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this is my first question for this site and I made this account specifically for help with the following topic. I am doing a research presentation on the Koch Snowflake, specifically, the area. So far, I have been attempting to generalize a formula for finding the area of the snowflake at n iterations, and I am now trying to find the limit as n tends toward infinity.

So, basically, what is the limit for the following?:

$$\lim_{n\to\infty} \sum_{r=2}^{n} \frac{3 \cdot 4^{r-2}}{9^{r-1}} \cdot \frac{s^2 \sqrt{3}}{4} $$

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  • $\begingroup$ is your formula right? $\endgroup$
    – Dr. Sonnhard Graubner
    Jan 15, 2017 at 16:19
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    $\begingroup$ Hint. Can you find a geometric series in your expression? $\endgroup$
    – Ethan Bolker
    Jan 15, 2017 at 16:19

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$$\lim_{n\to\infty} \sum_{r=2}^{n} \frac{3 \cdot 4^{r-2}}{9^{r-1}} \cdot \frac{s^2 \sqrt{3}}{4}=\frac{s^2 \sqrt{3}}{4}\cdot \frac 13\lim_{n\to\infty} \sum_{k=0}^{n-2}\left(\frac 49\right)^k$$ Where $k=r-2$. Now you have a geometric series to sum.

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  • $\begingroup$ I see how you put the \frac{3\cdot\sqrt{3}}{4} on the left side of the limit, but would you mind explaining where the \frac{1}{3} came from? edit: I guess there is no LaTeX in comments, I apologize. $\endgroup$
    – Cade
    Jan 15, 2017 at 17:32
  • $\begingroup$ There was a $3$ in the numerator and one more $9$ than $4$ in the powers, so I pulled out $\frac 39$ You get MathJax in comments by using dollar signs, but it doesn't render as you are typing. I just do the best I can, then check how it renders after I post it and fix any problems. If you want to "sandbox" it, you can type the question as a new answer, verify that it renders as you want, then copy/paste into a comment box. $\endgroup$
    – Ross Millikan
    Jan 15, 2017 at 17:40
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The following forms a GP with $a=\frac{1}{9}$ and $r=\frac{4}{9}$. $$\begin{align} &\lim_{n\to\infty} \sum_{r=2}^{n} \frac{3 \cdot 4^{r-2}}{9^{r-1}} \cdot \frac{s^2 \sqrt{3}}{4}\\ =\ &\frac{ 3\sqrt{3}\cdot s^2}{4}\lim_{n\to\infty} \sum_{r=2}^{n} \frac{ 4^{r-2}}{9^{r-1}} \\ =\ &\frac{ 3\sqrt{3}\cdot s^2}{4}\cdot\frac{\frac{1}{9}}{1-\frac{4}{9}}\\ =\ &\frac{ 3\sqrt{3}\cdot s^2}{4}\cdot\frac{1}{5}\\ =\ &\frac{ 3\sqrt{3}\cdot s^2}{20} \end{align}\\ $$

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  • $\begingroup$ How can you tell that the formula is wrong? $\endgroup$
    – Cade
    Jan 15, 2017 at 17:26
  • $\begingroup$ The final expression should have been $\frac{2\sqrt3}{5}\cdot s^2$. If you need reference, check this out. $\endgroup$
    – 8hantanu
    Jan 15, 2017 at 19:47

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