# Cauchy geometric seq. proof

1. May 28, 2012

### jaqueh

1. The problem statement, all variables and given/known data
Suppose the sequence (Sn) is defined as:
|Sn+1-Sn|<2-n
show that this is a cauchy sequence

2. Relevant equations
hint: prove the polygon identity such that
d(Sn,Sm)≤d(Sn,Sn+1)+d(Sn+1,Sn+2)...+d(Sm-1,Sm)

3. The attempt at a solution
I have defined Sm and Sn and created the inequality that:
d(Sn,Sm)=|2-n-2-m|>||Sn+1-Sn|-|Sm+1-Sm||

2. May 28, 2012

### tt2348

Using the hint, you'd get |Sn-Sm|<|Sn-Sn+1|+...+|Sm-1-Sm|<2-n+...+2-m
Is there any type of cauchy sequence (perhaps of partial sums) that would fit the latter half?

3. May 28, 2012

### jaqueh

ok I think I get what I can do now with the polygon ineq. except I thought that it would only be less than 2-n+...+2m-1also I don't know if I need to prove the polygon identity or not. if I did how would I go about it? would I use a fact of geometric series?

4. May 29, 2012

### tt2348

Proving the "polygon" identity comes as a direct result of the definition of a metric, and the triangle inequality. A simple proof by induction would suffice.
While it is true you could remove the factor of 2-m from the inequality, you can add it on and still have the inequality hold true (plus it's better looking notation wise).
We know Ʃ2-n is a geometric series, and thus converges. Convergent sequences of partial sums (partial sums of a geometric series) are cauchy sequences. So Ʃ2-k from 1 to n minus Ʃ2-k from 1 to m would be less than ε