coefficient $[q^n]\sum_{m\ge1}\frac{q^{\alpha m}}{1-q^{\beta m}}$ where $0<\alpha<\beta$ Problem:
I am looking for a finite-sum expression for the coefficient $c_n=c_n(\alpha,\beta)$, where
$$C(\alpha,\beta;q)=\sum_{m\ge1}\frac{q^{\alpha m}}{1-q^{\beta m}}=\sum_{n\ge1}c_n(\alpha,\beta)q^n,$$
with $\alpha,\beta\in\Bbb Z$, $0<\alpha<\beta$, and $|q|<1$.
Context: I know such $c_n$ exist, because $|q|<1$ ensures that $\frac{q^{\alpha m}}{1-q^{\beta m}}$ is analytic about $q=0$ for $m\ge1$.
I was motivated to attempt this because of the well-known
$$\begin{align}
\vartheta_3^2(q)&=1+4\sum_{m\ge1}\frac{q^m}{1+q^{2m}}\\
&=1+4\left(C(3,4;q)-C(1,4;q)\right),
\end{align}$$
which implies that $r_2(n)=4c_n(3,4)-4c_n(1,4)$ for $n\ge1$. I know that finite-sum expressions for $c_n(3,4)$ and $c_n(1,4)$, along with $c_n(\alpha,\beta)$ exist because the following (more general) result is true:
$$A(a,b;q)=\sum_{m\ge1}\frac{(aq)^m}{1-bq^m}=\sum_{n\ge1}q^n\sum_{d|n}a^db^{n/d-1}.\tag1$$
My Attempts:
Let $i(n,m)=1$ when $n|m$ and $i(n,m)=0$ otherwise, and note that for positive integer $v$,
$$\frac{x^v}{1-x^v}=\sum_{r\ge v}x^ri(v,r).$$
Then let $u=\beta-\alpha$ so that
$$\begin{align}
C(q)&=\sum_{m\ge1}\frac{q^{\alpha m}}{1-q^{\beta m}}\\
&=\sum_{m\ge1}q^{-um}\frac{q^{\beta m}}{1-q^{\beta m}}\\
&=\sum_{m\ge1}q^{-um}\sum_{r\ge\beta}q^{mr}i(\beta,r)\\
&=\sum_{r\ge\beta}i(\beta,r)\sum_{m\ge1}q^{(r-u)m}\\
&=\sum_{t\ge\alpha}i(\beta,t+u)\sum_{m\ge1}q^{tm}\\
&=\sum_{t\ge\alpha}i(\beta,t+u)\sum_{s\ge t}q^{s}i(t,s)\\
&=\sum_{t\ge\alpha}\sum_{s\ge t}q^{s}i(\beta,t+u)i(t,s)\\
&=\sum_{t\ge\alpha}\sum_{j\ge 0}q^{j+t}i(\beta,t+u)i(t,j+t)\\
&=\sum_{j\ge0}\sum_{t\ge \alpha}q^{j+t}i(\beta,t+u)i(t,j+t)\\
&=\sum_{j\ge0}\sum_{l\ge 0}q^{j+l+\alpha}i(\beta,l+\beta)i(l+\alpha,j+l+\alpha).
\end{align}$$
Then write $n=j+l+\alpha$. Since $C(0)=0$ we know that $n\ge1$. Thus
$$\begin{align}
C(q)&=\sum_{n\ge1}\sum_{\,\,\,j,l\ge0\\ j+l=n-\alpha}q^{j+l+\alpha}i(\beta,l+\beta)i(l+\alpha,j+l+\alpha)\\
&=\sum_{n\ge1}\sum_{\,\,\,j,l\ge0\\ j+l=n-\alpha}q^{n}i(\beta,l+\beta)i(l+\alpha,n)\\
&=\sum_{n\ge1}q^n\sum_{\,\,\,j,l\ge0\\ j+l=n-\alpha}i(\beta,l+\beta)i(l+\alpha,n).
\end{align}$$
Since $j,l\ge0$ and $j+l=n-\alpha$, we have $0\le l\le n-\alpha$. Then upon writing $d=l+\alpha$, we have
$$\begin{align}
C(q)&=\sum_{n\ge1}q^n\sum_{l=0}^{n-\alpha}i(\beta,l+\beta)i(l+\alpha,n)\\
&=\sum_{n\ge1}q^n\sum_{d=\alpha}^{n}i(\beta,d+u)i(d,n)\\
&=\sum_{n\ge1}q^n\sum_{d|n,\,d\ge\alpha}i(\beta,d+u).\tag{*}
\end{align}$$
This would imply that $$c_n(\alpha,\beta)=\sum_{d|n,\,d\ge\alpha}i(\beta,d+u).$$

Question:
Is $(*)$ correct? If not, what is the expression for $c_n$ that I'm looking for?

Thanks :)
 A: Yes, that's fine. I think your arrangement is a bit overcomplicated and less intuitive than it needs to be. Here's a more direct approach that fills in a little gap.
For $|q| < 1$, we have
$$\left|\frac{q^{\alpha m}}{1-q^{\beta m}}\right| \leq \frac{1}{1-|q|^{\beta m}} \leq \frac{1}{(1-|q|)^{bm}}.$$
By comparison with a geometric series, it follows from the Weierstrass M-test that indeed $\sum_{m \geq 1} \frac{q^{\alpha m}}{1-q^{\beta m}}$ converges uniformly on compact subsets of $|q|<1$, and your limit is indeed analytic on $|q|<1$. By the Cauchy integral formula and the geometric series,
\begin{align*}
c_n(\alpha, \beta)
&:= [q^n] \sum_{m \geq 1} \frac{q^{\alpha m}}{1-q^{\beta m}} \\
&= \sum_{m \geq 1} [q^n] \frac{q^{\alpha m}}{1-q^{\beta m}} \\
&= \sum_{m \geq 1} [q^n] (q^{\alpha m} + q^{\alpha m + \beta m} + q^{\alpha m + 2\beta m} + \cdots) \\
&= \#\{m \geq 1 : \exists k \geq 0 \text{ s.t. }n=\alpha m + \beta m k\}\\
&= \#\{m \geq 1 : m \mid n \text{ and }\beta \mid n/m - \alpha \geq 0\}\\
&= \#\{d \mid n : \beta \mid d - \alpha \geq 0\}.\\
\end{align*}
