The intuition becomes clear when we evaluate the series by bare hands.
For $r\in\{1,\dotsc,A\}$ we have $$\sum_{n=0}^\infty \frac{1}{A^n(An+r)}= \sum_{\substack{m\geq 1\\m\equiv r\pmod{A}}}\frac{1}{A^{(m-r)/A}m}=\frac{1}{A^{1-r/A}}\sum_{k=1}^A\exp^{-2\pi i kr/A}\sum_{m=1}^\infty\frac{\exp^{2\pi ikm/A}}{A^{m/A}m}.$$ The inner sum on the right-hand side equals $-\log(1-\exp^{2\pi ik/A} A^{-1/A})$, and dropping the $n=0$ term is easy.