The discussion revolves around the properties of Cauchy sequences and their relationship to boundedness, specifically examining the partial sums of the series (sigma,n->infinity)(1/n). The original poster questions the validity of their assertion that the sequence of partial sums is Cauchy despite the series diverging, leading to an unbounded sequence.
Some participants have provided clarifications on the definition of Cauchy sequences and pointed out potential misunderstandings in the original poster's reasoning. There is an ongoing exploration of the relationship between the terms of a sequence and the Cauchy condition.
Participants note the use of cryptic notation and the importance of clear communication in mathematical discussions. There is also a mention of the original poster's previous use of similar notation in an exam context.
Yes, that's not the definition of Cauchy. A sequence a(n) is Cauchy if for all E > 0, there exist N natural such that for all m and n > N, |a(n) - a(m)| < E. All you've shown is that for all E > 0, there exists some natural N such that |a(N) - a(N+1)| < E, or something like that.pivoxa15 said:Homework Statement
Thm: If a sequence is Cauchy than that sequence is bounded.
However Take the partial sums of the series (sigma,n->infinity)(1/n). The partial sums form a series which is Cauchy. But the series diverges so the sequence of partial sums is unbounded.
Sequence of partial sums is Cauchy b/c d(1/x,1/(x+1))=1/(k(k+1)) -> 0 as k->infinity
Have I done something wrong?
siddharth said:Is b/c d(.., "because" , the "distance" ...?
As matt suggested, I think you should be less cryptic in your notation.
pivoxa15 said:I think I have. Each term in the sequence is a series summed over a finite number of terms. However the larger the number of terms, the larger the sequence gets.
Very much unlike Cauchy seqences where a recquirement is that each term in the seqence decreases in magnitude the further out the sequence gets.