Countably many APs partitioning the natural numbers Suppose there are countably many APs partitioning  $\mathbb{N}$, with starting points $a_i$ and common difference $d_i$. For the finite case it is clear that $\sum\frac 1 d_i =1$, and  $\sum\frac 1 d_i \leq 1$, even for the infinite case. We also know that there exist partitions where $\sum\frac 1 d_i \neq1$, for example, take common differences $4,8,$ and so on.
Can we prove that $\theta(\{a_i\})+\sum\frac 1 d_i=1$, where $\theta(A)=\lim_{n\to \infty} \dfrac{A\cap\{1,2\dots n\}}{n}$?
 A: The key insight here is that, if we let $A=\{a_i\}$, then $\sum\frac{1}{d_i}$equals $\theta(A^C)$, and obviously, $\theta(A)+\theta(A^C)=1$ where either density exists.
To show this, note that if we let $S_i$ be the set of elements, except the first, in the $i^{th}$ progression - that is
$$S_i=\{a_i+d_i,a_i+2d_i,a_i+3d_i\}$$
and so on, it should be clear that $\{S_i\}\cup\{A\}$ is a partition of $\mathbb{N}$ - as it contains every element of every progression in a partition. Note that, letting $\theta_k(A)=\frac{A\cap \{1,2,\ldots,k\}}{k}$, we have that, for $k\geq a_i$, that
$$\theta_k(S_i)=\frac{1}k\left\lfloor\frac{k-a_i}{d_i} \right\rfloor $$
which is clearly subject to the bounds
$$\frac{k-a_i-d_i}{kd_i}=\frac{1}{d_i}-\frac{a_i+d_i}{kd_i}\leq\theta_k(S_i)<\frac{1}{d_i}.$$
Thus, if we choose any finite set $P\subset I$ of indices, it should be clear that, since all $S_i$ would be contained in $A^C$ and be disjoint, we'd have that, for all large enough $k$,
$$\sum_{i\in P}\frac{1}{d_i}+\frac{1}k\sum_{i\in P}\frac{a_i+d_i}{d_i}\leq\theta_k(A^C)$$
which, by choosing a large enough $P$, shows that for any $\varepsilon$, large enough $k$, that $\theta_k(A^C)$ is always at least $\left(\sum_{i\in I} \frac{1}{d_i}\right)-\varepsilon$. Moreover, it is clear that $\theta_k(A^C)$ may never exceed $\sum_{i\in I}\frac{1}{d_i}$, being a union of sets of density at most $\frac{1}{d_i}$. Therefore, we may conclude that $$\theta(A^C)=\sum_{i\in I}\frac{1}{d_i}$$
which converges, implying that we have $\theta(A)+\theta(A^C)=1$ as the sum.
