# Show seq. $x_n$ with $|x_{n+1} - x_n| < \epsilon$ is Cauchy

1. Dec 5, 2011

### Damidami

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

The problem is longer but the part I'm stuck is to show that $\{x_n\}$ is convergent (I thought showing it is Cauchy) if I know that for all $\epsilon > 0$ exists $n_0$ such that for all $n \geq n_0$ I have that
$|x_{n+1} - x_n| < \epsilon$

2. Relevant equations

A sequence is Cauchy if for all $\epsilon > 0$ and for all $n,m \geq n_0$ one has
$|x_m - x_n| < \epsilon$

3. The attempt at a solution

I called $m = n+p$ (for $p$ an arbitrary positive integer)
Then
$|x_m - x_n| = |x_{n+p} - x_n|$
But (and I think there is some mistake here):
$|x_{n+1} - x_n| < \epsilon/p$
$|x_{n+2} - x_{n+1}| < \epsilon/p$
$\vdots$
$|x_{n+p} - x_{n+p-1}| < \epsilon/p$

So
$|x_{n+p} - x_n| < \underbrace{|x_{n+1} - x_n|}_{< \epsilon/p} + \underbrace{|x_{n+2} - x_{n+1}|}_{< \epsilon/p} + \ldots + \underbrace{|x_{n+p} - x_{n+p-1}|}_{< \epsilon/p} < \epsilon$

Any help on why it's wrong (if it is) and how to solve it correctly?
Thanks!

2. Dec 5, 2011

### Office_Shredder

Staff Emeritus
Re: Show seq. $x_n$ with $|x_{n+1} - x_n| < \epsilon$ is Ca

This isn't true. For example the sequence
$$x_n = \sum_{i=1}^{n} 1/i$$

3. Dec 5, 2011

### Damidami

Re: Show seq. $x_n$ with $|x_{n+1} - x_n| < \epsilon$ is Ca

You are right, thanks.

I suppose I have to write the full problem: Given $\{x_n\}$ a sequence of real numbers, and $S_n = \Sigma_{n=1}^n |x_{k+1} - x_k|$, with $S_n$ bounded, prove that $\{ x_n \}$ converges.

My attempt at a proof:
Clearly $\{ S_n \}$ converges as it is a series of positive terms and it is bounded.
So I define $a_k = | x_{k+1} - x_k|$, and now I know that $\lim_{n \to \infty} a_n = 0$ (because the series $S_n$ converges).

From there I really didn't know how to continue, I thoght proving $x_n$ was Cauchy, but didn't work. Any help?