# proof of Cauchy’s root test

If for all $n\ge N$

$$ |

then

$$ |

Since ${\sum}_{i=N}^{\mathrm{\infty}}{k}^{i}$ converges so does ${\sum}_{i=N}^{\mathrm{\infty}}{a}_{n}$ by the comparison test^{}. If $\sqrt[n]{{a}_{n}}>1$ then by comparison with ${\sum}_{i=N}^{\mathrm{\infty}}1$ the series is divergent. Absolute convergence in case of nonpositive ${a}_{n}$ can be proven in exactly the same way using $\sqrt[n]{|{a}_{n}|}$.

Title | proof of Cauchy’s root test^{} |
---|---|

Canonical name | ProofOfCauchysRootTest |

Date of creation | 2013-03-22 13:23:43 |

Last modified on | 2013-03-22 13:23:43 |

Owner | mathwizard (128) |

Last modified by | mathwizard (128) |

Numerical id | 5 |

Author | mathwizard (128) |

Entry type | Proof |

Classification | msc 40A05 |