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Suppose one is given the following visual proof that

$$\lim\limits_{n \to \infty} \sum_{k=1}^n \frac{1}{2^k} = 1$$

which is the following construction over $[0,1]\times[0,1]$

enter image description here

What this is suggesting is that the sum of the areas of the triangles, which is $1/2,1/4,1/8,\dots$ indeed covers the unitary square with area 1.

What I want to know if it is possible to give any sense to this construction, or if it is possible to give a geometric/analytic proof, by showing the sequence of points converges to $(1,1)$. The sequence is defined as, $x(P)$ and $y(P)$ being the $x$ and $y$ entry of the point $P$:

$$P_1 = (1,0)$$ $$P_2 = \left(\frac{x(P_1)+y(P_1)}{2},\frac{x(P_1)+y(P_1)}{2} \right)$$ $$P_3 = (1, x(P_2))$$ $$P_{2n} = \left(\frac{x(P_{2n-1})+y(P_{2n-1})}{2},\frac{x(P_{2n-1})+y(P_{2n-1})}{2} \right) $$ $$P_{2n+1} = \left(1,x(P_{2n}) \right) $$

this is to say, we take the arithmetic mean of the $x,y$ entries, and then move the point to the line $x=1$. The idea is that since the limit/fixed point is $1$ then it must be the case that the square is covered by the infinitely many triangles produced. I'm not sure what other mathematical notion is needed to interpret this, so I hope you can understand what I'm striving for and help me out.

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I don't understand, we can find a simple explicit expression for the coordinates of the $P_i$. – André Nicolas Apr 16 '12 at 17:56
@AndréNicolas What is it that you don't understand? – Pedro Tamaroff Apr 16 '12 at 17:59
Since one can find explicit expression, the limit of the $P_i$ is easy to compute. – André Nicolas Apr 16 '12 at 18:01
Maybe that would help ?'s_method – Student Apr 16 '12 at 18:06
@AndréNicolas What I want to know is what is needed to give sense to the proof. I have no formal interpretation of the fact the triangle "covers" the unit square, which I thing are pertinent to measures, topology, or something of the kind. – Pedro Tamaroff Apr 16 '12 at 18:06
up vote 6 down vote accepted

It's a two-dimensional visualization of the fact that $$\Bigl[0,{1\over2}\Bigl[\ \cup\ \Bigl[{1\over2},{3\over4}\Bigl[ \ \cup\ \Bigl[{3\over4},{7\over8}\Bigl[\ \cup\ \ldots=\ [0,1[$$ where the successive intervals have lengths ${1\over2}$, ${1\over4}$, $\ldots\ $. So it has to do with $\sigma$-additivity of measure. If you are uneasy with this it is simpler to consider the one-dimensional situation.

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Very nice ! (the best I ever seen ) – Khosrotash Mar 25 at 9:40

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