Cauchy Sequences: What it Means to be $|x_{n+1}-x_n|_p< \epsilon$

[evinda]

Hi! (Wave)

I am looking at the following exercise:

If $\{ x_n \}$ is a sequence of rationals, then this is a Cauchy sequence as for the p-norm, $| \cdot |_p$, if and only if :

$$\lim_{n \to +\infty} |x_{n+1}-x_n|_p=0$$

That's what I have tried:

$\lim_{n \to +\infty} |x_{n+1}-x_n|_p=0$ means that $\forall \epsilon>0, \exists n_0$ such that $\forall n \geq n_0$:

$$|x_{n+1}-x_n|_p< \epsilon$$

Is it right so far? (Thinking) How could I continue? (Thinking)

 

[Euge]

Hi! (Wave)

I am looking at the following exercise:

If $\{ x_n \}$ is a sequence of rationals, then this is a Cauchy sequence as for the p-norm, $| \cdot |_p$, if and only if :

$$\lim_{n \to +\infty} |x_{n+1}-x_n|_p=0$$

That's what I have tried:

$\lim_{n \to +\infty} |x_{n+1}-x_n|_p=0$ means that $\forall \epsilon>0, \exists n_0$ such that $\forall n \geq n_0$:

$$|x_{n+1}-x_n|_p< \epsilon$$

Is it right so far? (Thinking) How could I continue? (Thinking)

Yes, right so far. By the strong triangle inequality, we have that for all $m$ and $n$ with $m > n \ge n_0$,

\(\displaystyle |x_m - x_n|_p = |(x_m - x_{m - 1}) + (x_{m-1} - x_{m-2}) + \cdots + (x_{n+1} - x_n)|_p \le \max\{|x_m - x_{m-1}|_p, |x_{m-1} - x_{m-2}|_p,\ldots, |x_{n+1} - x_n|_p\} < \epsilon.\)

 

[evinda]

Yes, right so far. By the strong triangle inequality, we have that for all $m$ and $n$ with $m > n \ge n_0$,

\(\displaystyle |x_m - x_n|_p = |(x_m - x_{m - 1}) + (x_{m-1} - x_{m-2}) + \cdots + (x_{n+1} - x_n)|_p \le \max\{|x_m - x_{m-1}|_p, |x_{m-1} - x_{m-2}|_p,\ldots, |x_{n+1} - x_n|_p\} < \epsilon.\)

So, from this we conclude that $\{ x_n \}$ is a Cauchy sequence, right? (Thinking)

In order to show the other direction, we suppose that $\{ x_n \}$ is a Cauchy sequence, that means that $\forall \epsilon>0 \exists n_0 \geq 0$, such that $\forall m>n \geq n_0: \ |x_m-x_n|< \epsilon$, right? (Thinking)
If I am right, how could we continue? (Thinking)

 

[Euge]

So, from this we conclude that $\{ x_n \}$ is a Cauchy sequence, right? (Thinking)

In order to show the other direction, we suppose that $\{ x_n \}$ is a Cauchy sequence, that means that $\forall \epsilon>0 \exists n_0 \geq 0$, such that $\forall m>n \geq n_0: \ |x_m-x_n|< \epsilon$, right? (Thinking)
If I am right, how could we continue? (Thinking)

Yes, so for $n \ge n_0$, choose $m = n + 1$.

 

[evinda]

Yes, so for $n \ge n_0$, choose $m = n + 1$.

So can we say it like that? (Thinking)

We suppose that $\{ x_n \}$ is a Cauchy sequence as for $p-$norm, that means that $\forall \epsilon>0 \exists n_0 \geq 0$, such that $\forall m>n \geq n_0: \ |x_m-x_n|_p< \epsilon$.

We mean $m=n+1$, so we have that $\forall n \geq n_0:$

$$|x_{n+1}-x_n|_p< \epsilon$$

Does this mean that $\lim_{n \to +\infty} |x_{n+1}-x_n|_p=0$ ? (Thinking)

 

[Euge]

So can we say it like that? (Thinking)

We suppose that $\{ x_n \}$ is a Cauchy sequence as for $p-$norm, that means that $\forall \epsilon>0 \exists n_0 \geq 0$, such that $\forall m>n \geq n_0: \ |x_m-x_n|_p< \epsilon$.

We mean $m=n+1$, so we have that $\forall n \geq n_0:$

$$|x_{n+1}-x_n|_p< \epsilon$$

Does this mean that $\lim_{n \to +\infty} |x_{n+1}-x_n|_p=0$ ? (Thinking)

Yes (Smile)

 

[evinda]

Yes (Smile)

Do I have to write the limit of the definition of $\lim_{n \to \infty} |x_{n+1}-x_n|_p=0$, in order to show that when $|x_{n+1}-x_n|_p<\epsilon$, then we have that $\lim_{n \to \infty} |x_{n+1}-x_n|_p=0$ ? Or can we just say that it is like that? (Thinking)

 

[Euge]

No, it's fine the way you have it. You can just change the phrase "we mean $m = n + 1$" to "we take $m = n + 1$".

 

[evinda]

No, it's fine the way you have it. You can just change the phrase "we mean $m = n + 1$" to "we take $m = n + 1$".

A ok... From $\lim_{n \to +\infty} |x_{n+1}-x_n|_p=0$, do we get that $|x_{n+1}-x_n|_p< \epsilon$ or that $||x_{n+1}-x_n|_p|_p< \epsilon$ ? (Thinking)

 

[Euge]

You get that $|x_{n+1}-x_n|_p < \epsilon$ for $n$ sufficiently large, just like you had in post #1.

 

[evinda]

You get that $|x_{n+1}-x_n|_p < \epsilon$ for $n$ sufficiently large, just like you had in post #1.

Could you explain me why? (Thinking)

 

[Euge]

Could you explain me why? (Thinking)

It's by definition of the limit of a sequence. Let $a_n = |x_{n+1} - x_n|_p$. Then $\lim_{n\to \infty} a_n = 0$ if and only if to every $\epsilon > 0$, there corresponds a positive integer $n_0$ such that $|a_n| < \epsilon$ for all $n \ge n_0$. Since $a_n \ge 0$ for all $n$, $|a_n| = a_n$ for all $n$. Thus $|a_n| < \epsilon$ if and only if $|x_{n+1} - x_n|_p < \epsilon$.

Now we know that $\lim_{n\to \infty} |x_{n+1} - x_n|_p < \epsilon$ if and only if to every $\epsilon > 0$, there corresponds a positive integer $n_0$ such that $|x_{n+1} - x_n|_p < \epsilon$ for all $n \ge n_0$.