# Cauchy => bounded?

1. Jun 9, 2007

### pivoxa15

1. The problem statement, all variables and given/known data
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?

Last edited: Jun 9, 2007
2. Jun 9, 2007

### matt grime

Why do you think the sequences of partial sums are cauchy? They aren't.

I cannot begin to decipher what b/c d(..... means.

If S_n means the sum to n terms, what makes you think that given e>0, there is an n(e) such that for all n,m>n(e) we have |S_n - S_m| < e?

It is easy to show this is impossible, using the same idea as to show that the series itself diverges. Remember that

1+1/2+1/3+1/4+1/5+... > 1+(1/2+1/2)+(1/4+1/4+1/4+1/4)+(1/8+1/8+1/8+...) > 1+1+1+1+...

Last edited: Jun 9, 2007
3. Jun 9, 2007

### siddharth

Is b/c d(.., "because" , the "distance" ...?

As matt suggested, I think you should be less cryptic in your notation.

4. Jun 9, 2007

### AKG

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.

5. Jun 9, 2007

### pivoxa15

oh... I thought it was obvious. I actually used b/c in my exam recently to indicate 'because'. Although I used it in the middle of a sentence so hopefully might be less confusing than here.

6. Jun 10, 2007

### matt grime

If you'd've put a 'The' at the start of the sentence to actually make it into a sentence that might have helped. Anyway, have you understood the explanation of your mistake?

7. Jun 10, 2007

### pivoxa15

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.

8. Jun 10, 2007

### matt grime

it is perfectly possible to be strictly increasing and still bounded/cauchy

nope, there is nothing that states the terms have to decrease in magnitude. It is the difference between terms in the sequence that is important.