Perimeter and area of the Koch snowflake in the limit

Start with an equilateral triangle of side $s$. At each step, on the middle third of every edge erect an outward equilateral triangle whose side is one-third of that edge, and remove the middle third. Repeating forever produces the Koch snowflake. What happens to the perimeter, and what is the limiting enclosed area?

The perimeter is infinite; the area tends to $\tfrac85$ of the original triangle, i.e. $\tfrac{2\sqrt3}{5}s^2$. Each step multiplies the perimeter by $\tfrac43$, so it diverges. Step $k$ adds $3\cdot4^{k-1}$ triangles each with area $9^{-k}$ of the original $A_0=\tfrac{\sqrt3}{4}s^2$, so the total added is $\tfrac{A_0}{3}\sum_{k\ge0}(4/9)^k=\tfrac35A_0$.