I am working on one of the fractals and finding its convergent area.
$$\begin{align} S & = 1+3\left(\frac{1}{9}+4(\frac{1}{9^2})+4^2(\frac{1}{9^3})+...\right)\\ & = 1+3*\sum_{i=0}^{\infty} \left[\frac{4^i}{9^{i+1}}\right] \end{align}$$
I want to say that I am looking at a geometric series, but I'm not 100% sure.
It is a geometric series. Pull out a factor $\frac19$ from the sum and you should see that the general term under the sum is $\frac{4^i}{9^i}$.