$p$ 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:

$${\int }_1^{\mathrm{\infty }}\frac{1}{x^p}𝑑x=\underset{n\to \mathrm{\infty }}{lim}{\left[\frac{x^{1-p}}{1-p}\right]}_1^n=\underset{n\to \mathrm{\infty }}{lim}\frac{n^{-p+1}}{1-p}-\frac{1}{1-p}.$$

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.∎