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Prove that $$\sum_{n=1}^{\infty}\log \cos \left (\frac{1}{n}\right )$$ converges absolutely.

The answer here suggests to use the Limit Comparison Test but it works for $a_n \geq 0$ while $\ln(\cos (1/n))<0$. Also the limit given in the answer is $-\frac{1}{2}$ while the test gives results for positive limit values only. That post is $6$ years old so I didn't leave this as a comment.

Bernard
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ZSMJ
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    As you note, all the summands are negative, so apply your tests to $\sum-\log\cos(1/n)$, which has positive terms. – Gerry Myerson Jun 09 '19 at 13:11
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    @GerryMyerson Then the LCT gives $1/2$ and proves absolute convergence. I feel silly. – ZSMJ Jun 09 '19 at 13:20
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    You've learned something. Feel happy! – Gerry Myerson Jun 09 '19 at 13:23
  • By the way, a useful thing to convince yourself that the series indeed converges is to use Taylor approximations: $\log \cos \frac{1}{n} \approx \log \left(1-\frac{1}{2n^2} \right) \approx -\frac{1}{2n^2}$, and the latter one is well-known to converge (https://en.wikipedia.org/wiki/Basel_problem) – lisyarus Jun 09 '19 at 13:25

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As $\;0<\cos\frac 1n<1$, we have $\;\Bigl|\log\cos\frac1n\biggr|=\log\biggl(\frac 1{\cos\tfrac1n}\biggr)$.

Now $\cos\frac1n=1-\frac1{2n^2}+o\bigl(\frac1{n^2}\bigr)$, so $$\frac 1{\cos\tfrac1n}=\frac 1{1-\tfrac1{2n^2}+o\bigl(\frac1{n^2}\bigr)}=1+\frac1{2n^2}+o\bigl(\frac1{n^2}\bigr)\sim_\infty1+\frac1{2n^2},$$ and ultimately $$\Bigl|\log\cos\frac1n\biggr|\sim_\infty \log\Bigl(1+\frac1{2n^2}\Bigr)\sim_\infty \frac1{2n^2}, $$ which converges.

Bernard
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