Convergence of circumcircles sequence

Question

I define a sequence as following:

A[0] = 0
A[1] = A[2] = 1
A[n] = A[n-1] / cos (pi/n)

My question is - does this sequence have a finite limit? If yes, what that is?

Where does this come from: I try to understand if there is a limit to how circumcircles can grow if I keep building on a regular n-gon step by step, each step increasing the count of sides of the n-gon by 1. So it goes like this:

For 0 we just assign radius as 0
For 1 and 2 there are no n-gons, we assign 1 (this is conventional)
For 3 we are looking on an excircle of a triangle. This is our A[3]
For 4 we are building a square around the circumcircles of a triangle (thus that circle will be an inscribed circle for that square). Then we build circumcircles for that square. This is our A[4]
...

My reasoning for the sequence goes like this:

I wrote a simple recursive function that calculates the circle radius. And it looks like it converges:

(outputting last 10 members)
9999998-th rad is: 8.7000323319286
9999999-th rad is: 8.700032331929

but I was unable to progress in proving that the sequence even converges, let alone to find the limit.

"Intuitively" I feel as there must be a finite limit because n-gon gets closer and closer to a circle eventually becoming its own circumcircles. But then again we all have this awkward situation with the harmonic sum which "feels like should converge"

Related (duplicate?): Calculate $\lim_{n\to\infty}\left(\prod_{i=3}^n\sec\frac{\pi}{i}\right)$ — Blue, Dec 11 '19 at 15:47
@Blue yes, this is exactly what I'm doing. Looks like there (and in linked articles) I can find all the answers. This should be closed as a duplicate I guess — Alma Do, Dec 11 '19 at 15:49
Then I'll vote to close your question as a duplicate. (Don't delete this question, as it has upvoted answers.) — Blue, Dec 11 '19 at 15:50

score 2 · Answer 1 · answered Dec 11 '19 at 15:38

2

Maybe this information will be helpful. The recurrent form can be easily reduced:

$$A_n=\frac{A_{n-1}}{\cos {\frac{\pi}{n-1}}}=\frac{A_{n-2}}{\cos {\frac{\pi}{n-1}} \cos {\frac{\pi}{n-2}}}= \dots = \frac{1}{\prod_{i=3}^{n-1} \cos{\frac{\pi}{i}}}$$

answered Dec 11 '19 at 15:38

Andronicus

3,436

score 1 · Answer 2 · answered Dec 11 '19 at 15:36

1

The sequence indeed has a finite limit: the Polygon circumscribing constant.

answered Dec 11 '19 at 15:36

Guus B

648

Is there some "neat" formula for it? – Alma Do Dec 11 '19 at 15:41
Not that I know of. But note that this is also the case for constants like $\pi$ and $e$, which can also be expressed as infinite sum/product of a sequence. – Guus B Dec 11 '19 at 15:52

Convergence of circumcircles sequence

2 Answers2