A constrained double summation

Question

This is a question that I asked on Math StackExchange (see here), but I believe it is better to ask if any number theorist has encountered it before.

Consider two positive integer $(k,l)$ and they are coprime, $(k,l) = 1$. Let $p, q$ be two real numbers satisfying $0 < p, q< 1$. Consider the double summation \begin{equation} {\sum_{n_1, n_2 = 0}^{+\infty}}'\, p^{n_1} q^{n_2}, \end{equation} where the dummy indices in the summand is constrained to be $ln_1 - n_2 \equiv 0 \mod k$. When $l =1$ evaluation is straightforward. How to calculate the sum for $l > 1$? Does it admit a closed form?

Answer 1

Consider the following sum for $|z|=1$: $$f(z)=\sum z^{ln_1-n_2}p^{n_1}q^{n_2}=\sum (pz^l)^{n_1}(qz^{-1})^{n_2}=\frac1{(1-pz^l)(1-qz^{-1})}.$$ We need only the terms with exponent of $z$ divisible by $k$, thus our sum equals $\frac1k\sum_{z:z^k=1} f(z)$.