Absolute convergence of the series $\sum\limits_{n=1}^{\infty} (-1)^n \ln\left(\cos \left( \frac{1}{n} \right)\right)$ 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? 
 A: 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.
A: 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.
A: 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.
A: 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 $$\ln\left(\cos { \dfrac{1}{n} }\right)\underset{n\to{\infty}}{\sim}{-\dfrac{1}{2n^2}}.$$
