Koch snowflake versus $\pi=4$ The only proof I could find of the Koch snowflake having infinite perimeter was by calculating the perimeter $P_n$ after the $n$th iteration $$P_n = 3s\left(\frac{4}{3}\right)^n,$$
where $s$ is the length of each side of the original equilateral triangle, and then taking the limit as $n$ approaches infinity (which is obviously infinity).
I was satisfied with this proof until I remembered the infamous "proof" that $\pi$ is equal to $4$ (see the link below). The length of the limit curve (a circle) is $\pi$ which is not the limit of the perimeter of the zig-zaged curve after the $n$th iteration because that's always $4$. This explanation http://qntm.org/trollpi of the false proof states that "the limit of a sequence need not necessarily share any properties with the members of that sequence" and that "we've seen a sequence of curves of length 4, whose limit does not have length 4".
So how come I can use this argument for proving that the Koch snowflake has an infinite perimeter? Is one of these derivations wrong or am I just missing something?
 A: That is a very good question.
Perimeter is quite weird concept. With polygons, everything seems simple, but even if you wanted to determine the perimeter of circle, whatever that means, you run into bit of a trouble. Like Archimedes did, you could try approximating the curve by polygons, but that's not very satisfying. With more obscure shapes it's not very clear where we are heading to.
A bit easier concept is the arc length. What is the arc length of a circle? One could pick points on a circle. Then the length of the circle should be at least as big as the length of the resulting polygon, right? So you could ask: By doing this, how "long" polygon could you get? Everything works out nicely and the upper bound for those arc lengths of polygons will be $2\pi r$. Why? Well, one could prove that this definition leads to the normal integral definition for "nice" curves. Or do some approximation, essentially, it's all about
$$
\lim_{x \to 0}\frac{\sin(x)}{x} = 1.
$$
Now the arc length of the Koch snowflake kinda makes sense. How long polygonal curve could you get by joining points on the curve? Well, if you consider the corner points of the iterates of snowflake, they will stay on the curve till the end. So joining those points will be joining points on the curve. Since choosing points on $n$:th iterate we can get arbitrarily long polygons, there is no upper bound on the arc length of Koch snowflake so it has, in some sense, infinite arc length.
What goes wrong with the $\pi = 4$ proof? The points won't be fixed like in the snowflake, except for the points on the circle. We could join them and note that the arc length of the resulting curve, if it exists, is at least $2\pi r$. Sure.
