I want to know whether the series $\sum \log\left(\cos\frac{1}{n}\right)$ converges or diverges. I have made some attempts to solve this problem, and I work out them here:
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$\cos(2\theta) = 1-2\sin^{2}(\theta)$
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So $\cos\frac{1}{n} = 1-2\sin^{2}\left(\frac{1}{2n}\right)$
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Now I know that $\log(1-x) = -\left(x + \frac{x^{2}}{2} + \cdots \right)$
Hence what I have with me is $\log\left(1-2\sin^{2}\frac{1}{2n}\right)$ which is I think is less than $2\sin^{2}\frac{1}{2n}$. Hence the series $$\sum\log\left(\cos\frac{1}{n}\right) < 2 \cdot \sum \sin^{2}\frac{1}{2n} < \sum \frac{1}{2n^{2}}$$ and so the series converges. Is my argument correct.
Best Answer
Apply the limit comparison test. What is $\displaystyle \lim_{n \to \infty} \frac{ \log \cos (1/n)}{1/n^2}$?