Calculate $\lim_{n\to\infty}\left(\prod_{i=3}^n\sec\frac{\pi}{i}\right)$

Question

This question arises from below:

Initially you are given a circle $C_2$ with radius $r=1$. You construct the regualr 3-gon (equilateral triangle) with its incircle as $C_2$, and its circumscribed circle $C_3$. Then use $C_3$ as incircle and repeat this process, so you can construct the regualr $n$-gon with its circumscribed circle $C_n$. Does the limit of the radius of $C_n$ exist as $n\to\infty$ ? If it exists, what is the value?

My attempt

I have worked out that the radius of $C_n$ is

$$r_n=\prod_{i=3}^n\sec\frac{\pi}{i}$$

Since $\lim_{n\to\infty}\sec\frac{\pi}{n}=1$, I concluded that $\lim_{n\to\infty}\frac{r_{n+1}}{r_n}=1$, and the limit of $r_n$ exists. But I cannot find its limiting value. MATHEMATICA says $r_{10^4}\approx8.69574$. Do you have any suggestions on how to proceed?

Thank you in advanced!

Generally speaking, $r_{n+1}/r_n \to 1$ doesn't imply the convergence of $r_n$. But in your case $\cos \frac{\pi}{n} \sim 1-\frac{\pi^2}{2n^2}$, therefore $r_n$ converges. — njguliyev, Sep 01 '13 at 10:45
@njguliyev why $r_{n+1}/r_n \to 1$ doesn't imply the convergence? Can you give a counterexample? — zytsang, Sep 01 '13 at 10:50
If you take $r_n=n$ then $1+\frac{1}{n} \to 1$ but $r_n \to \infty$. — njguliyev, Sep 01 '13 at 10:52

Raymond Manzoni · Accepted Answer · 2013-09-01T12:27:14.627

2

At the limit we get the inverse of the Kepler–Bouwkamp constant : $$\rho:=\prod_{i=3}^\infty\cos\frac{\pi}{i}\approx 0.1149420448532962$$

with these results from Bouwkamp : $$\rho=\frac 2{\pi}\prod_{m=1}^\infty\prod_{n=1}^\infty1-\frac 1{m^2(n+1/2)^2}=\frac 2{\pi}\exp\left[-\sum_{k=1}^\infty\frac{\zeta(2k)(\lambda(2k)-1)4^k}k\right]$$

$\qquad\qquad$( with $\lambda(x)=\left(1-2^{-x}\right)\zeta(x)$ )$\quad$ as found in Finch $6.3$ with more informations.

See too $(12)$ from Chamberland and Straub's paper 'On gamma quotients and inﬁnite products'.

edited Sep 01 '13 at 12:27

answered Sep 01 '13 at 12:00

Raymond Manzoni

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Thank you for your information! – zytsang Sep 01 '13 at 12:09
@YorkTsang: Glad it was helpful (I'll have to edit since the second formula was for the other product over $\sin$ : $\rho$). – Raymond Manzoni Sep 01 '13 at 12:13

score 1 · Answer 2 · answered Sep 01 '13 at 12:10

Let

$$P = \prod_{k=3}^{\infty} \cos{\frac{\pi}{k}}$$

We want $1/P$. Then

$$\log{P} = \sum_{k=3}^{\infty} \log{\cos{\frac{\pi}{k}}}$$

Use the fact that

$$\log{\cos{\frac{\pi}{k}}} = -\int_0^{\pi/k} dt \, \tan{t} = -\frac{\pi}{k} \int_0^1 du \, \tan{\frac{\pi u}{k}}$$

and get

$$\log{P} = -\pi \int_0^1 du \, \sum_{k=3}^{\infty} \frac{1}{k} \tan{\frac{\pi u}{k}}$$

I do not know of an analytical expression for the sum, but the expression is an integral over a smooth function of $u$ and may be approximated using standard numerical integration techniques.

Calculate $\lim_{n\to\infty}\left(\prod_{i=3}^n\sec\frac{\pi}{i}\right)$

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