# Convergence of a sequence of integers

1. Mar 9, 2010

### Fizz_Geek

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

Given a Cauchy sequence of integers, prove that the sequence is eventually constant.

2. Relevant Definitions and Theorems

Definition of Cauchy sequences and convergence
Monotone convergence
Every convergent sequence is bounded
Anything relevant to integers

3. The attempt at a solution

I can see why the theorem is true. I thought that, since the sequence in nonempty and bounded, the supremum and infimum of the set containing the sequence both exist and that the limit of the sequence must be a number between the infimum and the supremum. But I got stuck trying to prove that the limit is contained within that set (that the limit is also an integer). I don't know if it's my approach that's leading me to a dead end or if there's a theorm I've overlooked or something else entirely. Maybe I'm making it overly complicated? Any help would be appreciated.

2. Mar 9, 2010

### Tinyboss

Look at the definition of Cauchy sequence. Choose epsilon<1. Conclude.

3. Mar 9, 2010

### Mathnerdmo

Start from the definition of a Cauchy sequence.

It usually starts "For all epsilon greater than zero, [...]"
Can you choose an epsilon that gives your conclusion?

4. Mar 9, 2010

### Fizz_Geek

I'm afraid I don't see what directon either of you are going in. As far as I know, the definition of a Cauchy sequence allows me to make the distance between members of the sequence small--it doesn't make it zero, which is what I need for the sequence to be constant for large n.

5. Mar 9, 2010

### Staff: Mentor

The point Tinyboss and Mathnerdmo are making is that the sequence consists of integers.

6. Mar 9, 2010

### Fizz_Geek

And what does that mean for my proof?

7. Mar 9, 2010

### Staff: Mentor

It says that there is some minimum distance between elements in the sequence.

8. Mar 9, 2010

### Fizz_Geek

Oh! That's right! Thanks so much!