20. Convergence of Positive Series

f.1. The Root Test

a. Proof

If \(\rho=1\) the Root Test FAILS and says nothing about \(\displaystyle \sum_{n=n_o}^\infty a_n\).

The statement \(\displaystyle \lim_{n\to\infty}\sqrt[\small n]{a_n}=\rho\) means:

For all   \(\varepsilon \gt 0\),   there is a positive integer,   \(N\),   such that

if   \(n \gt N\)   then   \(\rho-\varepsilon \lt \sqrt[\small n]{a_n} \lt \rho+\varepsilon\).

1. First suppose \(\rho< 1\). We want to apply the Simple Comparison Test by comparing the series \(\displaystyle \sum_{n=n_o}^\infty a_n\) to a geometric series whose ratio \(r\) is slightly larger than \(\rho\).
   
   So pick \(\varepsilon \gt 0\) such that \(\rho+\varepsilon \lt 1\) and let \(r=\rho+\varepsilon\). Then there is an \(N \gt 0\), such that for all \(n \ge N\) we have \(\sqrt[\small n]{a_n} \lt \rho+\varepsilon=r\) or \(a_n \lt r^n\). Since \(r \lt 1\), the geometric series \(\displaystyle \sum_{n=N}^\infty r^n\) converges. By the Simple Comparison Test the series \(\displaystyle \sum_{n=n_o}^\infty a_n\) also converges.
2. Now suppose \(\rho \gt 1\). We want to apply the Simple Comparison Test by comparing the series \(\displaystyle \sum_{n=n_o}^\infty a_n\) to a geometric series whose ratio \(r\) is slightly smaller than \(\rho\).
   
   So pick \(\varepsilon \gt 0\) such that \(\rho-\varepsilon \gt 1\) and let \(r=\rho-\varepsilon\). Then there is an \(N \gt 0\), such that for all \(n \ge N\) we have \(\sqrt[\small n]{a_n} \gt \rho-\varepsilon=r\) or \(a_n \gt r^n\). Since \(r \gt 1\), the geometric series \(\displaystyle \sum_{n=N}^\infty r^n\) diverges. By the Simple Comparison Test the series \(\displaystyle \sum_{n=n_o}^\infty a_n\) also diverges.