Nested Regular Polygons I've created a problem that I do not know how to answer without a huge amount of effort.  If you have an elegant solution to it, please share!
Take an equilateral triangle of side length 1 and inscribe the largest square.  In that square, inscribe the largest regular pentagon.  In that pentagon, inscribe the largest regular hexagon.  Continue this pattern indefinitely, and record the sequence of side lengths of the inscribed polygons.  The sequence begins
$$1,\dfrac{\sqrt 3}{4},\dfrac12\sqrt{\dfrac32\left(4-\sqrt{2(5+\sqrt5)}\right)},\ldots\approx1,0.43301,0.27095,\ldots$$
(1) Find a formula for the general term of the sequence.
(2) Determine if the sum of the sequence converges or diverges.  If convergent, find the exact value of the sum.
 A: For 2, the sum diverges.  We will do a similar construction that reduces the side lengths and show the sum diverges.  We will inscribe a circle in the $n$-gon, then inscribe an $n+1$-gon in that circle.  At each stage, our side will be smaller than the original construction.  Let $s(n)$ be the side length of the polygon with $n$ sides, $R(n)$ the radius of the circumscribed circle of the polygon with $n$ sides, $r(n)$ the radius of the inscribed circle with $n$ sides.  Wikipedia gives the formula $$R(n)=\frac {s(n)}{2\sin \frac \pi n}=\frac {r(n)}{\cos \frac \pi n}$$ Then using the fact that $R(n+1)=r(n)$ 
$$s(n+1) = 2R(n+1)\sin \frac \pi{n+1}=2r(n)\sin \frac \pi{n+1}=s(n)\cos \frac \pi n\frac {\sin \frac \pi {n+1}}{\sin \frac \pi {n}}\\ \gt s(n)\cos \frac \pi n \frac {n}{n+1}$$ where the greater than comes from playing with the Taylor series for the sines.  We get $$s(n)\gt s(3)\frac 3n\prod_{k=3}^{n-1}\cos \frac \pi k$$  When we extend the product of the cosines to infinity, it converges to a finite value greater than $0.1149$ so we can say $$\sum_{n=3}^\infty s(n)\gt \sum_{n=3}^\infty s(3)\frac 3n 0.1149$$ which diverges because of the harmonic series.
