# Cauchy Sequence

## Homework Statement

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.

## Homework 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 $m, n > N$.

## 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
$1 = \frac{1}{x_1^2} > \frac{1}{x_2^2} > \frac{1}{x_3^2} > \dotsb.$

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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 :)

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

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still stuck

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LCKurtz
Homework Helper
Gold Member
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 $\lim_{x\rightarrow \infty} = L$?

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 $\lim_{x\rightarrow \infty} = L$?
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 $n$?

LCKurtz
Homework Helper
Gold Member
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 $n$?
It's easier than that. If the sequence has a limit, what happens if you take the limit of both sides of your recursion?

It's easier than that. If the sequence has a limit, what happens if you take the limit of both sides of your recursion?
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 $\lim_{n \to \infty} \frac{1}{x_n^2} = 0$?

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 $\lim_{n \to \infty} \frac{1}{x_n^2} = 0$?
But then doesn't this imply $\lim_{n \to \infty} x_n^2 = \infty$? completing the problem?

Matterwave