Cauchy sequences and convergent sequences [closed]

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I am very confused with a little problem.

What is the difference between a Cauchy sequence and a convergent sequence?

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3 Answers

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Every convergent sequence is Cauchy but not every Cauchy sequence is convergent depending on which space you are considering.

A Cauchy sequence is a sequence where the terms of the sequence get arbitrarily close to each other after a while.

A convergent sequence is a sequence where the terms get arbitrarily close to a specific point.

Formally a convergent sequence $\{x_n\}_{n}$ converging to $x$ satisfies:

$$\forall \varepsilon>0, \exists N>0, n>N\Rightarrow |x_n-x|<\varepsilon.$$

A Cauchy sequence $\{x_n\}_n$ satisfies:

$$\forall \varepsilon>0, \exists N>0, n,m>N\Rightarrow |x_n-x_m|<\varepsilon.$$

For real numbers with the euclidean metric the properties are equivalent but not for every metric space. Consider for instance in $\mathbf{Q}$ the sequence 

$$x_1 = 3,\ x_2 = 3.1,\ x_3 = 3.14,\ x_4 = 3.141, x_5 = 3.1415\ \dotsc$$

whic gets arbitrarily close to the real number $\pi$. Now $\pi\notin \mathbf{Q}$ so the sequence does not converge in $\mathbf{Q}$. It is however a Cauchy sequence

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You can easily find the definitions. Every convergent sequence is necessarily Cauchy but not every Cauchy sequence converges. The notions can be defined in any metric space. The notions are tied to the notion of completeness: A space is complete if, and only if, a sequence converges precisely when it is Cauchy. Also, a sequence is Cauchy if, and only if, it converges in the completion of the space.

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Let $(X,d)$ be a metric space (or a normed vector space, if you want. In this case we define $d(x,y) := ||x-y||$). We call $(x_n)$ convergent with limit $x$ iff $\forall \varepsilon>0\;\exists N\in\mathbb{N}\;\forall n\geq N: d(x_n,x) < \varepsilon$ (or, intuitively speaking, if the distance between $x_n$ and $x$ will eventually approach zero as $n \rightarrow \infty$). We say $(x_n)$ is a cauchy sequence iff $\forall \varepsilon > 0\;\exists N\in\mathbb{N}\;\forall m,n\geq N: d(x_m,x_n) < \varepsilon$ (intuitively, if the distance between two arbitrary sequence elements will eventually approach zero as $n \rightarrow \infty$). One can show that any convergent sequence is a cauchy sequence (you can try to prove this, it's a very easy exercise). If for any metric space or normed vector space $X$ the converse holds true (i.e. every cauchy sequence converges to some limit $x$), we call that space complete. We can show that any normed vector space with finite dimension is complete (though this requires some work, you could start with proving the Bolzano Weierstraß theorem, first for $\mathbb{R}$, then for $\mathbb{R}^n$ via induction and then for $\mathbb{C}^n$ via $\mathbb{C}^n \cong \mathbb{R}^{2n}$, then by showing that cauchy sequences are bounded and that every cauchy sequence with a convergent subsequence is in fact convergent you can show that $\mathbb{R}^n$ and $\mathbb{C}^n$ are complete, then you can generalize this result for all normed vector spaces of finite dimension because all vector spaces of the same finite dimension are isomorphic to $\mathbb{R}^n$ or $\mathbb{C}^n$ and all norms are equivalent on $\mathbb{R}^n$ and $\mathbb{C}^n$ (the equivalence of norms part is really difficult to prove and requires some additional work so you should skip that if you really want to prove everything else).

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