Does the mean of sum of the reciprocal of the distinct prime divisors of a number converge to the prime zeta function $P(2)$

Question

Let $k \ge 2$ and $f(k)$ be the sum of the reciprocal of the distinct prime divisors of $k$. Is it true that

$$ \lim_{n \to \infty}\frac{1}{n}\sum_{k = 2}^{n}f(k) = \sum_{p}\frac{1}{p^2} \approx 0.452247 $$

Note that $P(s) = \sum_{p}\frac{1}{p^s}$ is the prime zeta function.

Answer 1

Sure $$\sum_{k\le n} \sum_{p| k} \frac1p = \sum_{p\le n} \frac1p \lfloor n/p \rfloor = \sum_{p\le n} \frac1p (\frac{n}p + O(1))= n \sum_{p\le n}\frac1{p^2}+O(\log n)$$