$$\prod_{n=3 }^{+\infty} \cos\frac{\pi}{n}$$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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Sign up to join this community$$\prod_{n=3 }^{+\infty} \cos\frac{\pi}{n}$$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.
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.
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.