Let \(a_n = \frac{1}{n^p}\) and let \(f(x) = \frac{1}{x^p} \). This means \(f(n) = a_n\) for all integers \(n \ge 1\).
If \(p \le 0\), then \(\sum_{n=1}^\infty \frac{1}{n^p}\) diverges as \(a_{n+1} \ge a_n\).
If \(p \gt 0\), then \(f(x)\) is a monotone decreasing function after \(x=1\). This means that according to the integral test either both \(\int^\infty_{x=1} \frac{1}{x^p} dx\) and \(\sum_{n=1}^\infty \frac{1}{n^p}\) converge or both diverge.
If \(0 \lt p \lt 1\), then \(q = 1-p\) is a positive number:
Therefore, \(\int^\infty_{x=1} \frac{1}{x^p} dx\) diverges when \(p \lt 1\), which means \(\sum_{n=1}^\infty \frac{1}{n^p}\) also diverges when \(p \lt 1\). If \(p=1\), then \(\sum_{n=1}^\infty \frac{1}{n^p}\) would be a harmonic series, which diverges. So far, we concluded that \(\sum_{n=1}^\infty \frac{1}{n^p}\) for all cases where \(p \le 1\). If \(p \gt 1\), then \(x^{1-p}\) would go towards 0 as \(x\) goes to infinity:
Since \(\int^\infty_{x=1} \frac{1}{x^p} dx\) converges to \(\frac{1}{1-p}\) if \(p \gt 1\), \(\sum_{n=1}^\infty \frac{1}{n^p}\) also converges.