$\sum_{n=1}^\infty (-1)^n$ - I know this doesn't converge and I want to prove it. I am using the null sequence test as the limit as $n$ tends to $\infty$ of $(-1)^n$ doesn't equal 0 which I am trying to prove.
$1.$ First I thought I could let $a_n := (-1)^n$ and note that the two subsequences $a_{ 2k}$ and $a_{2k-1}$ have two distinct limits where $k$ is a natural number, them being $1$ and $-1$, so $a_n$ clearly diverges. Hence, if it diverges it doesn't converge to $0$, meaning the series diverges by null sequence test. $\square$
$2.$ Assume for sake of contradiction that $a_n = (-1)^n$ converges to a limit $l$ where $l$ is a real number. Let $\epsilon = 1$. Now, for all $\epsilon > 0$, there exists a natural number $N$ such that $|(-1)^n-l|<1$.
Note that when $n$ is even, $(-1)^n = 1$, so we have $|1-l| < 1$. When $n$ is odd, we have $|-1-l| < 1$. Now $2 = |1-l + 1+l| <= |1-l| + |1+l| = |1-l| + |-1-l| < 1 + 1 = 2$, so $2<2$ which is a contradiction. Hence, $a_n$ diverges so doesn't converge to $0$, meaning the series diverges by null sequence test. $\square$