Does the mean of the ratio of the number of distinct prime factors to the number of divisors of natural numbers converge? Let $d(n)$ and $\omega(n)$ be the number of divisors and the number of distinct prime factors of $n$ respectively. What is the limiting value of
$$
\lim_{n \to \infty} \frac{1}{n}\sum_{r=1}^n \frac{\omega(r)}{d(r)}
$$
For $n \le 23275000000 $, the value is approximately $0.275967$.

I don't know but I think I should point out the following: It's known that
\[ \sum _{r=1}^n\omega (r)\sim n\log \log n\]
and
\[ \sum _{r=1}^nd(r)\sim n\log n,\]
so that "$\omega (r)\approx \log \log r$ on average and $d(r)\approx \log r$ on average", so I guess we'd expect
\[ \sum _{r=1}^n\frac {\omega (r)}{d(r)}\]
to be "like"
\[ \sum _{r=1}^n\frac {\log \log r}{\log r}\approx \int _{1}^n\frac {\log \log rdt}{\log r}\ll \log \log n \cdot \frac {n}{\log n}\]
so we'd expect
\[ \frac {1}{n}\sum _{r=1}^n\frac {\omega (r)}{d(r)}\approx \frac {\log \log n}{\log n}\rightarrow 0.\]
