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$$\prod_{n=3 }^{+\infty} \cos\frac{\pi}{n}$$is convergent or not?

Note that $$\lim_{n \to \infty} \ln{\cos\frac{\pi}{n}}=0,$$I try to turn it into $$\sum_{n=3}^{+\infty}\ln{\cos\frac{\pi}{n}} $$but have no idea how to do next.

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  • $\begingroup$ Hint: Compute $\lim_{x\rightarrow 0}\frac{\ln\left(\cos(\pi x)\right)}{x^2}$. $\endgroup$
    – Matthew H.
    Apr 27, 2022 at 3:57

2 Answers 2

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Alternative approach:

For $r \in \{3,4,5,\cdots\},$ let $a_r$ denote $\displaystyle \prod_{n=3}^r \left[\cos\left(\frac{\pi}{n}\right)\right].$

So, the question is whether the infinite sequence $\langle a_r\rangle$ is a convergent sequence.

The sequence is strictly decreasing, and bounded below by $0$. Therefore, the sequence must be convergent.

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let $a_n=\ln\cos\frac{\pi}{n}$, $a_n<0$.

hence $$\ln\cos\frac{\pi}{n}=\ln\left(1+\cos\frac{\pi}{n}-1\right)\\ \sim\cos\frac{\pi}{n}-1\sim-\frac{\pi^2}{2n^2}\quad(n\rightarrow \infty ) $$

since we have known $$\sum_{n=1}^{\infty} -\frac{\pi^2}{2n^2}$$is convergent. $$\sum_{n=3}^{+\infty}\ln{\cos\frac{\pi}{n}}$$is convergent.

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