I am really confused. I will appreciate if somebody can help me to define the difference between Cauchy and convergent sequence.

Many thanks.

  • $\begingroup$ If you are only dealing with $\mathbb R$, you SHOULD be confused. $\endgroup$ – user99914 Mar 27 '14 at 8:42
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    $\begingroup$ One important difference is the way the notion is defined: the notion of Cauchy sequence only refers to the terms of the sequence itself, while the notion of convergent sequence refers to (the existence of) a limit value of the sequence. $\endgroup$ – Marc van Leeuwen Mar 27 '14 at 14:56

When you have a convergent sequence: $x_n\to\xi$, then for large $n$ all points $x_n$ are near $\xi$, and by the triangle inequality they are then near each other as well – in formal language: The $x_n$ automatically form a Cauchy sequence; no big deal.


In some metric spaces $X$, among them ${\mathbb R}$, ${\mathbb R}^n$, and ${\mathbb C}$, the converse also holds: A Cauchy sequence is automatically convergent. (Such spaces are called complete.) This property of spaces is all-important for the following reason: In order to test a sequence $(x_n)_{\geq1}$ for Cauchy-ness you only have to look at the given $x_n$ themselves, and you don't have to know the limit beforehand. When the sequence passes this "finitary" test you know for sure that it has a limit $\xi\in X$.

In all, the so-called Cauchy criterion (for real sequences, say) is not a little proposition about convergence, but a deep theorem about the fine structure of ${\mathbb R}$.

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    $\begingroup$ "a deep theorem about the fine structure of ${\mathbb R}$" -- or an axiom of ${\mathbb R}$, or a fairly straightforward consequence of the construction of ${\mathbb R}$, depending how you do your foundations :-) $\endgroup$ – Steve Jessop Mar 27 '14 at 11:14
  • $\begingroup$ @Christian Blatter May I ask, does the converse also hold for $\mathbb Q$? $\endgroup$ – Dole Feb 26 '18 at 10:38
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    $\begingroup$ @Dole: No, it doesn't. The finite decimal approximations to $\sqrt{2}$ converge to $\sqrt{2}$, hence they form a Cauchy sequence of rational numbers. Nevertheless this sequence is divergent in ${\mathbb Q}$. $\endgroup$ – Christian Blatter Feb 26 '18 at 10:54

Let $(X,d)$ be a metric space.

Definition. A sequence $(x_n)_{n\in\mathbb N}$ with $x_n\in X$ for all $n\in\mathbb N$ is a Cauchy sequence in $X$ if and only if for every $\varepsilon > 0$ there exists $N\in\mathbb N$ such that $d(x_n,x_m)<\varepsilon$ for all $n,m>N$.

Informally speaking, a Cauchy sequence is a sequence where the terms of the sequence are getting closer and closer to each other.

Definition. A sequence $(x_n)_{n\in\mathbb N}$ with $x_n\in X$ for all $n\in\mathbb N$ is convergent if and only if there exists a point $x\in X$ such that for every $\varepsilon > 0$ there exists $N\in\mathbb N$ such that $d(x_n,x)<\varepsilon$. In this situation we say $x$ is a limit of the sequence $(x_n)_{n\in\mathbb N}$, or $(x_n)_{n\in\mathbb N}$ converges to $x$.

Informally speaking, a sequence is convergent if the terms of the sequence are getting closer and closer to some point $x\in X$.

A metric space $X$ is said to be complete if every Cauchy sequence is convergent. This is the case for the spaces $\mathbb R^n$, which is the reason why you might not see the difference of the concepts at first glance.

Let's take a look at a familiar metric space which is not complete, so we have Cauchy sequence that are not convergent. Let $X=\mathbb Q$ be the set of rational numbers with the usual metric given by $d(x,y)=\left|x-y\right|$. Consider the sequence defined by $$ x_1 = 1,\quad x_{n+1} = 1+\frac{1}{1+x_n}. $$ We can show that this sequence is a Cauchy sequence in $\mathbb Q$, but there is no $x\in \mathbb Q$ that is a limit of the sequence.

Why is that? Well, consider the same sequence as a sequence in the complete metric space $X=\mathbb R$, we can now show that $(x_n)_{n\in\mathbb N}$ converges to $\sqrt 2\in\mathbb R$, but $\sqrt 2$ is not in $\mathbb Q$. So while the sequence converges in $\mathbb R$, it doesn't converge in $\mathbb Q$. Nevertheless it is a Cauchy sequence in both spaces.

  • $\begingroup$ I know this is an old post , thanks for your answer helped me loads , but just wondering how could one show the limit is root2 I know we could just do limxn=limxn+1 and find L but we cant just assume liman=L because it may not converge , so how do I show it converges in the first place? Thanks $\endgroup$ – Rivaldo Nov 22 '18 at 17:56
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    $\begingroup$ See for example this thread. $\endgroup$ – Christoph Nov 22 '18 at 19:41

A convergent sequence is also a Cauchy sequence.

A Cauchy sequence is not necessarily a convergent sequence. For example if our space is $X=\mathbb Q$, then $$ x_n=\frac{\lfloor n\sqrt{2}\rfloor}{n}, $$
is a Cauchy sequence which DOES NOT converge is $\mathbb Q$. It DOES converge is $\mathbb R$ but not in $\mathbb Q$.

A metric space where every Cauchy sequence converges is called complete. The space $\mathbb Q$ with metric the distance $|a-b|$ is not complete, while $\mathbb R$ with the same metric is complete.


Every convergent sequence is a cauchy sequence. The converse may however not hold. For sequences in $\mathbb{R}^k$ the two notions are equal. More generally we call an abstract metric space $X$ such that every cauchy sequence in $X$ converges to a point in $X$ a complete metric space. One can for instance show that the space of continuous functions on a compact set with the uniform metric is complete.

As an example of a space that is not complete, consider the interval $(0,1)$ and the sequence $1/n$, clearly $1/n$ is cauchy, but $1/n$ is not convergent in $(0,1)$ as $0$ is not in $(0,1)$.

In contrast $[0,1]$ is complete, since it is a closed subset of the complete space $\mathbb{R}$.

  • $\begingroup$ "Every convergent sequence is a Cauchy sequence." That's what I was looking for! I couldn't think of any counterexamples. $\endgroup$ – Mateen Ulhaq May 8 '17 at 10:19

Let's start in metric space $\mathbb{R}$ equipped with its normal metric.

A sequence $\left(x_{n}\right)$ in $\mathbb{R}$ is convergent if some $x\in\mathbb{R}$ exists such that for every $\varepsilon>0$ some $n_{0}\in\mathbb{N}$ can be found such that $\left|x-x_{n}\right|<\varepsilon$ for each $n\geq n_{0}$. If that is the case then it can be shown that this $x$ is unique and the sequence is said to converge to $x$.

A sequence $\left(x_{n}\right)$ in $\mathbb{R}$ is a Cauchy-sequence if for every $\varepsilon>0$ some $n_{0}\in\mathbb{N}$ can be found such that $\left|x_{n}-x_{m}\right|<\varepsilon$ for each pair $n,m\in\mathbb{N}$ with $n,m\geq n_{0}$.

It can be shown that a convergent sequence is a Cauchy-sequence. This in any metric space.

In the case we are dealing with (metric space $\mathbb{R}$) the opposite is also true: every Cauchy-sequence in $\mathbb{R}$ is a convergent sequence.

However, there are metric spaces in wich the opposite is not true. An example is $\mathbb{Q}$. Start with a sequence $\left(x_{n}\right)$ in $\mathbb{R}$ that converges to some $x\in\mathbb{R}-\mathbb{Q}$ and let it be that $x_{n}\in\mathbb{Q}$ for each $n$. It is a Cauchy-sequence in $\mathbb{R}$ and is convergent in $\mathbb{R}$. However, if we look at it as a sequence in $\mathbb{Q}$ then it is still a Cauchy-sequence, but it is not convergent, simply because $x\notin\mathbb{Q}$.

Spaces in which the opposite is true are so-called 'complete spaces'.


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