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 $0.275967$.
 A: Just to expand on my comment on tomos's answer:  we always have $d(r) \geq 2^{\omega(r)}$, so when $\omega(r)$ is large, we have
$$
\frac{\omega(r)}{d(r)} \leq \frac{\omega(r)}{2^{\omega(r)}} \to 0.
$$
But recall that for all $m$, asymptotically $100\%$ of positive integers $r$ have $\omega(r) > m$.  Now let $\varepsilon > 0$.  By the previous statement, there exists some $N$ such that for all $n > N$, the proportion of integers $r$ in $[1, n]$ with $\frac{\omega(r)}{2^{\omega(r)}} \geq \frac{\varepsilon}{2}$ is less than $\frac{\varepsilon}{2}$.  Then the sum
$$
\sum_{r=1}^n \frac{\omega(r)}{d(r)}
$$
can be split into two pieces:  one containing at most $n$ terms that are each less than $\frac{\varepsilon}{2}$, and the other containing fewer than $\frac{n \varepsilon}{2}$ terms, each of them at most $1$.  This yields
$$
\sum_{r=1}^n \frac{\omega(r)}{d(r)} < n \cdot \frac{\varepsilon}{2} + \frac{n \varepsilon}{2} \cdot 1 = n \varepsilon.
$$
So the average is less than $\epsilon$ whenever $n$ is large enough; i.e. it converges to $0$.
A: 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
\[ \frac {1}{n}\sum _{r=1}^n\frac {\omega (r)}{d(r)}\approx \frac {\log \log n}{\log n}\rightarrow 0.\]
