How can I prove that $\omega(n)= o(\log{n})$ when $x\to\infty$?
Here $\omega(n)= \sum_{p|n}1,$ i.e. if $n = \prod_{i=1}^{r} p_i^{v_i}$, then $\omega(n)=r.$
I want $$\lim_{n\to\infty}\frac{\omega(n)}{\log{n}}=0.$$
I have tried using the fact that $2^{w(n)}\leq \tau(n)$, but I am stuck. Can anyone help me?
The largest $\omega(n)$ occurs at $$n = \prod_{p < k} p$$ (a primorial)
The prime number theorem shows that $$\ln n = \sum_{p < k} \ln p \sim k$$ Now $$\omega(n) = \sum_{p < k} 1 = \pi(k) \sim \frac{k}{\ln k}$$ again by the PNT.
Hence at those $n$ primorials $$\omega(n) \sim \frac{\ln n}{\ln \ln n}$$
And with $n$ arbitrary : $$\omega(n) < C \frac{\ln n}{\ln \ln n}$$ for some $C$ that $\to 1$ when restricting to $n $ large enough .