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 $\pm1$. 

(this first part just added)
Here's as general as I can make it:
Let $(a_n)_{n=0}^{\infty}$
be a sequence of reals
(or complex numbers)
such that
$L = \lim a_n$
exists
and
$\sum_{n=0}^{\infty} |a_n-L|
$ converges.
Let $(c_k)_{k=0}^{m-1}$
be real or complex numbers
such that
$\sum_{k=0}^{m-1} c_k
= 0
$
and
$\sum_{k=0}^j c_k
\ne 0
$
for
$0 \le j < m-1$.
Then
the sum
$\sum_{k=0}^{\infty} 
a_k c_{k\bmod m}
$
can have $m$
distinct values,
each gotten
by first summing the
first $j$ terms of the sum
for $j = 1$ to $m$,
and the remaining terms of the sum
being grouped in sets
of $m$ terms.
That is,
for $j = 1$ to $m$,
$S_j
=\sum_{k=0}^{j-1} a_k c_{k}
+\sum_{n=0}^{\infty}\sum_{k=0}^{m-1} a_{mn+j+k} c_{(k+j)\bmod m}
$
exists 
and is a possible value for
$\sum_{k=0}^{\infty} 
a_k c_{k\bmod m}
$.
This is done 
by showing that,
if
$b_{n, j}
=\sum_{k=0}^{m-1} a_{mn+j+k} c_{(k+j)\bmod m}
$,
then the limit conditions on
the $a_n$
and the fact that
the sum of the
$c_m$ is zero
implies that
$\sum_{n=0}^{\infty} b_{n,j}$
converges.
(now back to the original answer)
If $\lim a_n =L$ exists and is
non-zero,
$\sum_{n \ge 1} (-1)^n a_n$
can be written in two ways:
$\sum_{n \ge 1}(- a_{2n-1}+a_{2n})$
and
$-a_1+\sum_{n \ge 1}(- a_{2n+1}+a_{2n})$.
If those last two sums exist,
there may be two possible values
for 
$\sum_{n \ge 1} (-1)^n a_n$.
An example is
$a_n = 1-\frac1{n}$.
(added)
An even simpler example
is the traditional
$a_n = 1$,
where the two sums
are always zero,
so the possible answers
are $0$ and $1$.
I believe
there is an article in a recent MAA
magazine about this.
