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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!

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    Yeah, I'm just $n$ot sure how to answer in a formal way.2011-10-24

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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.

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I think you may say that the Limits of subsequences {1,1,1,...} and {-1,-1,-1,...} are not equal.

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    Of course everything you can prove about limits follows from the definition. But your statement doesn't show the limit does not exist directly. It's presupposing the theorem which says every convergent sequence has a unique limit.2011-10-24
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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. 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.