Question:
Give an example of a divergent sequence that satisfies the following conditions:
(i) For every , there exists an such that for infinitely many ,
(ii) There exists an and a such that for all ,
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I have absolutely no idea how to do these For the infinitely many x>X one, I'm thinking this could involve a trig function since it doesn't converge and repeats every . Please help.
Thanks for the reply.
But if I had a sequence like that: for n is odd, and for n is even....
...how does this satisfy the criteria of being less than ? I mean, if a_n = 0, then you have |-3| = 3 < and what if epsilon was 0.6, or 2 etc.?
Sorry if I'm being stupid.
Hey, after spending the best part of an hour jotting stuff down on paper, I've come up with this for part (i)
is that a divergent sequence? I think it is. And it satisfies the conditions, right? I mean if you let X = -1, then are infinitely many x>X.
no idea about (ii) though Any ideas?
Indeed it does, thanks a lot mate!Originally Posted by Plato
Yep, I did. Sorry about that.
But if j is odd we have:
- but what if epsilon was 2? That's the question I was trying to ask. There doesn't seem to be a way around it.
EDIT: Nevermind I understand it now. I've still been misreading my question. I think I get it now. Thanks Plato and ThePerfectHacker