Convergence of shapes Let $a_n$ be a sequence of shapes where:
$a_1$ is an equilateral triangle
$a_2$ is a square.
$a_3$ is a regular pentagon and in general $a_n$ is a regular $n-$gon(if we say or define that the equilateral triangle  is a regular $3-$gon the the $n$ term of the sequence will be a regular $(n+2)-$gon.)
Intuitively i believe that this sequence converges to a circle.
We can assume that all the terms of the sequence, i.e the shapes are inscribed in the same circle(which i believe will be the limit)
What methods of mathematics does this problem require to be proven formally.
I would like to know how can this be solved with analysis(using a kind of epsilon definition?.
At a first thought we can define an order relation in the set of the areas of the terms of  the sequence.
Excuse the pontential naivity of my question and i hope it is not a duplicate of another question.
I would appreciate if someone could help me.
Thank you in advance.
 A: First of all we need to define the "distance" between two figures. All figures are compact. Let
$$
d(\Phi_1,\Phi_2)=\max(\rho(\Phi_1,\Phi_2),\rho(\Phi_2,\Phi_1)),
$$
where $\rho(\Phi_1,\Phi_2)=\sup_{A\in \Phi_1}\inf_{B\in \Phi_2} |AB|$.


*

*It's easy to see that if $\rho(\Phi_1,\Phi_2)=0$ iff $\Phi_1\subset \Phi_2$. So, $d(\Phi_1,\Phi_2)\geqslant 0$ and $d(\Phi_1,\Phi_2)=0$ iff $\Phi_1=\Phi_2$.

*It's obvious that $d(\Phi_1,\Phi_2)=d(\Phi_2,\Phi_1)$.

*Uff.. The triangle inequality.
Lemma. $\rho(\Phi_1,\Phi_2)+\rho(\Phi_2,\Phi_3)\geqslant \rho(\Phi_1,\Phi_3)$
Proof. Let $\rho(\Phi_1,\Phi_2)=a$, $\rho(\Phi_2,\Phi_3)=b$, $A\in \Phi_1$. Since $a$ is $\sup$ there exist a point $B\in \Phi_2$ that $|AB|\leqslant a$. Since $b$ is $\sup$, there exist a point $C\in \Phi_3$ that $|BC|\leqslant b$. So $|AC|\leqslant |AB|+|BC|\leqslant a+b$.
Then $\inf_{X\in\Phi_3} |AX|\leqslant |AC|\leqslant a+b$ and $\rho(\Phi_1,\Phi_3)\leqslant a+b$. $\square$
$$
d(\Phi_1,\Phi_2)+d(\Phi_2,\Phi_3)\geqslant \rho(\Phi_1,\Phi_2)+\rho(\Phi_2,\Phi_3)\geqslant \rho(\Phi_1,\Phi_3)\\
d(\Phi_1,\Phi_2)+d(\Phi_2,\Phi_3)\geqslant \rho(\Phi_2,\Phi_1)+\rho(\Phi_3,\Phi_2)\geqslant \rho(\Phi_3,\Phi_1)\\
d(\Phi_1,\Phi_2)+d(\Phi_2,\Phi_3)\geqslant \max(\rho(\Phi_1,\Phi_3),\rho(\Phi_3,\Phi_1))=d(\Phi_1,\Phi_2).
$$
So, in your problem, $d(a_n,a)=R\left(1-\cos\frac{2\pi}{2n}\right)\to 0$.
