Consider the series \(\sum^{\infty}_{n=1}a_n\). Let \(p\) be defined as:
\[ p = \lim_{n \to \infty} \sqrt[n]{| a_n |} \]
If \(p \lt 1\), then there exists some integer \(N\) such that \(\sqrt[n]{ | a_n |} \lt 1\) for all \(n \le N\). Let there be an integer \(k\) such that \(\sqrt[n]{ | a_n | } \le k \lt 1\). This means \(| a_n | \le k^n\) Since the geometric series \(\sum^{\infty}_{n=N}k^n\) converges, then, by the comparison test, \(\sum^{\infty}_{n=N} |a_n|\) converges as well. Thus, \(\sum^{\infty}_{n=1}a_n\) converges absolutely.
If \(p \gt 1\), then there exists an integer \(N\) such that \(\sqrt[n]{| a_n |} \gt 1\) for all \(n \ge N\):
\[\begin{gather} \sqrt[n]{| a_n |} \gt 1 \\ | a_n | \gt 1^n \\ |a_n| \gt 1 \end{gather}\]
If \(|a_n| \gt 1\) for all \(n \ge N\), then \(|a_n|\) (and \(a_n\) as well) does not converge to 0.