1

Consider an initial square of side length $s$. Construct an infinite sequence of decreasing squares:

ii

For each square, the radius of the incircle ($r_{\text{incircle}}$) and the circumcircle ($r_{\text{circumcircle}}$) can be calculated as follows:

  • For the first square: $r_{\text{incircle1}} = \frac{s}{2}$, $r_{\text{circumcircle1}} = \frac{s}{\sqrt{2}}$
  • For the second square: $r_{\text{incircle2}} = \frac{s}{2\sqrt{2}}$, $r_{\text{circumcircle2}} = \frac{s}{2}$
  • For the third square: $r_{\text{incircle3}} = \frac{s}{4}$, $r_{\text{circumcircle3}} = \frac{s}{2\sqrt{2}}$

The area of each annulus is then the difference between the area of the circumcircle and the area of the incircle:

  • $A_{\text{annulus1}} = \pi (r_{\text{circumcircle1}}^2) - \pi (r_{\text{incircle1}}^2) = \frac{\pi s^2}{4}$
  • $A_{\text{annulus2}} = \pi (r_{\text{circumcircle2}}^2) - \pi (r_{\text{incircle2}}^2) = \frac{\pi s^2}{8}$
  • $A_{\text{annulus3}} = \pi (r_{\text{circumcircle3}}^2) - \pi (r_{\text{incircle3}}^2) = \frac{\pi s^2}{16}$

The total area is

\begin{align*} A_{\text{total}} &= A_{\text{annulus1}} + A_{\text{annulus2}} + A_{\text{annulus3}} + \ldots\\ &= \frac{\pi s^2}{4} + \frac{\pi s^2}{8} + \frac{\pi s^2}{16} + \ldots\\ &= \frac{\pi s^2}{4}\cdot\sum_{n=0}^{\infty} \left(\frac{1}{2}\right)^n \end{align*}

The area of the initial circumcircle

$$A_{\text{circle}} = \pi \cdot r_{\text{circumcircle1}}^2 = \frac{\pi s^2}{2}$$

So

$$A_{\text{total}} = A_{\text{circle}}$$

$$\frac{\pi s^2}{4}\cdot\sum_{n=0}^{\infty} \left(\frac{1}{2}\right)^n = \frac{\pi s^2}{2}$$

Thus

$$\sum_{n=0}^{\infty} \left(\frac{1}{2}\right)^n=2$$

Question

Is that a correct proof?

Reference: Why is $\sum_{n=0}^{\infty }\left ( \frac{1}{2} \right )^{n}= 2$?

vengy
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    Depends on what you consider more fundamental. For instance, the modern route of making the concept of "area" rigorous uses measure theory. And at some point in developing measure theory, the geometric series formula already gets proven by some other means and used extensively later on. So in that framework your proof is at best circular. On the other hand, if you start from synthetic geometry like the ancients did and found all of calculus on that, your proof maybe fine. – balddraz May 31 '23 at 18:58
  • @0XLR since we're just working with circles here you could probably get a more elementary proof if you use Jordan measure as your definition of area (as opposed to Lebesgue). – csch2 May 31 '23 at 19:08

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