The polygon circumscribing constant is found by: $$ \prod _3 ^\infty \sec \left( \frac\pi n \right)$$

I am trying to find a proof that this product converges. I know it is equal to:

$$\exp \left( \sum _{3} ^{\infty} \ln\left(\sec\left(\frac\pi n\right)\right) \right)$$

So I just need to show that sum converges. I do not see an easy way to use any of the convergence tests. In particular, I spent way to long trying to integrate this function to no avail.

So what convergence test is useable in this case?

  • $\begingroup$ $\ln\sec x=O(x^2)$ as $x\to0$. $\endgroup$ Apr 26, 2019 at 16:27

2 Answers 2


Since $\log(\sec x)=\frac{x^2}2+o(x^3)$,$$\lim_{n\to\infty}\frac{\log\left(\sec\left(\frac\pi n\right)\right)}{\frac1{n^2}}=\frac{\pi^2}2.$$So, since the series $\sum_{n=1}^\infty\frac1{n^2}$ converges, your series converges too, by the comparison test.


This can be solved by the comparison test. As $\sum_{n=1}^\infty\frac1{n^2}$ converges, it only needs to be shown that $\lim_{n\to\infty}\frac{\log\left(\sec\left(\frac\pi n\right)\right)}{\frac1{n^2}}$ is finite and non-zero.

First, this limit can be simplified by letting $x= \frac{1}{n}$:

$$\lim_{n\to\infty}\frac{\log\left(\sec\left(\frac\pi n\right)\right)}{\frac1{n^2}}= \lim_{x\to 0}\frac{\log(\sec(\pi x))}{x^2}$$

Applying L'Hôpital's rule:

$$\lim_{x\to 0}\frac{\log(\sec(\pi x))}{x^2}= \lim_{x\to0}\frac{\frac{\pi \sec(\pi x)\tan(\pi x)}{\sec(\pi x)}}{2x}= \lim_{x\to0}\frac{\pi \tan(\pi x)}{2x}$$

Applying L'Hôpital's rule again:

$$\lim_{x\to0}\frac{\pi \tan(\pi x)}{2x}= \lim_{x\to0}\frac{\pi^2 \sec^2(\pi x)}{2}= \frac{\pi^2}{2}$$

This is finite and non-zero, therefore $\prod _3 ^\infty \sec \left( \frac\pi n \right)$ converges.


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