The intuition becomes clear when we evaluate the series by bare hands. 

For $r\in\{1,\dotsc,A\}$ we have
\begin{align*}\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^{-r/A}}\frac{1}{A^{m/A}} \sum_{k=1}^A \frac{\mathrm{e}^{2\pi i (-r+m)k/A}}{Am}\\  &=\frac{1}{A^{1-r/A}}\sum_{k=1}^A\mathrm{e}^{-2\pi i kr/A}\sum_{m=1}^\infty\frac{\mathrm{e}^{2\pi ikm/A}}{A^{m/A}m}.\end{align*}
The inner sum on the right-hand side equals $-\log(1-\mathrm{e}^{2\pi ik/A} A^{-1/A})$, and dropping the $n=0$ term is easy.