Background
I recently had the following idea. We define $k(n)$ as the product of the first $n$ primes where $p_n$ is the $n$'th prime.
$$k(n) = \prod_{i=1}^n p_i$$
For example, $k(2) = 2 \cdot 3 = 6$. We write $n$'th integer as:
$$ n = \underbrace{1+ 1+ \dots}_{n \text{ terms}}$$
Inserting $n(n-1)$ terms of $0$'s into series, which will repeat as:
$$ n = \underbrace{0+ 0+ \dots + 0 +1}_{n^2 \text{ terms}}$$
Now consider, the expansion of $p_n \lambda_n$ as an example:
$$ 2 \lambda_1 = \underbrace{0+1+0+1 + \dots}_{2^2 \lambda_1 \text{ terms}} $$
$$ 3 \lambda_2 = \underbrace{0+0+1+0+0+1 +0+0+1+ \dots}_{3^2 \lambda_2 \text{ terms}} $$
$$ 5 \lambda_3 = \underbrace{0+0+0+0+1+0+0+0+0+1+ \dots}_{5^2 \lambda_3 \text{ terms}} $$
Now, we choose $\lambda_1$, $\lambda_2$, $\dots$, $\lambda_n$. Such that:
$$ 2^2 \lambda_1 = 3^2 \lambda_2 = 5^2 \lambda_3 = k(3)^2$$
Hence, when we add $2\lambda_1$, $3\lambda_2$, $5 \lambda_3$ together to get:
$$k(3)^2 (\frac{1}{2} + \frac{1}{3} + \frac{1}{5})= \sum_{r=1}^{k(3)^2}\sum_{5 \geq p_i} \omega_{p_i}(r)$$
where we vertically add for the R.H.S with $\omega_{p_i}(n)$ being the number of distinct primes of $p_i$ in $n$. Hence, in general:
$$ k(n)^2 (\sum_{r=1}^{n} \frac{1}{p_r}) = \sum_{r=1}^{k(n)^2} \sum_{p_n \geq p_i} \omega_{p_i}(r) $$
Question
Is the formula valid in the limit $n \to \infty$
$$ k(n)^2 (\sum_{r=1}^{n} \frac{1}{p_r}) \sim \sum_{r=1}^{k(n)^2} \omega(r) $$
where $\omega(r)$ is then number of distinct primes of $r$?
I believe what you are asking is equivalent to this statement. Given $x>2$, define $K=\prod_{p\le x} p$ and $R=\sum_{p\le x} \frac1p$ (where both sums are over primes only). You are asking whether $K^2R \sim \sum_{r=1}^{K^2} \omega(r)$.
By standard results, $R\sim \log\log x$, while $\sum_{r=1}^{K^2} \omega(r) \sim K^2\log\log(K^2) \sim K^2\log\log K$. Since $\log K = \theta(x) \sim x$, we have $K^2R \sim K^2\log\log x$ while $\sum_{r=1}^{K^2} \omega(r) \sim K^2\log x$; thus the two are not asymptotic.