# $p$ test

The following is an immediate corollary of the integral test^{}.

###### Corollary ($p$-Test).

A series of the form ${\mathrm{\sum}}_{n\mathrm{=}\mathrm{1}}^{\mathrm{\infty}}\frac{\mathrm{1}}{{n}^{p}}$ converges if $p\mathrm{>}\mathrm{1}$ and diverges if $p\mathrm{\le}\mathrm{1}$.

###### Proof.

The case $p=1$ is well-known, for ${\sum}_{n=1}^{\mathrm{\infty}}\frac{1}{n}$ is the harmonic series^{}, which diverges (see this proof (http://planetmath.org/ProofOfDivergenceOfHarmonicSreies)). From now on, we assume $p\ne 1$ (notice that one could also use the integral test to prove the case $p=1$). In order to apply the integral test, we need to calculate the following improper integral:

$${\int}_{1}^{\mathrm{\infty}}\frac{1}{{x}^{p}}\mathit{d}x=\underset{n\to \mathrm{\infty}}{lim}{\left[\frac{{x}^{1-p}}{1-p}\right]}_{1}^{n}=\underset{n\to \mathrm{\infty}}{lim}\frac{{n}^{-p+1}}{1-p}-\frac{1}{1-p}.$$ |

Since ${lim}_{n\to \mathrm{\infty}}{n}^{t}$ diverges when $t>0$ and converges for $t\le 0$, the integral above converges for $$, i.e. for $p>1$ and diverges for $$ (and also diverges for $p=1$). Therefore, the corollary follows by the integral test. ∎

Title | $p$ test |
---|---|

Canonical name | PTest |

Date of creation | 2013-03-22 15:08:51 |

Last modified on | 2013-03-22 15:08:51 |

Owner | alozano (2414) |

Last modified by | alozano (2414) |

Numerical id | 6 |

Author | alozano (2414) |

Entry type | Corollary |

Classification | msc 40A05 |

Synonym | p-test |

Synonym | $p$-test |

Synonym | p test |

Synonym | p series test |

Synonym | $p$-series test |

Synonym | $p$ series test |

Related topic | ExamplesUsingComparisonTestWithoutLimit |

Related topic | ASeriesRelatedToHarmonicSeries |