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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.

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

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.

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.

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

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.

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.