Corollary 1 ($p$ -Test)   A series of the form $\sumn \frac{1}{n^p}$ converges if $p>1$ and diverges if $p\leq 1$ .

Proof. The case $p=1$ is well-known, for $\sumn \frac{1}{n}$ is the harmonic series, which diverges (see this proof). 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: $$\int_1^\infty \frac{1}{x^p} dx=\lim_{n\to \infty}\left[ \frac{x^{1-p}}{1-p} \right]_1^n= \limn \frac{n^{-p+1}}{1-p}-\frac{1}{1-p}.$$ Since $\limn n^t$ 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.