Consider the truncated sum $$ S(x):=\sum_{\substack{{d\mid P(\sqrt{x})}\\{d\leq x}}}\mu(d)/d, $$ where $P(z)$ is the product of all primes less than or equal to $z$, and $\mu(d)$ is the Möbius function. My question is simply: >Does $S(x)$ have a provable asymptotic limit? The question is motivated by the following plot: ![Plot of truncated Euler product][1] It is well known from the prime number theorem and Merten's product theorem that $$ \frac{x}{\log x \cdot \pi(x)} \sim 1 \quad \textrm{and} \quad \log x \cdot \sum_{\substack{{d\mid P(\sqrt{x})}}}\mu(d)/d \sim 2 \textrm{e}^{-\gamma}. $$ And in seeing the plot, I got curious of $$ \log x \cdot S(x) \sim ? $$ It is not possible to state from the numerical example what the asymptotic limit of $\log x \cdot S(x)$ will be. But note that so is the case also of $x/(\log x \cdot \pi(x))$, which provable tends to 1. What I'm curious about therefore, is whether $\log x \cdot S(x)$ will go all the way to 1 or stagnate before that. I did check out chapter 4 of Opera de Cribro by Friedlander and Iwaniec. There it appears that at least in terms of the elementary Legendre sieve, it is not possible to prove the asymptotic limit of $S(x)$. It might be that an affirmative answer lies other places in that book, with more powerful sieving methods. I would be glad if anyone could point me to the right place in that case, as it is a rather chunky piece of literature to just browse through. [1]: https://i.sstatic.net/aZ3yR.png