Find the sequence of partial sums for the series $a_n = (-1)^n$ Does this series converge? Find the sequence of partial sums for the series $$ \sum_{n=0}^\infty (-1)^n = 1 -1 + 1 -1 + 1 - \cdots$$
Does this series converge ?
My answer is that the sequence $= 0.5 + 0.5(-1)^n$. This makes a sequence that alternates between $1$ and $0$.
I know that the sequence does not converge since it is not monotone. But how can I prove this?

$$ \sum\limits_{n=0}^\infty (-1)^n = \lim\limits_{m\to\infty}\sum\limits_{n=0}^m (-1)^n $$
$$ = \lim\limits_{m\to\infty} \frac12\left((-1)^m + 1\right) $$
Note that the sequence 
$$a_m=\frac12\left((-1)^m + 1\right) $$ 
is convergent if
$$ \lim\limits_{m\to\infty} a_{2m} = \lim\limits_{m\to\infty} a_{2m+1} = L$$
However, in this case we have
$$ \lim\limits_{m\to\infty} \frac12\left((-1)^{2m}+1\right) =\frac22= 1 $$
And
$$ \lim\limits_{m\to\infty} \frac12\left((-1)^{2m+1}+1\right) =\frac02= 0 $$
Therefore $a_m$ is a divergent sequence.
