# proof of comparison test

Assume $|{a}_{k}|\le {b}_{k}$ for all $k>n$. Then we define

$${s}_{k}:=\sum _{i=k}^{\mathrm{\infty}}|{a}_{i}|$$ |

and

$${t}_{k}:=\sum _{i=k}^{\mathrm{\infty}}{b}_{i}.$$ |

Obviously ${s}_{k}\le {t}_{k}$ for all $k>n$. Since by assumption $({t}_{k})$ is http://planetmath.org/node/601convergent^{} $({t}_{k})$ is bounded and so is $({s}_{k})$. Also $({s}_{k})$ is monotonic and therefore . Therefore ${\sum}_{i=0}^{\mathrm{\infty}}{a}_{i}$ is absolutely convergent.

Now assume ${b}_{k}\le {a}_{k}$ for all $k>n$. If ${\sum}_{i=k}^{\mathrm{\infty}}{b}_{i}$ is divergent then so is ${\sum}_{i=k}^{\mathrm{\infty}}{a}_{i}$ because otherwise we could apply the test we just proved and show that ${\sum}_{i=0}^{\mathrm{\infty}}{b}_{i}$ is convergent, which is is not by assumption.

Title | proof of comparison test |
---|---|

Canonical name | ProofOfComparisonTest |

Date of creation | 2013-03-22 13:22:06 |

Last modified on | 2013-03-22 13:22:06 |

Owner | mathwizard (128) |

Last modified by | mathwizard (128) |

Numerical id | 4 |

Author | mathwizard (128) |

Entry type | Proof |

Classification | msc 40A05 |