Finding the limit of the area of a Koch Snowflake

Question

this is my first question for this site and I made this account specifically for help with the following topic. I am doing a research presentation on the Koch Snowflake, specifically, the area. So far, I have been attempting to generalize a formula for finding the area of the snowflake at n iterations, and I am now trying to find the limit as n tends toward infinity.

So, basically, what is the limit for the following?:

$$\lim_{n\to\infty} \sum_{r=2}^{n} \frac{3 \cdot 4^{r-2}}{9^{r-1}} \cdot \frac{s^2 \sqrt{3}}{4} $$

Answer 1

$$\lim_{n\to\infty} \sum_{r=2}^{n} \frac{3 \cdot 4^{r-2}}{9^{r-1}} \cdot \frac{s^2 \sqrt{3}}{4}=\frac{s^2 \sqrt{3}}{4}\cdot \frac 13\lim_{n\to\infty} \sum_{k=0}^{n-2}\left(\frac 49\right)^k$$ Where $k=r-2$. Now you have a geometric series to sum.

Answer 2

The following forms a GP with $a=\frac{1}{9}$ and $r=\frac{4}{9}$. $$\begin{align} &\lim_{n\to\infty} \sum_{r=2}^{n} \frac{3 \cdot 4^{r-2}}{9^{r-1}} \cdot \frac{s^2 \sqrt{3}}{4}\\ =\ &\frac{ 3\sqrt{3}\cdot s^2}{4}\lim_{n\to\infty} \sum_{r=2}^{n} \frac{ 4^{r-2}}{9^{r-1}} \\ =\ &\frac{ 3\sqrt{3}\cdot s^2}{4}\cdot\frac{\frac{1}{9}}{1-\frac{4}{9}}\\ =\ &\frac{ 3\sqrt{3}\cdot s^2}{4}\cdot\frac{1}{5}\\ =\ &\frac{ 3\sqrt{3}\cdot s^2}{20} \end{align}\\ $$