# Cauchy Sequence

1. Feb 9, 2012

### tylerc1991

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

Show that the sequence of real numbers defined by $x_{n + 1} = x_n + \frac{1}{x_n^2}, \, x_1 = 1$ is not a Cauchy sequence.

2. Relevant equations

A sequence $\{ p_n \}$ is Cauchy if and only if, for all $\varepsilon > 0$, there exists an $N > 0$ such that $d(p_n, p_m) < \varepsilon$ for all $m, n > N$.

3. The attempt at a solution

We can assume that $d$ is the usual metric on $\mathbb{R}$. I don't even see where to begin. I see that the sequence is monotonically increasing, so that
$1 = \frac{1}{x_1} > \frac{1}{x_2} > \frac{1}{x_3} > \dotsb.$
So
$1 = \frac{1}{x_1^2} > \frac{1}{x_2^2} > \frac{1}{x_3^2} > \dotsb.$
To me it looks like the sequence is in fact Cauchy. Please help!

2. Feb 9, 2012

### Kindayr

Well if you're trying to show that it is not Cauchy, state what it means for a sequence to not be Cauchy. That is where I would start :)

3. Feb 9, 2012

### tylerc1991

The sequence isn't Cauchy if there exists an $\varepsilon > 0$ such that for all $N > 0$ there exists $m, n > 0$ such that $d(x_m, x_n) \geq \varepsilon$.

...

still stuck

...

4. Feb 9, 2012

### LCKurtz

Do you have the theorem that a sequence is Cauchy if and only if it is convergent? And if so, what happens if you suppose $\lim_{x\rightarrow \infty} = L$?

5. Feb 9, 2012

### tylerc1991

Yes, I can assume the sequence is Cauchy if and only if it is convergent. The definition of limits that we are using states that $\lim_{n \to \infty} x_n = L$ if and only if
$\forall \varepsilon > 0 \, \exists N > 0 \, s.t. \, \forall n > N \quad d(L, x_n) < \varepsilon$.

Now can I somehow use the fact that $x_n$ is increasing to say that $d(L, x_n)$ is always increasing? And hence it is greater than or equal to $\varepsilon$ for some $n$?

6. Feb 9, 2012

### LCKurtz

It's easier than that. If the sequence has a limit, what happens if you take the limit of both sides of your recursion?

7. Feb 9, 2012

### tylerc1991

The limit of both sides of the recurrence should then equal the same thing, namely $L$. Then I would have that

$\lim_{n \to \infty} x_{n + 1} = \lim_{n \to \infty} x_n + \frac{1}{x_n^2} = L + \lim_{n \to \infty} \frac{1}{x_n^2} = L$.

Doesn't this simply prove that $\lim_{n \to \infty} \frac{1}{x_n^2} = 0$?

8. Feb 9, 2012

### tylerc1991

But then doesn't this imply $\lim_{n \to \infty} x_n^2 = \infty$? completing the problem?

9. Feb 9, 2012

### Matterwave

Can that second limit be 0 while the first limit exists? Is there a number L such that 1/L^2=0?

10. Feb 9, 2012

### tylerc1991

Exactly. Thank you LCKurtz for your patience! :)