p-series test

Statement

Let $p$ be a real number. The series $\sum_{n=1}^{\infty} \frac{1}{n^p}$ converges if and only if $p > 1$.

Remarks

For $p = 1$ it is the harmonic series, which diverges. For $p \le 0$ the terms do not tend to $0$. One proof uses the Cauchy condensation test. The condensed series is $\sum 2^n \cdot 2^{-np} = \sum (2^{1-p})^n$, a geometric series with ratio $2^{1-p}$. Another proof uses the integral test with $f(x) = x^{-p}$. For $p > 1$ the sum is the value $\zeta(p)$ of the Riemann zeta function. For example, $\zeta(2) = \pi^2/6$, which is the Basel problem.

Explanations

L1 human

The terms $1/n^p$ shrink faster when $p$ is larger. The value $p = 1$ is the border. Above it the sum is finite, and at it or below it the sum is infinite.