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Using the law of cosines to find each side length and multiplying by the number of sides, I see a pattern, which should approach $~2\pi~$ as the number of sides grows to infinity.

How can $~2\pi~$ be calculated, when it's infinity multiplied by an infintesimal?

Using the law of cosines, the missing side of each perimeter section can be found, where $~x~$ is the number of sides the polygon has. In this, the distance from each vertex on the polygon to the center $~= 1~$ unit.

The formula for each polygon's perimeter based on its number of sides $(x)$ is as follows:

perimeter $= x * \sqrt{2 - 2 ~\cos\left(\frac{360}{x}\right)}~.$

So, what is $~2\pi~$ , using this infinite by infinitesimal product? Calculus anyone?

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Using Taylor series for large $x$

$$\cos \left(\frac{2 \pi }{x}\right)=1-\frac{2 \pi ^2}{x^2}+\frac{2 \pi ^4}{3 x^4}+O\left(\frac{1}{x^6}\right)$$ $$2-2\cos \left(\frac{2 \pi }{x}\right)=\frac{4 \pi ^2}{x^2}-\frac{4 \pi ^4}{3 x^4}+O\left(\frac{1}{x^6}\right)$$ $$\sqrt{2-2\cos \left(\frac{2 \pi }{x}\right)}=\frac{2 \pi }{x}-\frac{\pi ^3}{3 x^3}+O\left(\frac{1}{x^5}\right)$$ $$x\sqrt{2-2\cos \left(\frac{2 \pi }{x}\right)}=2 \pi -\frac{\pi ^3}{3 x^2}+O\left(\frac{1}{x^4}\right)$$

Use it for $x=6$; the exact perimeter is $6$ for sure and the above formula gives $2 \pi -\frac{\pi ^3}{108}\approx 5.99609$.

If you want a still better approximation, using Padé approximants, $$x\sqrt{2-2\cos \left(\frac{2 \pi }{x}\right)}=2 \pi -\frac{20 \pi ^3}{3 \left(20 x^2+\pi ^2\right)}$$ which, for $x=6$, would give $5.99997$.

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\begin{align*} \text{If}\;\,& f(x) = x \sqrt{ 2- 2\cos \bigl( {\small{\frac{2\pi}{x}}} \bigr) } \\[12pt] \text{then}\;&\lim_{x\to\infty}(f(x))^2\\[8pt] &= \lim_{x\to\infty} x^2 \left( 2- 2\cos \bigl( {\small{\frac{2\pi}{x}}} \bigr) \right) \\[8pt] &= 4\pi^2\lim_{x\to\infty} \frac { 2- 2\cos \bigl( \frac{2\pi}{x} \bigr) } { \left(\frac{2\pi}{x}\right)^2 } \\[8pt] &= 4\pi^2\lim_{t\to 0^+} \frac{2-2\cos(t)}{t^2} \\[8pt] &= 4\pi^2\lim_{t\to 0^+} \frac{\sin(t)}{t} \qquad\text{[by L'Hopital's Rule]} \\[8pt] &= (4\pi^2)\,(1) \\[8pt] &= 4\pi^2 \\[12pt] \text{hence}\;&\lim_{x\to\infty}f(x)=2\pi \\[4pt] \end{align*}

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