truncation error of removing the floor brackets from the sum.

Since $\log \log x$ grows quite slowly, perhaps this extra term is what gave you the impression that $c \log \log x$ was not adequate.