I'm studying a process where I am having to take a sum of integrals across parallelograms with various offsets based on sums of powers of $\frac{1}{3}$. So for instance, at Level $k=2$ of the process, I'm having to add up integrals whose bounds correspond to 4 different parallelograms that need to be offset by 0, $\frac{1}{3}$, $\frac{1}{9}$, and $\frac{1}{3}+\frac{1}{9}$, respectively. At Level $k=3$ of the process, there's 8 parallelograms that need to be offset by 0, $\frac{1}{3}$, $\frac{1}{9}$, $\frac{1}{27}$, $\frac{1}{3}+\frac{1}{9}$, $\frac{1}{3}+\frac{1}{27}$, $\frac{1}{9}+\frac{1}{27}$, and $\frac{1}{3}+\frac{1}{9}+\frac{1}{27}$, respectively, and the number of parallelograms keeps doubling like that as $k$ increases.
One way I've thought of notating this process would be to declare that for a set of numbers $T$, we can say $\displaystyle\frac{1}{3^T} := \sum_{t \in T} \frac{1}{3^t}$ (and also define $\frac{1}{3^{\emptyset}} := 0$). So if for example $T = \{1,4,8\}$, we would get that $\frac{1}{3^T} = \frac{1}{3^1}+\frac{1}{3^4}+\frac{1}{3^8}$. To describe the parallelograms from the beginning, I would use $\sum_{T \subseteq \{1,2,\dots,k\}} [\text{offsets using $\frac{1}{3^T}$}]$.
My ultimate goal is to let $k\rightarrow\infty$, but I'm not sure how to do that. I tried to turn the offsets into a double- or triple-sum, but it seems that a triple-sum like $\sum_{i=1}^n \sum_{j=1}^n \sum_{k=1}^n \text{(stuff)}$ can handle the (stuff) changing at most $n^3$ times. Is there a better way to allow for sums to take on the exponential $2^n$-type complexity that subsets allow? How can I get the behavior of a process like this when we go to $\infty$?