# Prove Cauchy sequence & find bounds on limit

1. Sep 11, 2005

### *melinda*

Here's the problem statement:

Prove that $x_1,x_2,x_3,...$ is a Cauchy sequence if it has the property that $|x_k-x_{k-1}|<10^{-k}$ for all $k=2,3,4,...$. If $x_1=2$, what are the bounds on the limit of the sequence?

Someone suggested that I use the triangle inequality as follows:

let $n=m+l$
$$|a_n-a_m|=|a_{m+l}-a_m|$$
$$|a_{m+l}-a_m|\leq |a_{m+l}-a_{m+l-1}|+|a_{m+l-1}-a_{m+l-2}|+...+|a_{m+1}-a_m|$$

Now by hypothesis, $|a_k-a_{k-1}|<10^{-k}$, so

$$|a_{m+l}-a_m|<10^{-(m+l)}+10^{-(m+l-1)}+...+10^{-(m+1)}$$.

It looks like we have an $\epsilon$ such that $|a_n-a_m|<\epsilon$. Before we get to the bounds on the limit, is that correct? Is anything missing?

2. Sep 12, 2005

### Tom Mattson

Staff Emeritus
You might take it a little further:

$$|a_{m+l}-a_m|<\sum_{i=0}^l 10^{m+l-i}$$

$$|a_n-a_m|<\sum_{i=0}^{n-m} 10^{n-i}$$

I'll let a real mathematician help you the rest of the way.

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