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The statement is:

Let $A=\{(a_s(n_s)\}^k_{s=1}$ be an exact cover of $\mathbb{Z}$ with $1<n_1\leq\ldots\leq n_{k-1}\leq n_k$. Then we must have $n_{k-1}=n_k$.

The only proof I could find was here. But I have some difficulties understanding it. It is very short and it goes like this:

Without loss of generality we assume that $0\leq a_s<n_s$ for all $s\in[1,k]$. For $|z|<1$ we have $$\sum_{s=1}^k\frac{z^{a_s}}{1-z^{n_s}}=\sum_{s=1}^k\sum_{q=0}^\infty z^{a_s+qn_s}=\sum_{n=0}^\infty z^n=\frac{1}{1-z}.$$ If $n_{k-1}<n_k$ then $$\infty=\lim_{\substack{z\to e^{2\pi i/n_k}\\ |z|<1}}\frac{z^{a_k}}{1-z^{n_k}}=\lim_{\substack{z\to e^{2\pi i/n_k}\\ |z|<1}}\left(\frac{1}{1-z}-\sum_{s=1}^{k-1}\frac{z^{a_s}}{1-z^{n_s}}\right)<\infty,$$ a contradiction!

What I don't get is the following: $$\sum_{s=1}^k\sum_{q=0}^\infty z^{a_s+qn_s}=\sum_{n=0}^\infty z^n.$$ It just doesn't make sense to me. So what am I misunderstanding?

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up vote 1 down vote accepted

That equation states that $A$ is an exact cover. On the left, you have all the terms $z^m$ where $m$ satisfies $m\equiv a_s\pmod{n_s}$, with coefficient the number of such congruences $m$ satisfies. On the right, you have $z^n$ for every $n$, with coefficient $1$. So, the equality is saying that every non-negative integer satisfies exactly one congruence.

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Oh... I think I get it now. While we're at it, can you please explain the last equation/inequality as well? I don't understand why we take the limit $z\to e^{2\pi i/n_k}$. – Carolus Nov 15 '12 at 12:55
The point of the proof is that exactly one of the terms $z^{a_s}/(1-z^{n_s})$ has a pole at $z=e^{2\pi i/n_k}$, so the sum on $s$ must have a pole there --- but the sum on $s$ equals $1/(1-z)$, which manifestly has no pole there. – Gerry Myerson Nov 15 '12 at 22:23
I see now, thank you very much. – Carolus Nov 16 '12 at 7:33

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