# Covering $\mathbb{N}$ with disjoint arithmetic sequences

Suppose that we have a collection $S_1 = \{a_1 + kr_1\}_{k=0}^{\infty}$, $\cdots$, $S_n = \{a_n + kr_n\}_{k=0}^{\infty}$ of disjoint arithmetic sequences where $a_i$, $r_i$ are nonnegative integers for all $1\le i \le n$. Furthermore, assume that these sequences cover $\mathbb{Z}_{\ge 0}$: $$\mathbb{N} \cup \{0\} = S_1 \cup S_2 \cup \cdots \cup S_n$$

Then, one has the formal equality of power series $$\frac{1}{1-x} = \sum_{k=0}^{\infty} x^k = \sum_{i=1}^{n} \sum_{k=0}^{\infty} x^{a_i + k r_i} = \sum_{i=1}^{n} \frac{x^{a_i}}{1-x^{r_i}}$$ Multiplying both sides by $1-x$ and taking $x\to 1^{-}$ proves that $$\frac{1}{r_1} + \frac{1}{r_2} + \cdots + \frac{1}{r_n} = 1 \tag{1}$$ Differentiating both sides and taking $x\to 1^{-}$ again shows $$\frac{a_1}{r_1} + \frac{a_2}{r_2} + \cdots + \frac{a_n}{r_n} = \frac{n-1}{2} \tag{2}$$

The first of these identities has a direct combinatorial/number theoretic proof. Let $N = r_1 r_2 \cdots r_n$. Then $S_i$ accounts for precisely $N/r_i$ of the residue classes modulo $N$. Since there are $N$ residue classes, the equation $$\frac{N}{r_1} + \cdots + \frac{N}{r_n} = N$$ yields identity $(1)$. Is there a similar direct counting argument for identity $(2)$?

The equality $2$ can be determined using the observation that the sum of all elements belonging to each set $S_i$ is the sum of all residues so: $$\overbrace{\frac{N}{r_1}a_1+r_1.\frac{N}{2r_1}(\frac{N}{r_1}-1)}^{\text{The sum of residus }\in S_1}+\cdots\cdots+\overbrace{\frac{N}{r_n}a_n+r_n.\frac{N}{2r_n}(\frac{N}{r_1}-1)}^{\text{The sum of residus }\in S_n}=\overbrace{\frac{N(N-1)}{2}}^{\text{The sum of all residus }}$$
with some simplification you get $(2)$