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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.

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  • $\begingroup$ Yes, thank you. I've edited. And regarding "It is monotonic downward and bounded below by 0," I think that is referring to the sequence itself, and not its sum. The harmonic sequence has the same characteristic, but its sum diverges. $\endgroup$ Commented Jun 10, 2015 at 15:15
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    $\begingroup$ You are right that I was referring to the sequence, not the sum. $\endgroup$ Commented Jun 10, 2015 at 16:33

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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.

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  • $\begingroup$ Very nice. I suspected a connection to the harmonic series. I also suspect that the formula for (1) is quite complicated. But I may be surprised. The trick is to find (n+1) points that lie on the sides of a regular n-gon and also on a circle, but are spaced in equal intervals along the circle.. From there we can use your Wikipedia formulas to calculate the sides of the (n+1)-gon from its n-gon. Some specific cases are discussed at THIS LINK. $\endgroup$ Commented Jun 10, 2015 at 19:41
  • $\begingroup$ Link in previous comment is dead but there is a French-language version at mathafou.free.fr/pbg/sol118.html $\endgroup$ Commented Jul 14, 2022 at 2:33

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