# Is being a Cauchy sequence equivalent to $\lim_{n\to+\infty}d(x_{n+k},x_n)=0$ for every $k$?

Is this statement true?

In a metric spase $(E,d)$, a sequence $(x_n)$ is Cauchy if and only if $\forall k\in \mathbb{N}, \lim_{n\rightarrow+\infty}d(x_{n+k},x_n)=0$

I proved that $\Rightarrow$ is true,indeed if $(x_n)$ is Cauchy then for all $\varepsilon>0,\exists n_0\in\mathbb{N}, \forall p,q\geq n, d(x_p,x_q)<\varepsilon$ it still right for all $n\geq n_0$ and $m=n+k>n\geq n_0$

But I don't know if $\Leftarrow$ is true ?

• Please show your proof for the $\implies$ direction. I don't think the implication holds, so I'd like to see if your proof is logically sound. Here is a counter example: Suppose $E = \mathbb{R}$ and $d(a,b) = |a - b|$. Let $(x_{n}) = (\frac{1}{n})$. This sequence is Cauchy but for each $n$, $d(x_{n + k}, x_{n}) > 0$ for all $k$. – layman Jan 3 '15 at 14:55
• Maybe $...\to 0$ as $n \to \infty$ instead of $...=0$ – user35603 Jan 3 '15 at 14:56
• Your condition states that all the terms are equal. This certainly implies that the sequence is Cauchy, but certainly there are Cauchy sequences whose terms are not all equal. In other words $(\Rightarrow)$ is false. – GPerez Jan 3 '15 at 15:08
• Ah, never mind then (the way you had it, every term was at distance zero from all the others i.e. they're all the same). – GPerez Jan 3 '15 at 15:15
• What is written in the question: "$(x_n)$ is Cauchy $\Longleftrightarrow \forall k\in \mathbb{N}, \lim\limits_{n\rightarrow+\infty}d(x_{n+k},x_n)=0$" is wrong, as the example of $x_n=\log n$ for every $n$ shows (for the Euclidean metrics). – Did Jan 3 '15 at 15:28

The equivalence is not correct.

To see this, consider any divergent series of real numbers $\sum a_n$ such that $a_n\to 0$ as $n\to \infty$; for example $a_n=\frac1n$. Then define $x_n=a_1+\dots +a_n$. This is a non-Cauchy sequence (as any non-convergent sequence of real numbers), but $x_{n+k}-x_n=a_{n+1}+\dots +a_{n+k}\to 0$ as $n\to \infty$ for any fixed $k\in\mathbb N$, since this is the sum of $k$ terms tending to $0$; that is, $d(x_n,x_{n+k})\to 0$ as $n\to\infty$. So the implication $\Leftarrow$ does not hold.

If you prefer, you can take $x_n=\log n$ or $x_n=\sqrt{n}$, and check that $x_{n+k}-x_n\to 0$ as $n\to \infty$, for any $k\in\mathbb N$.

On the other hand, your proof of the other implication is correct.

Last remark : in fact the assumption "$\forall k\in\mathbb N\; d(x_{n+k},x_n)\to 0$" is equivalent to the seemingly weaker "$d(x_{n+1},x_n)\to 0$", as you can easily check.

• why $\sqrt{n+k}-\sqrt{n} \rightarrow 0$ ? – Vrouvrou Jan 3 '15 at 15:49
• For example, because it is equal to $\frac{k}{\sqrt{n+k}+\sqrt{n}}\cdot$ – Etienne Jan 3 '15 at 16:06
• Ok that's right and why is not a Cauchy sequence – Vrouvrou Jan 3 '15 at 16:22
• Please can you tel me if $\forall \varepsilon>0, \exists n_0\in \mathbb{N}, \forall (n,k)\in\mathbb{N}^2,(n\geq n_0\Rightarrow d(x_n,x_{n+k})\leq \varepsilon)$ is the same as $\forall k\in \mathbb{N}, \lim_{n\rightarrow+\infty}d(x_{n+k},x_n)=0$ please i nee your help – Vrouvrou Jan 4 '15 at 17:26
• No, this is not the same : the first property says "$(x_n)$ is Cauchy". – Etienne Jan 4 '15 at 17:42

of course that it's wrong !! Consider $x_n=\frac{1}{n}$, the sequence converge therefore it's a Cauchy sequence, but $d(x_{n+k},x_n)\neq 0$ for all $k$.

The correct statement would be

$(x_n)$ is cauchy $\iff \forall \varepsilon>0, \exists N\in\mathbb N:\forall n,k\in\mathbb N, n\geq N\implies d(x_{n+k},x_n)<\varepsilon$

and this is the definition of a Cauchy sequence as well !

• what is the difference between $\forall k\in\mathbb{N}, \lim_{n\rightarrow+\infty}d(x_{n+k},x_n)=0$ and your statement ? please – Vrouvrou Jan 3 '15 at 15:04
• I don't see why $\lim_{n\to\infty}d(x_{n+k},x_n)\ne0$. – user84413 Jan 3 '15 at 18:46
• @user84413 : you cannot see it, since this is a wrong answer. – Watson Aug 20 at 10:00

With the way you edited it, you actually just wrote the definition of a cauchy sequence in a different manner.

$\lim_{n\to+\infty}d(x_{n+k},x_n)=0$ means that for every $\varepsilon>0$ there exists $N$ such that $d(x_{n+k},x_n)<\varepsilon\ \forall\ n>N$.

But since $k$ can be any arbitrary natural number, this means that $d(x_n,x_m)<\varepsilon\ \forall\ n>N,m>N$.

Thus, you didn't really change anything there.

• so this equivalence is true right ? – Vrouvrou Jan 3 '15 at 15:20
• It is true, but it isn't really any different from the definition itself. – Hasan Saad Jan 3 '15 at 15:21
• That was BEFORE you put the limit there. Without the limit, that guy is right, it's wrong. But when you put the limit as $n\to\infty$ there, the equivalence is established. – Hasan Saad Jan 3 '15 at 15:24
• "With the way you edited it, you actually just wrote the definition of a cauchy sequence in a different manner." Not at all. – Did Jan 3 '15 at 15:29
• @HasanSaad Write down carefully the statements "for all $k$ $(d(x_n,x_{n+k})\to 0$" and "$(x_n)$ is Cauchy" with quantifiers. You'll see that the order of the quantifiers is not exactly the same. – Etienne Jan 3 '15 at 16:10