How can I show that $a_n$ = $(-1)^n$ is Cauchy?
I know that a definition of Cauchy is $\forall$ $\epsilon$ > 0, $\exists$ $N \in$ $\mathbb N$ such that if $x,y$ $\ge$ $N$, then $|a_x - a_y|$ < $\epsilon$.
This was my informal attempt.
We know that x,y can either be even or odd. If x and y are both even and greater than 0, then $|a_x - a_y|$ = 0, which is not less than 0. Therefore no such $N$ exists, and $a_n$ = $(-1)^n$ is Cauchy? is not Cauchy.
Let's prove that $\left \{ a_{n} \right \}_{n\geq 0}\ $ is NOT Cauchy:
To do this we must find an $\epsilon >0$ such that for all $N>0$ there are integers $m,n\ \geq N$ with the property that $\vert a_n-a_m\vert >\epsilon$.
So we set $\epsilon =1$ and let $N>0$ be given.
Now we just observe that if $n=N+1$ and $m=N,\ $then $\vert (-1)^{N+1}-(-1)^N\vert =2>\epsilon$ and we are done.