Using the sum of a geometric series with ratio ( r = \frac{4}{9} < 1 ), as ( n \to \infty ): [ A_{\infty} = A_0 \left[ 1 + \frac{1}{3} \times \frac{1}{1 - \frac{4}{9}} \right] = A_0 \left[ 1 + \frac{1}{3} \times \frac{9}{5} \right] = A_0 \left[ 1 + \frac{3}{5} \right] = \frac{8}{5} A_0 ]
Since ( A_0 = \frac{\sqrt{3}}{4} ), the final area is: [ A_{\infty} = \frac{8}{5} \cdot \frac{\sqrt{3}}{4} = \frac{2\sqrt{3}}{5} ] snowflake by haese mathematics
For each side, remove the middle third and replace it with two segments of the same length (forming an equilateral "bump"). The number of sides increases. Using the sum of a geometric series with