Theorem
20
[21C] Assume that \(a_ n\neq 0\). Let \(๐ผ=\limsup _{nโโ}\frac{|a_{n+1}|}{|a_{n}|}\) then
if \(๐ผ{\lt}1\) the series \(โ_{n=1}^โ a_ n\) converges absolutely;
if \(๐ผโฅ 1\) nothing can be concluded.
Proof
If \(๐ผ{\lt}1\), taken \(Lโ(๐ผ,1)\) you have eventually \(\frac{|a_{n+1}|}{|a_{n}|}{\lt}L\) so there is a \(N\) for which \(\frac{|a_{n+1}|}{|a_{n}|}{\lt}L\) for each \(nโฅ N\), by induction it is shown that \(|a_ n|โค L^{n-N} |a_ N|\) and ends by comparison with the geometric series.
Letโs see some examples. For the two series \(1/n\) and \(1/n^ 2\) you have \(๐ผ=1\).
Defining
\begin{equation} a_ n= \begin{cases} 2^{-n} & n~ \text{even}\\ 2^{2-n} & n~ \text{odd}\\ \end{cases} \label{eq:f3422p3sa} \end{equation}21we obtain a convergent series but for which \(๐ผ=2\).