A circle with radius $r_1=1$ is inscribed in an equilateral triangle. The triangle itself is inscribed in a larger circle and this larger circle is inscribed in a square. If we continue this process of alternatively circumscribing between a circle and a regular $n$-gon, theoretically, we will end up with a circle whose radius is $r_\infty$ and we cannot circumscribe that circle with any larger regular polygon (since regular $n$-gon will become a circle when $n$ tends toward infinity). Find the value of $r_\infty$.
I figure that because the relationship between the apothem $a$ and circumradius $r$ of a regular $n$-gon is: $r=\frac{a}{\cos\frac{\pi}{n}}$, I would have to calculate the infinite product: $$r_\infty=\prod_{n=3}^{\infty} \frac{1}{\cos(\frac{\pi}{n})}$$
Does this infinite product converge? If it is, what is the value of $r_\infty$ and how to compute it? Please help, thank you all in advance.
Since $\frac{1}{\cos(\frac{\pi}{n})} \geq 1$ the product $\prod_{n=3}^{\infty} \frac{1}{\cos(\frac{\pi}{n})}$ converges if and only if $\sum_{n=3}^{\infty}\left( \frac{1}{\cos(\frac{\pi}{n})}-1 \right)$ converges.
Now $$\lim_{n\to \infty} \frac{\frac{1}{\cos(\frac{\pi}{n})}-1}{\frac{1}{n^2}}= \lim_{n\to \infty} \frac{1}{\cos(\frac{\pi}{n})}\frac{1-\cos(\frac{\pi}{n})}{\frac{1}{n^2}}= \lim_{n\to \infty} \frac{1-\cos(\frac{\pi}{n})}{\frac{1}{n^2}}\\=\pi^2\lim_{n\to \infty} \frac{1-\cos(\frac{\pi}{n})}{\frac{\pi^2}{n^2}}=\pi^2\lim_{x \to 0} \frac{1-\cos(x)}{x^2}=\frac{\pi^2}{2}$$
Since $\sum_{n=3}^\infty \frac{1}{n^2}$ converges, by the lmit comparison theorem $\sum_{n=3}^{\infty}\left( \frac{1}{\cos(\frac{\pi}{n})}-1 \right)$ converges. Therefore, your product converges.