Use the definition of a limit/triangle inequality to show divergence I just asked a question about this kind of stuff so I feel bad asking again, but I could use some help. This is a homework question that reads:
Use the definition of limit to prove that the sequence $(-1)^{(n-1)}$ diverges. Hint: Use the triangle inequality.
I do not really understand how the triangle inequality relates to divergence. I am not necessarily looking for a direct answer to a homework question, but a push in the right direction would be nice. Thanks!
 A: If you must use the triangle inequality, maybe you can do something like this:
Define $a_n=(-1)^{n-1}$ for $n\in\mathbb{N}$. Suppose $(a_n:n\in\mathbb{N})$ is converging with limit $a$. Let $1>\epsilon>0$ then there exists $N>0$ such that for all $n>N$, $|a_n-a|<\epsilon$ but since
$$
|a_{n+1}-a|\geq \big||a_{n+1}-a_n|-|a_n-a|\big| > 2 -\epsilon > \epsilon
$$
for all $n>N$ this leads to a contradiction from which we conclude that $(a_n)$ diverges.
A: I think you may say that the Limits of subsequences {1,1,1,...} and {-1,-1,-1,...} are not equal.
A: You need to show that for all $L$ there exists an $\epsilon>0$ such that for each $N>0$ there exists some $n \geq N$ such that $|(-1)^{(n-1)}-L|>\epsilon$. Thus $L$ is not the limit and so no limit exists.
The distance between $1$ and $-1$ is $2$. Picking an $\epsilon$ less than half of that should do the trick. 
Suppose the limit is $L$. Let $\epsilon=1/2$. Let $N>0$ (replace $N$ with the next integer up if you allow $N$ to be real). Then $2N$ is an even integer and $2N+1$ is odd and both are larger than $N$. $|(-1)^{2N-1}-L|=|-1-L|=|L+1|$ and $|(-1)^{2N+1-1}-L|=|1-L|=|L-1|$.
If $|L-1|<\epsilon=1/2$ then $1/2 < L < 3/2$. If $|L+1|<\epsilon=1/2$ then $-3/2<L<-1/2$. Since both cannot be true, $L$ cannot be the limit of our sequence. Therefore, no limit exists.
Not quite sure how the triangle inequality is suppose to help. I'd just ignore the hint.
