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This sum

$$\sum_{n=1}^{\infty} (-1)^n \ln\left(\cos \left( \frac{1}{n} \right)\right)$$

apparently converges absolutely, but I'm having trouble understanding how so.

First of all, doesn't it already fail the alternating series test? the $B_{n+1}$ term is greater than the $B_n$ term, correct?

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1  
@Kaish Why did you edit something to misspell absolutely, and remove absolute convergence from the title? –  Ishan Banerjee Apr 5 '13 at 13:10
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@Ishan: Sometimes one person starts editing before another edit is applied. I don't think he intentionally added the typo back to the text. In any case, the title is now a combination of the two previous titles... –  TMM Apr 5 '13 at 13:36
    
@IshanBanerjee I started editing, then when I'd already clicked to submit my edit, that's when I found out it had already been edited. I asked on META if I could change this but didn't get an answer quickly before my edit was accepted. –  Kaish Apr 5 '13 at 14:37
    
@Kaish Sorry about that. –  Ishan Banerjee Apr 5 '13 at 15:30

4 Answers 4

Just use an asymptotic expansion: $$ \ln (\cos(1/n)) = \ln(1-1/2n^2 + o(1/n^2)) \sim -\frac{1}{2n^2} $$


By the way, for $0 < x < \frac{\pi}{2}$, we have $$ \frac{d}{dx} \ln(\cos(x)) = -\frac{\sin x}{\cos x} < 0 $$ so that $-\ln(\cos(1/n))$ is decreasingly converging to $0$. Hence, the alternating series test does apply.

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Is there any other way to go about doing this? Is this function itself, Bn, decreasing? –  Billy Thompson Apr 5 '13 at 13:13
    
Why would it be useful? Te alternating series test does not provide absolute convergence. –  Siméon Apr 5 '13 at 13:18
    
I understand that, but doesn't absolute convergence apply conditional convergence also? Or can a function can be absolutely convergent, but not conditionally convergent? –  Billy Thompson Apr 5 '13 at 13:21
    
It is true that absolutely convergent $\implies$ convergent, but you should remember that a test is a sufficient condition but not necessary. –  Siméon Apr 5 '13 at 13:25

As the other answerers mentioned, $\ln(\cos(x)) \sim {x^2 \over 2}$ as $x \rightarrow 0$. This can stated formally and proved using L'hopital's rule: $$\lim_{x \rightarrow 0} {\ln(\cos(x)) \over x^2} = \lim_{x \rightarrow 0} {(\ln(\cos(x)))' \over (x^2)'}$$ $$= \lim_{x \rightarrow 0} -{\sin(x) \over 2 x\cos(x)} $$ $$ =\lim_{x \rightarrow 0} {\sin(x) \over x}\times\lim_{x \rightarrow 0} -{1 \over 2\cos(x)}$$ $$ = -{1 \over 2}$$ As a result, $$\lim_{n \rightarrow \infty} {|\log\cos({1 \over n})| \over {1 \over n^2}} = {1 \over 2}$$ So the series converges absolutely by the limit comparison test, since we know $\sum_{n=1}^{\infty} {1 \over n^2}$ converges.

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This is a good approach to explain the problem to a Calc II student +1! The issue with this, though, is that once you know the right series to compare it to, you really know the answer already. :) –  Mark McClure Apr 5 '13 at 15:09
    
I agree. So either the student would be suggested to use the ${1 \over n^2}$ series to compare with, or alternatively this can be considered to be a way of making a guess based on Taylor series rigorous. –  Zarrax Apr 5 '13 at 16:52

Note that

$$\ln(\cos(x)) = \ln(1-\frac{x^2}{2}+O(x^4)) = -\frac{x^2}{2} + O(x^4)$$

Thus, ignoring the $(-1)^n$, the general term looks like $1/n^2$ yielding convergence.

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Since $$1-\cos{x}\underset{x\to{0}}{\sim}{\dfrac{x^2}{2}}\;\; \Rightarrow \;\; \cos{\dfrac{1}{n}}={1-\dfrac{1}{2n^2}} +o\left(\dfrac{1}{n^2} \right),\;\; n\to\infty$$ and $$\ln(1+x)\underset{x\to{0}}{\sim}{x},$$ thus $$\ln\left(\cos { \dfrac{1}{n} }\right)\underset{n\to{\infty}}{\sim}{-\dfrac{1}{2n^2}}.$$

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