# Convergent series proof

1. Nov 24, 2012

### bedi

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

Let x_i be integers. Prove that $\sum{x_i}$ converges iff x_i=0 for all i>I.

2. Relevant equations

3. The attempt at a solution

I need to show that the partial sums converge. That is, they are Cauchy. So for any $\epsilon >0$, $|s_n - s_m|<\epsilon$ holds.
Now we have
$\sum^n_{k=1}{x_i} - \sum^m_{k=1}{x_i} = \sum^n_{k=m}{x_i} < \epsilon$, for all n>m>N. Now assume $x_i \neq 0$ for all i. And as the partial sums are finite we can choose a minimal element from the sum S_n, say x_a. Obviously (n-m)x_a < S_n but by the Archimedean property $(n-m)x_a > \epsilon$ for some n. Thus either n=m, which implies that the sequence is stationary, or x_a=0. I think this is not valid but can't find the proper solution. help

2. Nov 24, 2012

### LCKurtz

If you are going to give a careful proof, it would be best to state what the problem is carefully:

Let x_i be integers. Prove that $\sum{x_i}$ converges iff there exists a number I such that x_i=0 for all i>I.
That isn't the correct denial of the statement $x_i=0$ for all $i>I$

Note that you have an "if and only if" statement to prove. You need to state what you are assuming and what you need to prove for both. I don't think you need the Cauchy criterion in either case. One case is trivial and the other perhaps you can show $x_i$ doesn't go to zero, which is necessary for convergence.

3. Nov 24, 2012

### bedi

Thank you :) I'm not allowed to use that last theorem (x_i goes to zero) because I'm not there in the book yet. I need to use some other tools, could you suggest something?

4. Nov 24, 2012

### LCKurtz

Well, I'm sure you can do a proof using the Cauchy criterion. But your argument seems to assume the $x_i\ge 0$. Is that given? It's hard to help you with your argument until you state whether you are working on the "if" or "only if" part, write down what you are assuming and what you have to prove. And you state the correct denial I mentioned earlier.