Talk:Cauchy Sequence is Bounded/Real Numbers

Why do you need to prove anything for less than $N$? The definition of a Cauchy sequence says nothing about the behaviour of a sequence for $m, n$ less than $N$, so there should be no need to address that case, surely? What am I missing? --prime mover (talk) 08:57, 26 August 2017 (EDT)

It's just a completeness thing, without taking for granted that unboundedness can only occur on the infinite end of the sequence. So just to construct an explicit bound for the absolute values and prove it bounded explicitly. — Lord_Farin (talk) 10:15, 26 August 2017 (EDT)

I have same opinion with Lord_Farin and I also think it is better to introduce $\epsilon$ rather than 1. Because by introducing $\epsilon$ we can show explicitly the tight bound for the given sequence. --Bltzmnn.k (talk) 11:51, 26 August 2017 (EDT)

The $\epsilon$ will make it unnecessarily complicated. If you have theorem in mind with more precise estimates that are needed somewhere or you think are interesting, you're welcome to add it, but it should not be the same page. In fact, and this is a general outstanding question on theorems in analysis, we need some kind of policy about how to present similar theorems with stronger estimates. For example, the Prime Number Theorem. It is worth starting a help page on that. --barto (talk) 13:30, 30 August 2017 (EDT)