Editing Ratio test (section)

=== Tong's modification of Kummer's test===

A new version of Kummer's test was established by Tong.<ref name="Tong1994"/> See also <ref name="Samelson1995"/><ref name="Duris2018"/><ref name="Abramov2021">{{cite arXiv |last1=Abramov |first1=Vyacheslav, M. |date= 21 June 2021 |title=A simple proof of Tong's theorem |eprint=2106.13808 |class=math.HO }}</ref>
for further discussions and new proofs. The provided modification of Kummer's theorem characterizes
all positive series, and the convergence or divergence can be formulated in the form of two necessary and sufficient conditions, one for convergence and another for divergence.

* Series <math>\sum_{n=1}^\infty a_n</math> converges if and only if there exists a positive sequence <math>\zeta_n</math>, <math>n=1,2,\dots</math>, such that <math>\zeta_n\frac{a_n}{a_{n+1}}-\zeta_{n+1}\geq c>0.</math>

* Series <math>\sum_{n=1}^\infty a_n</math> diverges if and only if there exists a positive sequence <math>\zeta_n</math>, <math>n=1,2,\dots</math>, such that <math>\zeta_n\frac{a_n}{a_{n+1}}-\zeta_{n+1}\leq0,</math> and <math>\sum_{n=1}^{\infty}\frac{1}{\zeta_n}=\infty.</math>

The first of these statements can be simplified as follows:<ref name="Abramov2022">
{{cite journal |last1=Abramov |first1=Vyacheslav M.|date=May 2022|title=Evaluating the sum of convergent positive series|url=http://elib.mi.sanu.ac.rs/files/journals/publ/131/publn131p41-53.pdf | journal=Publications de l'Institut Mathématique |series=Nouvelle Série |volume=111 |issue=125 |pages=41–53 |doi=10.2298/PIM2225041A|s2cid=237499616 }}</ref>

* Series <math>\sum_{n=1}^\infty a_n</math> converges if and only if there exists a positive sequence <math>\zeta_n</math>, <math>n=1,2,\dots</math>, such that <math>\zeta_n\frac{a_n}{a_{n+1}}-\zeta_{n+1}=1.</math>

The second statement can be simplified similarly:
* Series <math>\sum_{n=1}^\infty a_n</math> diverges if and only if there exists a positive sequence <math>\zeta_n</math>, <math>n=1,2,\dots</math>, such that <math>\zeta_n\frac{a_n}{a_{n+1}}-\zeta_{n+1}=0,</math> and <math>\sum_{n=1}^{\infty}\frac{1}{\zeta_n}=\infty.</math>

However, it becomes useless, since the condition <math>\sum_{n=1}^{\infty}\frac{1}{\zeta_n}=\infty</math> in this case reduces to the original claim <math>\sum_{n=1}^{\infty}a_n=\infty.</math>