Show That \{x_n\} is Convergent: Step-by-Step Guide

[Damidami]

Homework Statement

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

Homework 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

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!

 

[Office_Shredder]

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

 

[Damidami]

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

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?