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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:

  • $\cos(2\theta) = 1-2\sin^{2}(\theta)$

  • So $\cos\frac{1}{n} = 1-2\sin^{2}\left(\frac{1}{2n}\right)$

  • 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.

ronnie.in
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  • At present your argument is wrong but you are close: to see the problem, note that $\log\cos(1/n)\lt0$ for every $n\geqslant1$ hence the upper bound you proved is trivial and does not help to show convergence. – Did Apr 29 '13 at 09:01
  • @BrianM.Scott Thanks – ronnie.in Apr 29 '13 at 09:03
  • @Did: What do you mean I am close. If I my upper bound is a trivial one, then I don't see how I am close :( – ronnie.in Apr 29 '13 at 09:04
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    @Did I think $\log(1+x) < x \implies \log(1-x) < -x$ and hence I can bound my series by $-\sum\frac{1}{2n^{2}}$ is this what you are trying to say. – ronnie.in Apr 29 '13 at 09:11
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    First, you wrote that $\log(u)\lt v$ for some $u\lt1$ and $v\gt0$. This is true and irrelevant. Second, using $\log(1-x)\lt-x$ (true) indeed yields $\sum\limits_n\log\cos(1/n)\lt-\sum\limits_n1/(2n^2)$ (also true)... and then what? This neither proves convergence nor divergence of $\sum\limits_n\log\cos(1/n)$. Note that $\log(n/(n+1))\lt-1/(2n^2)$ as well (I think) but $\sum\limits_n\log(n/(n+1))$ diverges. – Did Apr 29 '13 at 10:05
  • @Did: Did, doesn't $\sum_{n} \log\cos(1/n)<-\sum_{n}\frac{1}{2n^{2}}$ imply $\sum_{n}\log\cos\left(\frac{1}{n}\right)$ convergent. Since $\sum\frac{1}{n^2} \to \frac{\pi^{2}}{6}$. We can use comparison test here – ronnie.in Apr 29 '13 at 10:10
  • No, please see the last sentence of my last comment for a counterexample. Another counterexample: $\sum\limits_n-1/n\lt\sum\limits_n-1/n^2$ and... $\sum\limits_n-1/n$ is not convergent, is it? – Did Apr 29 '13 at 10:14
  • @Did Thanks for clarifying :) Then what else should I do :( – ronnie.in Apr 29 '13 at 10:17
  • @Did: But since $\sum -\frac{1}{2n^{2}} \to -\frac{\pi^2}{12}$ doesn't it imply $\sum \log\cos\frac{1}{n}$ converge. – ronnie.in Apr 29 '13 at 10:18
  • Find a lower bound $\log\cos(1/n)\gt-x_n$ with $x_n\gt0$ and $\sum\limits_nx_n$ finite. – Did Apr 29 '13 at 10:18
  • @Did: Because with my bound, i show that $\sum_{n} \log\cos\frac{1}{n} < -\frac{\pi^{2}}{12}$. – ronnie.in Apr 29 '13 at 10:19
  • doesn't it imply... No it doesn't. At all. In no way whatsoever. $\langle$Sigh$\rangle$... Please READ my previous comments. – Did Apr 29 '13 at 10:19
  • @Did: I read, but suddenly this thought came up. I understood why it can't work :) Sorry – ronnie.in Apr 29 '13 at 10:20

1 Answers1

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Apply the limit comparison test. What is $\displaystyle \lim_{n \to \infty} \frac{ \log \cos (1/n)}{1/n^2}$?

Umberto P.
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