Let $r_{n-2}$ be the the inradius and $r_{n-1}$ the circumradius of a regular n-gon. From this (1) and this (14) we get:
$$r_{n-1} = \frac{r_{n-2}}{\cos(\frac{180}{n})}$$
Basically what I'm doing is starting with the incircle of a regular triangle, draw the circumcircle and let that be the incircle of a square. Then draw the circumcircle of that square and let that be the incircle of a pentagon. When we follow these steps to infinity, does the "last" circumcircle drawn have a finite radius?
With each step the circumradius of the last drawn polygon grows by $r_{n-1} - r_{n-2}$. So to get an answer we could also ask if the following series converges:
$$\sum_{n=3}^\infty r_{n-1}-r_{n-2}$$
Basically you are asking whether $\lim_{n\to \infty} r_n$ exists which is equivalent to asking whether $$A=\frac{1}{\lim_{n\to \infty}\prod_{k=3}^n \cos(\pi/k) }$$ exists. Now, $$A=\lim_{n\to \infty}\frac{1}{\prod_{k=3}^n \left(1-2\sin^2(\pi/2k) \right)}$$ The sequence $\{a_n\}:=\{2\sin^2(\pi/2n)\}$ is upper bounded by $1$, monotone decreasing and convergent to $0$ $\left(\right.$Also the series $\sum_{n\ge 1}a_n$ is convergent since $\displaystyle\sum_{n\ge 1}a_n\le 2\sum_{n\ge 1}\frac{\pi^2}{4n^2}=\frac{\pi^4}{12}$$\left.\right)$. Since $a_n$ is decreasing, $\exists N$ such that $\forall n\ge N$, $a_n\le 1/m$ for any $m\in \mathbb{N}$. Thus, $$\prod_{n=N}^{N+m}(1-a_n)\ge (1-1/m)^m\implies \prod_{n=N}^{\infty}(1-a_n)\ge e^{-1}$$ Also, $a_n<1,\ \forall n\implies \prod_{n=1}^N(1-a_n)>0\ \forall N$. Thus, $\prod_{n\ge 3}(1-a_n)>0$ and hence $A=\lim_{n\to \infty}\frac{1}{\prod_{n\ge 3}(1-a_n)}$ is finite.