Cauchy geometric seq. proof

[jaqueh]

Homework Statement

Suppose the sequence (Sn) is defined as:
|S_n+1-S_n|<2^-n
show that this is a cauchy sequence

Homework Equations

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

The Attempt at a Solution

I have defined Sm and Sn and created the inequality that:
d(S_n,S_m)=|2^-n-2^-m|>||S_n+1-S_n|-|S_m+1-S_m||

 

[tt2348]

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

 

[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+...+2^m-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?

 

[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