Given let be such that .
Then for every (supose without loss of generality that m<n) we have that
what do you think?
Errrr WAIT something is not right with the last inequality
Let be a sequence such that for all . Prove that is a Cauchy sequence.
How would I exactly go about this? My professor really didn't go over this. All she defined was that a Cauchy sequence is such that given there exists such that if , then .
About how we got to we aplied the hypothesis for each term. How many terms we had? n-m terms.
What i mean is that every term has the form
About hte other issue, i am thinking maybe we need another bound there because i cant see how to continue form here.
Hmm I understand that there are terms, but I'm still not following how that inequality holds. For instance, if , we'd have
So we have that whole addition of those absolute values must be less than . But how is THAT less than ?
Edit: Nevermind, I see how each term must be less than or equal to which is added together times.
So I understand that, but yes, how does that lead to the inequality involving the epsilon?
I don't think so, but I am not sure how to actually show it (at least only based on the knowledge we have learned so far in class). Because then we'd have (if we assume WLOG that , , and in order for that to be Cauchy, that whole inequality must also be less than any given where exists and . But if that were true for ALL , then would have to be less than some number for ALL n,m, which doesn't look like the case...ugh, I have a vague idea but I have no idea how to formulate into coherent/logical/proof-worthy words.