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? 

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 $$\ln\left(\cos { \dfrac{1}{n} }\right)\underset{n\to{\infty}}{\sim}{-\dfrac{1}{2n^2}}.$$
