The sequence members of a Cauchy Sequence eventually become arbitrarily close to each other.

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__Show requirements__

Concept |
Content |

Sequences |
These object are needed to define limits later on. |

Convergence |
Convergent sequences have a well-defined limit. |

Bounded
Sequences |
Sequences can be bounded from above and from below. The values of a bounded sequence cannot become arbitrarily large. |

__Show consequences__

Concept |
Content |

Completeness |
Completeness says that Cauchy sequences must converge. |

Cauchy Criterion |
A series converges if its partial sums form a Cauchy sequence. As a consequence, for a convergent series, the underlying sequence necessarily needs to converge to zero. |

**Definition 1** (Cauchy sequences). *A sequence
\((a_n)_{n\in\mathbb{N}}\) in \(\mathbb{K}\) is called **Cauchy
sequence* if for all \(\varepsilon>0\), there exists some \(N\) such that for all \(n,m\geq N\) holds \[|a_n-a_m|<\varepsilon.\]

Now we show that convergent sequences are indeed Cauchy
sequences.

**Theorem 3**. *Let \((a_n)_{n\in\mathbb{N}}\) be a convergent
sequence. Then \((a_n)_{n\in\mathbb{N}}\) is a Cauchy
sequence.*

*Proof:* Let \(a=\lim_{n o\infty}a_{n}\) and \(\varepsilon>0\). Then there exists some
\(N\) such that for all \(k\geq N\) holds \(|a-a_k|<\frac{\varepsilon}2\). Hence,
for all \(m,n\geq N\) holds \[|a_n-a_m|=|(a_n-a)+(a-a_m)|\leq|a_n-a|+|a-a_m|<
\frac{\varepsilon}2+\frac{\varepsilon}2=\varepsilon.\] \(\Box\)

We know that convergent sequences are bounded. The following theorem
shows that this is also the case for Cauchy sequences.

**Theorem 4** (Cauchy sequences are bounded). *Let
\((a_n)_{n\in\mathbb{N}}\) be a Cauchy
sequence. Then \((a_n)_{n\in\mathbb{N}}\) is
bounded.*

*Proof:* Take \(\varepsilon=1\). Then there exists some
\(N\) such that for all \(n,m\geq N\) holds \(|a_n-a_m|<1\). Thus, for all \(n\geq N\) holds \[|a_n|=|a_n-a_N+a_N|\leq
|a_n-a_N|+|a_N|<1+|a_N|.\] Now choose \[c=\max\{|a_1|,|a_2|,\ldots,|a_{N-1}|,|a_N|+1\}\]
and consider some arbitrary sequence element \(a_k\).

If \(k<N\), we have that \(|a_k|\leq \max\{|a_1|,|a_2|,\ldots,|a_{N-1}|\}\leq
c\).

If \(k\geq N\), we have, by the above
calculations, that \(|a_k|<|a_N|+1\leq
c\).

Altogether, this implies that \(|a_k|\leq
c\) for all \(k\in\mathbb{N}\),
so \((a_n)_{n\in\mathbb{N}}\) is
bounded by \(c\).\(\Box\)

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