How do I prove the circumference of the Koch snowflake is divergent?

Question

How do I prove that the circumference of the Koch snowflake is divergent?

Let's say that the line in the first picture has a lenght of $3cm$. Since the middle part ($1cm$) gets replaced with a triangle with sidelenghts of $1cm$ each we can assume that the circumference increases by the $\frac{4}{3}$-fold.

I guess to calculate the circumference the following term should work, no?

$\lim\limits_{n \rightarrow \infty}{3cm\cdot\frac{4}{3}^n}$

I know that the limit of the circumference is divergent ( $+\infty$).

I also know that a divergent sequence is defined as :

But how do I prove syntactically and mathematically correct that the sequence diverges to $+\infty$ ?

Answer 1

You can use the binomial theorem to write

$$\begin{align*} \left(\frac43\right)^n&=\left(1+\frac13\right)^n=\sum_{k=0}^n\binom{n}k\left(\frac13\right)^k1^{n-k}\\\\ &=\sum_{k=0}^n\binom{n}k\left(\frac13\right)^k\tag{1}\\\\ &\ge\binom{n}0\left(\frac13\right)^0+\binom{n}1\left(\frac13\right)^1\\\\ &=1+\frac{n}3 \end{align*}$$

for $n\ge 1$, since all terms of $(1)$ are positive.

Answer 2

Hint:$$\frac{4^n}{3^n}=L\quad\Rightarrow\quad n=\log_{4/3}L$$