# Proof that every absolutely convergent series is unconditionally convergent

Suppose the series ${\sum}_{k=1}^{\mathrm{\infty}}{a}_{k}$ converges to a $A$ and that it is also absolutely convergent. So, as ${\sum}_{k=1}^{\mathrm{\infty}}|{a}_{k}|$ converges, we have $\forall \u03f5>0$:

$$ |

As the series converges to $A$, we may choose $n$ such that:

$$ |

Now suppose that ${\sum}_{k=1}^{\mathrm{\infty}}{b}_{k}$ is a rearrangement of ${\sum}_{k=1}^{\mathrm{\infty}}{a}_{k}$, that is, it exists a bijection $\sigma :\text{mathbb}N\to \text{mathbb}N$, such that ${b}_{k}={a}_{\sigma (k)}$

Let $J=max\{j:\sigma (j)\le n\}$. Now take $m\ge J$. Then:

$$\sum _{k=1}^{m}{b}_{k}={b}_{1}+\mathrm{\dots}+{b}_{m}={a}_{\sigma (1)}+\mathrm{\dots}+{a}_{\sigma (m)}$$ |

This sum must include the terms ${a}_{1},\mathrm{\dots},{a}_{n}$, otherwise $J$ wouldn’t be maximum. Thus, for $m\ge J$, we have that ${\sum}_{j=1}^{m}{b}_{j}-{\sum}_{k=1}^{n}{a}_{k}$ is a finite sum of terms ${a}_{k}$, with $k\ge n+1$. So:

$$ |

Finally, taking $m\ge J$, we have:

$$ |

So ${\sum}_{k=1}^{\mathrm{\infty}}{b}_{k}=A$

Title | Proof that every absolutely convergent series is unconditionally convergent |
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Canonical name | ProofThatEveryAbsolutelyConvergentSeriesIsUnconditionallyConvergent |

Date of creation | 2013-03-11 19:17:17 |

Last modified on | 2013-03-11 19:17:17 |

Owner | Filipe (28191) |

Last modified by | (0) |

Numerical id | 7 |

Author | Filipe (0) |

Entry type | Proof |

Classification | msc 40A05 |

Synonym | |

Related topic | |

Defines |