# Homework Help: Cauchy sequence; I need some help

1. Sep 1, 2010

### ╔(σ_σ)╝

1. The problem statement, all variables and given/known data
Let $$x_{n}$$ be a Cauchy sequence. Suppose that for every $$\epsilon>0$$ there is $$n > \frac{1}{\epsilon}$$ such that $$|x_{n}| < \epsilon$$. Prove that $$x_{n} \rightarrow 0$$.

2. Relevant equations

3. The attempt at a solution

My problem with the question is I do not understand it.

if,

$$|x_{n}| < \epsilon$$ when $$n > \frac{1}{\epsilon}$$ ;

Doesn't that mean that $$Lim_{n\rightarrow infinity} x_{n} = 0 ?$$

In which case $$x_{n} \rightarrow 0$$ because if the limit as n goes to infinity is zero, then the terms can be arbitrarily brought close to zero as n gets large enough.

What exactly is the point of me knowing that $$x_{n}$$ is a Cauchy sequence ?
It doesn't seem like I even need to know that it is a Cauchy sequence. After-all, every convergent sequence is a cauchy sequence.

Am I missing something ?

2. Sep 1, 2010

### Office_Shredder

Staff Emeritus
You only know that there is one n such that $$|x_n|<\epsilon$$. You don't know that for example $$|x_{n+1}|<\epsilon$$

If we take away the Cauchy condition then the sequence (1,0,1,0,1,0,1,0,...) would fit the criterion

3. Sep 1, 2010

### ╔(σ_σ)╝

Sorry I don't follow.

Doesn't the inequality hold for every epsilon?

And given that n is not a specific number can't I just make n to be n+1 ?

And given your sequence, if $$x_{n} = 1$$ the inequality would not hold.

I am a bit lost. Please explain further.

4. Sep 1, 2010

### Office_Shredder

Staff Emeritus
Ok, let's continue with my example.

Let's pick $$\epsilon=\frac{1}{100,000}$$. Then we need to find n larger than 100,000 such that $$x_n<\epsilon$$. Well that's easy, n=100,002 works, since we know that's zero.

The point is that it doesn't say there exists N, such that for all n>N, |xn|<epsilon, it only says there exists ONE n for which |xn|<epsilon

Basically, the statement of the problem is that a SUBSEQUENCE of xn converges to zero

5. Sep 1, 2010

### ╔(σ_σ)╝

Okay I understand the problem now. I will think more about the problem and request more help if I need to. Thanks for now. I'll repost if i don't get anywhere.

6. Sep 2, 2010

### ╔(σ_σ)╝

Okay, I thought about the question and I am still not sure what to do.

Now that I know a sub-sequence in $$x_{n}$$ converges. So this means the following are true:

1) The Cauchy sequence converges since we have found a convergent subsequence.

2) The Cauchy sequence converges to 0 also , since the terms of the Cauchy sequence can brought arbitrarily close to each other. This is a Lemma in my book anyway; I assume I can use it.
3) since the Cauchy sequence converges to zero this means $$x_{n} \rightarrow 0$$

How exactly do I go about proving this mathematically ?

7. Sep 2, 2010

### ╔(σ_σ)╝

Can someone please point me in the right direction?!

8. Sep 2, 2010

### Office_Shredder

Staff Emeritus
You basically have the right idea. If you want to get the sequence within $$\epsilon$$, pick N so that the elements of your subsequence are within $$\frac{\epsilon}{2}$$ and the elements of your sequence are within $$\frac{\epsilon}{2}$$ of each other when the indices are larger than N and see what you can do

9. Sep 2, 2010

### ╔(σ_σ)╝

UThe elements of the sequence are within epsilon of what? 0? Are you suggesting some sort of triangle inequality ? Something like
$$|x_{n}-x_{m}|\leq |x_{n}|+|x_{m}|< \frac{ \epsilon_{0}}{2}} + \frac{\epsilon_{0}}{2}$$.
EDIT

I am not sure how to get the terms in the subsequence close to those of the sequence. Can you help me with this?

Last edited: Sep 2, 2010