The statement is:
Let $A=\{(a_s(n_s)\}^k_{s=1}$ be an exact cover of $\mathbb{Z}$ with $1
. 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
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} 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?