written one way, the limit doesn't exist. written another, the limit equals 0? Hi I'd greatly appreciate some help/clarification with this problem, I'm also open to suggestions about how to be more articulate when writing problems out in general.
On my way to saying that the series $\sum_{n=1}^{\infty}(-1)^{n+1}cos(\pi/n)$ diverges, 
I figured I would show that $lim_{n\rightarrow\infty}(-1)^{n+1}cos(\pi/n)\neq0$
I tried re-writing the expression $(-1)^{n+1}cos(\pi/n)$ as $cos(\pi/(2n-1))-cos(\pi/2n)$,
because I figured I could just show that $lim_{n\rightarrow\infty}cos(\pi/(2n-1))-cos(\pi/2n)$.
But that limit equals 0. 
I suppose I wonder 
(1) why If $\sum a_n=\sum b_n$, why can't we assume that $lim_{n\rightarrow\infty} a_n=lim_{n\rightarrow\infty} b_n $  ?
and 
(2) Maybe (1) does hold, and I am confused about the first limit? I think it equals $\pm1$. 

The condition $a_n\to 0$ is necessary for the convergence of the series $\sum_{n\ge 1}a_n$. Therefore, you series does not converge - $|(-1)^n \cos (\pi/n)|\to 1$ as $n\to \infty$.
As a consequence, if both series $\sum a_n$ and $\sum b_n$ converge, then $\lim a_n = \lim b_n=0.$
