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

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

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

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

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

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

When we turn from ${\displaystyle {}K_{n}}$ to ${\displaystyle {}K_{n+1}}$, 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 ${\displaystyle {}s}$ is ${\displaystyle {}{\frac {\sqrt {3}}{4}}s^{2}}$. So in the step from ${\displaystyle {}K_{n}}$ to ${\displaystyle {}K_{n+1}}$ there are ${\displaystyle {}3\cdot 4^{n}}$ triangles added with area ${\displaystyle {}{\frac {\sqrt {3}}{4}}{\left({\frac {1}{3}}\right)}^{2(n+1)}r^{2}={\frac {\sqrt {3}}{4}}r^{2}{\left({\frac {1}{9}}\right)}^{n+1}}$. The total area of ${\displaystyle {}K_{n}}$ is therefore

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