# $p$ test

The following is an immediate corollary of the integral test.

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

A series of the form $\sum_{n=1}^{\infty}\frac{1}{n^{p}}$ converges if $p>1$ and diverges if $p\leq 1$.

###### Proof.

The case $p=1$ is well-known, for $\sum_{n=1}^{\infty}\frac{1}{n}$ is the harmonic series, which diverges (see this proof (http://planetmath.org/ProofOfDivergenceOfHarmonicSreies)). From now on, we assume $p\neq 1$ (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:

 $\int_{1}^{\infty}\frac{1}{x^{p}}dx=\lim_{n\to\infty}\left[\frac{x^{1-p}}{1-p}% \right]_{1}^{n}=\lim_{n\to\infty}\frac{n^{-p+1}}{1-p}-\frac{1}{1-p}.$

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

Title $p$ test PTest 2013-03-22 15:08:51 2013-03-22 15:08:51 alozano (2414) alozano (2414) 6 alozano (2414) Corollary msc 40A05 p-test $p$-test p test p series test $p$-series test $p$ series test ExamplesUsingComparisonTestWithoutLimit ASeriesRelatedToHarmonicSeries