## Putting It Together: Sequences and Series

### Finding the Area of the Koch Snowflake

Define a sequence of figures $\left\{{F}_{n}\right\}$ recursively as follows in the figure below. Let ${F}_{0}$ be an equilateral triangle with sides of length $1$. For $n\ge 1$, let ${F}_{n}$ be the curve created by removing the middle third of each side of ${F}_{n - 1}$ and replacing it with an equilateral triangle pointing outward. The limiting figure as $n\to \infty$ is known as Koch’s snowflake.

1. Find the length ${L}_{n}$ of the perimeter of ${F}_{n}$. Evaluate $\underset{n\to \infty }{\text{lim}}{L}_{n}$ to find the length of the perimeter of Koch’s snowflake.
2. Find the area ${A}_{n}$ of figure ${F}_{n}$. Evaluate $\underset{n\to \infty }{\text{lim}}{A}_{n}$ to find the area of Koch’s snowflake.

### Solution

1. Let ${N}_{n}$ denote the number of sides of figure ${F}_{n}$. Since ${F}_{0}$ is a triangle, ${N}_{0}=3$. Let ${l}_{n}$ denote the length of each side of ${F}_{n}$. Since ${F}_{0}$ is an equilateral triangle with sides of length ${l}_{0}=1$, we now need to determine ${N}_{1}$ and ${l}_{1}$. Since ${F}_{1}$ is created by removing the middle third of each side and replacing that line segment with two line segments, for each side of ${F}_{0}$, we get four sides in ${F}_{1}$. Therefore, the number of sides for ${F}_{1}$ is

${N}_{1}=4\cdot 3$.

Since the length of each of these new line segments is $\frac{1}{3}$ the length of the line segments in ${F}_{0}$, the length of the line segments for ${F}_{1}$ is given by

${l}_{1}=\frac{1}{3}\cdot 1=\frac{1}{3}$.

Similarly, for ${F}_{2}$, since the middle third of each side of ${F}_{1}$ is removed and replaced with two line segments, the number of sides in ${F}_{2}$ is given by

${N}_{2}=4{N}_{1}=4\left(4\cdot 3\right)={4}^{2}\cdot 3$.

Since the length of each of these sides is $\frac{1}{3}$ the length of the sides of ${F}_{1}$, the length of each side of figure ${F}_{2}$ is given by

${l}_{2}=\frac{1}{3}\cdot {l}_{1}=\frac{1}{3}\cdot \frac{1}{3}={\left(\frac{1}{3}\right)}^{2}$.

More generally, since ${F}_{n}$ is created by removing the middle third of each side of ${F}_{n - 1}$ and replacing that line segment with two line segments of length $\frac{1}{3}{l}_{n - 1}$ in the shape of an equilateral triangle, we know that ${N}_{n}=4{N}_{n - 1}$ and ${l}_{n}=\frac{{l}_{n - 1}}{3}$. Therefore, the number of sides of figure ${F}_{n}$ is

${N}_{n}={4}^{n}\cdot 3$

and the length of each side is

${l}_{n}={\left(\frac{1}{3}\right)}^{n}$.

Therefore, to calculate the perimeter of ${F}_{n}$, we multiply the number of sides ${N}_{n}$ and the length of each side ${l}_{n}$. We conclude that the perimeter of ${F}_{n}$ is given by

${L}_{n}={N}_{n}\cdot {l}_{n}=3\cdot {\left(\frac{4}{3}\right)}^{n}$.

Therefore, the length of the perimeter of Koch’s snowflake is

$L=\underset{n\to \infty }{\text{lim}}{L}_{n}=\infty$.

2. Let ${T}_{n}$ denote the area of each new triangle created when forming ${F}_{n}$. For $n=0$, ${T}_{0}$ is the area of the original equilateral triangle. Therefore, ${T}_{0}={A}_{0}=\frac{\sqrt{3}}{4}$. For $n\ge 1$, since the lengths of the sides of the new triangle are $\frac{1}{3}$ the length of the sides of ${F}_{n - 1}$, we have

${T}_{n}={\left(\frac{1}{3}\right)}^{2}{T}_{n - 1}=\frac{1}{9}\cdot {T}_{n - 1}$.

Therefore, ${T}_{n}={\left(\frac{1}{9}\right)}^{n}\cdot \frac{\sqrt{3}}{4}$. Since a new triangle is formed on each side of ${F}_{n - 1}$,

$\begin{array}{cc}\hfill {A}_{n}& ={A}_{n - 1}+{N}_{n - 1}\cdot {T}_{n}\hfill \\ & ={A}_{n - 1}+\left(3\cdot {4}^{n - 1}\right)\cdot {\left(\frac{1}{9}\right)}^{n}\cdot \frac{\sqrt{3}}{4}\hfill \\ & ={A}_{n - 1}+\frac{3}{4}\cdot {\left(\frac{4}{9}\right)}^{n}\cdot \frac{\sqrt{3}}{4}.\hfill \end{array}$

Writing out the first few terms ${A}_{0},{A}_{1},{A}_{2}$, we see that

$\begin{array}{}\\ {A}_{0}=\frac{\sqrt{3}}{4}\hfill \\ {A}_{1}={A}_{0}+\frac{3}{4}\cdot \left(\frac{4}{9}\right)\cdot \frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{4}+\frac{3}{4}\cdot \left(\frac{4}{9}\right)\cdot \frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{4}\left[1+\frac{3}{4}\cdot \left(\frac{4}{9}\right)\right]\hfill \\ {A}_{2}={A}_{1}+\frac{3}{4}\cdot {\left(\frac{4}{9}\right)}^{2}\cdot \frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{4}\left[1+\frac{3}{4}\cdot \left(\frac{4}{9}\right)\right]+\frac{3}{4}\cdot {\left(\frac{4}{9}\right)}^{2}\cdot \frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{4}\left[1+\frac{3}{4}\cdot \left(\frac{4}{9}\right)+\frac{3}{4}\cdot {\left(\frac{4}{9}\right)}^{2}\right].\hfill \end{array}$

More generally,

${A}_{n}=\frac{\sqrt{3}}{4}\left[1+\frac{3}{4}\left(\frac{4}{9}+{\left(\frac{4}{9}\right)}^{2}+\cdots +{\left(\frac{4}{9}\right)}^{n}\right)\right]$.

Factoring $\frac{4}{9}$ out of each term inside the inner parentheses, we rewrite our expression as

${A}_{n}=\frac{\sqrt{3}}{4}\left[1+\frac{1}{3}\left(1+\frac{4}{9}+{\left(\frac{4}{9}\right)}^{2}+\cdots +{\left(\frac{4}{9}\right)}^{n - 1}\right)\right]$.

The expression $1+\left(\frac{4}{9}\right)+{\left(\frac{4}{9}\right)}^{2}+\cdots +{\left(\frac{4}{9}\right)}^{n - 1}$ is a geometric sum. As shown earlier, this sum satisfies

$1+\frac{4}{9}+{\left(\frac{4}{9}\right)}^{2}+\cdots +{\left(\frac{4}{9}\right)}^{n - 1}=\frac{1-{\left(\frac{4}{9}\right)}^{n}}{1-\left(\frac{4}{9}\right)}$.

Substituting this expression into the expression above and simplifying, we conclude that

$\begin{array}{cc}\hfill {A}_{n}& =\frac{\sqrt{3}}{4}\left[1+\frac{1}{3}\left(\frac{1-{\left(\frac{4}{9}\right)}^{n}}{1-\left(\frac{4}{9}\right)}\right)\right]\hfill \\ & =\frac{\sqrt{3}}{4}\left[\frac{8}{5}-\frac{3}{5}{\left(\frac{4}{9}\right)}^{n}\right].\hfill \end{array}$

Therefore, the area of Koch’s snowflake is

$A=\underset{n\to \infty }{\text{lim}}{A}_{n}=\frac{2\sqrt{3}}{5}$.

### Analysis

The Koch snowflake is interesting because it has finite area, yet infinite perimeter. Although at first this may seem impossible, recall that you have seen similar examples earlier in the text. For example, consider the region bounded by the curve $y=\frac{1}{{x}^{2}}$ and the $x$ -axis on the interval $\left[1,\infty \right)$. Since the improper integral

${\displaystyle\int }_{1}^{\infty }\frac{1}{{x}^{2}}dx$

converges, the area of this region is finite, even though the perimeter is infinite.