Koch snowflake/Recursive/Description/Area convergence and length divergence/Example

The Koch snowflakes are given by the sequence of plane geometric shapes ${}K_{n}$, which are defined recursively in the following way: The starting object ${}K_{0}$ is an equilateral triangle. The object ${}K_{n+1}$ is obtained from ${}K_{n}$ by replacing in each edge of ${}K_{n}$ the third in the middle by the corresponding equilateral triangle showing outside.

Let ${}A_{n}$ denote the area and ${}L_{n}$ the length of the boundary of the ${}n$-th Koch snowflake. We want to show that the sequence ${}A_{n}$ converges and that the sequence ${}L_{n}$ diverges to ${}\infty$.

The number of edges of ${}K_{n}$ is ${}3\cdot 4^{n}$, since in each division step, one edge is replaced by four edges. Their length is ${}1/3$ of the length of a previous edge. Let ${}r$ denote the base length of the starting equilateral triangle. Then ${}K_{n}$ consists of ${}3\cdot 4^{n}$ edges of length ${}r{\left({\frac {1}{3}}\right)}^{n}$ and the length of all edges of ${}K_{n}$ together is

${}L_{n}=3\cdot 4^{n}r{\left({\frac {1}{3}}\right)}^{n}=3r{\left({\frac {4}{3}}\right)}^{n}\,.$

Because of ${}{\frac {4}{3}}>1$, this diverges to ${}\infty$.

When we turn from ${}K_{n}$ to ${}K_{n+1}$, there will be for every edge a new triangle whose side length is a third of the edge length. The area of an equilateral triangle with side length ${}s$ is ${}{\frac {\sqrt {3}}{4}}s^{2}$. So in the step from ${}K_{n}$ to ${}K_{n+1}$ there are ${}3\cdot 4^{n}$ triangles added with area ${}{\frac {\sqrt {3}}{4}}{\left({\frac {1}{3}}\right)}^{2(n+1)}r^{2}={\frac {\sqrt {3}}{4}}r^{2}{\left({\frac {1}{9}}\right)}^{n+1}$. The total area of ${}K_{n}$ is therefore

${}{\begin{aligned}&\,{\frac {\sqrt {3}}{4}}r^{2}{\left(1+3{\frac {1}{9}}+12{\left({\frac {1}{9}}\right)}^{2}+48{\left({\frac {1}{9}}\right)}^{3}+\cdots +3\cdot 4^{n-1}{\left({\frac {1}{9}}\right)}^{n}\right)}\\&={\frac {\sqrt {3}}{4}}r^{2}{\left(1+{\frac {3}{4}}{\left({\frac {4}{9}}\right)}^{1}+{\frac {3}{4}}{\left({\frac {4}{9}}\right)}^{2}+{\frac {3}{4}}{\left({\frac {4}{9}}\right)}^{3}+\cdots +{\frac {3}{4}}{\left({\frac {4}{9}}\right)}^{n}\right)}.\end{aligned}}$

If we forget the ${}1$ and the factor ${}{\frac {3}{4}}$, which does not change the convergence property, we get in the bracket a partial sum of the geometric series for ${}{\frac {4}{9}}$, and this converges.