Use the definition of a limit/triangle inequality to show divergence

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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Yes, the hint is baffling. The sequence goes -1,+1,-1,+1,...it clearly diverges. – tkr Oct 24 '11 at 17:48
Yeah, I'm just not sure how to answer in a formal way. – Logan Serman Oct 24 '11 at 17:49

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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Yes. $\limsup\limits_{n\to\infty}(-1)^n=1$ and $\liminf\limits_{n\to\infty}(-1)^n=-1$. A limit exists if and only if $\limsup$ and $\liminf$ exist and are equal. – robjohn Oct 24 '11 at 18:07
These things are true. But they certainly wouldn't qualify as "using the definition" to prove the limit does not exist. – Bill Cook Oct 24 '11 at 18:14
@BillCook To prove that limit of subsequence {1,1,...} is 1, one needs to use that definition of limit that's what I thought. – Ramana Venkata Oct 24 '11 at 18:21
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. – Bill Cook Oct 24 '11 at 18:26

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.

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