# Prove a Cauchy Sequence using Geometric Sums

• mathkiddi

## Homework Statement

Let {x_n} be a sequence. and let r be a real number 0<r<1. Suppose |x_(n+1) - x_n|<=r|x_n -x_(n-1)| for all n>1. Prove that {x_n} is Cauchy and hence convergent.

## Homework Equations

if |r|<1 then the sequence $$\sum r^k$$ from k=0 to n converges to 1/(1-r)

## The Attempt at a Solution

If I let {x_n}=r^2+r^3+...+r^n &
{x_(n+1)}=r^3+r^4+...+r^(n+1) &
{x_(n-1)}=r+r^2+..._r^(n-1)
I can then take |x_(n+1) - x_n|=|r^(n+1)-r^2|
and |x_n -x_(n-1)|=|r^n-r|

If I plug these into my given inequality I have |r^(n+1)-r^2|<=r||r^n-r|

I can then say that since we know 0<r<1 and from our relevat equations we know $$\sumr^k$$ converges then we know {x_n} is Cauchy and hence converges since every convergent sequence of real numbers is Cauchy.

I am not sure.

Use this
|x_(n+1) - x_n|<=r|x_n -x_(n-1)|
to show that

|x_(n+1) - x_n| <= r^(n-1)|x_2 - x_1|
which can be done by induction

By the way,
You cannot say this:
x_n=r^2+r^3+...+r^n &
x_(n+1)=r^3+r^4+...+r^(n+1) &
x_(n-1)=r+r^2+..._r^(n-1)

x_n is an ARBITRARY sequence with the property that |x_(n+1) - x_n|<=r|x_n -x_(n-1)|.