For any non-zero natural integer $n$, we set $$\omega(n) = \operatorname{Card}\{p \text{ prime} : p \mid n\} = \sum_{\substack{p \mid n \\ p \text{ prime}}} 1.$$ We define $\mathscr{S} = \left\{n \geqslant 3 : \left|\frac{\omega(n) - \ln_{2}(n)}{\sqrt{\ln_{2}(n)}}\right| \geqslant \left(\ln_{2}(n)\right)^{1/4}\right\}$. Show that $$\lim_{x \rightarrow +\infty} \frac{1}{x} \operatorname{Card}\{n \leqslant x : n \in \mathscr{S}\} = 0$$ One may begin by writing $\operatorname{Card}(\mathscr{S} \cap [1,x]) \underset{x \rightarrow +\infty}{=} \operatorname{Card}(\mathscr{S} \cap [\sqrt{x}, x]) + O(\sqrt{x})$ and note that in the sum on the right-hand side, the difference $\left|\ln_{2}(n) - \ln_{2}(x)\right|$ remains bounded.
For any non-zero natural integer $n$, we set
$$\omega(n) = \operatorname{Card}\{p \text{ prime} : p \mid n\} = \sum_{\substack{p \mid n \\ p \text{ prime}}} 1.$$
We define $\mathscr{S} = \left\{n \geqslant 3 : \left|\frac{\omega(n) - \ln_{2}(n)}{\sqrt{\ln_{2}(n)}}\right| \geqslant \left(\ln_{2}(n)\right)^{1/4}\right\}$. Show that
$$\lim_{x \rightarrow +\infty} \frac{1}{x} \operatorname{Card}\{n \leqslant x : n \in \mathscr{S}\} = 0$$
One may begin by writing $\operatorname{Card}(\mathscr{S} \cap [1,x]) \underset{x \rightarrow +\infty}{=} \operatorname{Card}(\mathscr{S} \cap [\sqrt{x}, x]) + O(\sqrt{x})$ and note that in the sum on the right-hand side, the difference $\left|\ln_{2}(n) - \ln_{2}(x)\right|$ remains bounded.