Editing Ratio test (section)

=====Proof of Kummer's test=====

If <math>\rho_n>0</math> then fix a positive number <math>0<\delta<\rho_n</math>.  There exists
a natural number <math>N</math> such that for every <math>n>N,</math>
:<math>\delta\leq\zeta_{n}\frac{a_{n}}{a_{n+1}}-\zeta_{n+1}.</math>
Since <math>a_{n+1}>0</math>, for every <math>n> N,</math>
:<math>0\leq \delta a_{n+1}\leq \zeta_{n}a_{n}-\zeta_{n+1}a_{n+1}.</math>
In particular <math>\zeta_{n+1}a_{n+1}\leq \zeta_{n}a_{n}</math> for all <math>n\geq N</math> which means that starting from the index
<math>N</math>
the sequence <math>\zeta_{n}a_{n}>0</math> is monotonically decreasing and
positive which in particular implies that it is bounded below by 0. Therefore, the limit
:<math>\lim_{n\to\infty}\zeta_{n}a_{n}=L</math> exists.
This implies that the positive  [[telescoping series]]
:<math>\sum_{n=1}^{\infty}\left(\zeta_{n}a_{n}-\zeta_{n+1}a_{n+1}\right)</math> is convergent,
and since for all <math>n>N,</math>
:<math>\delta a_{n+1}\leq \zeta_{n}a_{n}-\zeta_{n+1}a_{n+1}</math>
by the [[direct comparison test]] for positive series, the series
<math>\sum_{n=1}^{\infty}\delta a_{n+1}</math> is convergent.

On the other hand, if <math>\rho<0</math>, then there is an ''N'' such that <math>\zeta_n a_n</math> is increasing for <math>n>N</math>.  In particular, there exists an <math>\epsilon>0</math> for which <math>\zeta_n a_n>\epsilon</math> for all <math>n>N</math>, and so <math>\sum_n a_n=\sum_n \frac{a_n\zeta_n}{\zeta_n}</math> diverges by comparison with <math>\sum_n \frac \epsilon {\zeta_n}</math>.