Editing Ratio test (section)

==== 6. Kummer's test ====

This extension is due to [[Ernst Kummer]].

Let ζ<sub>''n''</sub> be an auxiliary sequence of positive constants. Define

:<math>\rho_n \equiv \left(\zeta_n \frac{a_n}{a_{n+1}} - \zeta_{n+1}\right)</math>

Kummer's test states that the series will:<ref name="Knopp"/><ref name="Tong1994"/><ref name="Duris2009"/><ref name="Duris2018"/>
* Converge if there exists a <math>c>0</math> such that <math>\rho_n \ge c</math> for all n>N. (Note this is not the same as saying <math>\rho_n > 0</math>)
* Diverge if <math>\rho_n \le 0</math> for all n>N and <math>\sum_{n=1}^\infty 1/\zeta_n</math> diverges.

For the limit version, the series will:<ref>{{mathworld|title=Kummer's Test|urlname=KummersTest}}</ref><ref name="Ali2008"/><ref name="Blackburn2012"/>
* Converge if <math>\lim_{n\to\infty}\rho_n>0</math> (this includes the case ''ρ'' = ∞)
* Diverge if <math>\lim_{n\to\infty}\rho_n<0</math> and <math>\sum_{n=1}^\infty 1/\zeta_n</math> diverges.
* Otherwise the test is inconclusive

When the above limit does not exist, it may be possible to use limits superior and inferior.<ref name="Bromwich1908"/> The series will
* Converge if <math>\liminf_{n \to \infty} \rho_n >0</math>
* Diverge if <math>\limsup_{n \to \infty} \rho_n <0</math> and <math>\sum 1/\zeta_n</math> diverges.

===== Special cases =====

All of the tests in De Morgan's hierarchy except Gauss's test can easily be seen as special cases of Kummer's test:<ref name="Bromwich1908"/>
* For the ratio test, let ζ<sub>n</sub>=1. Then:
::<math>\rho_\text{Kummer} = \left(\frac{a_n}{a_{n+1}}-1\right) = 1/\rho_\text{Ratio}-1</math>
* For Raabe's test, let ζ<sub>n</sub>=n. Then:
::<math>\rho_\text{Kummer} = \left(n\frac{a_n}{a_{n+1}}-(n+1)\right) = \rho_\text{Raabe}-1</math>
* For Bertrand's test, let ζ<sub>n</sub>=n ln(n). Then:
::<math>\rho_\text{Kummer} =  n \ln(n)\left(\frac{a_n}{a_{n+1}}\right)-(n+1)\ln(n+1)</math>
:Using <math>\ln(n+1)=\ln(n)+\ln(1+1/n)</math> and [[approximation|approximating]] <math>\ln(1+1/n)\rightarrow 1/n</math> for large ''n'', which is negligible compared to the other terms, <math>\rho_\text{Kummer}</math> may be written:
::<math>\rho_\text{Kummer} =  n \ln(n)\left(\frac{a_n}{a_{n+1}}-1\right)-\ln(n)-1 = \rho_\text{Bertrand}-1</math>
* For Extended Bertrand's test, let <math>\zeta_n=n\prod_{k=1}^K\ln_{(k)}(n).</math> From the [[Taylor series]] expansion for large <math>n</math> we arrive at the [[approximation]] 
::<math>\ln_{(k)}(n+1)=\ln_{(k)}(n)+\frac{1}{n\prod_{j=1}^{k-1}\ln_{(j)}(n)}+O\left(\frac{1}{n^2}\right),</math>
where the empty product is assumed to be 1. Then,
::<math>\rho_\text{Kummer} = n\prod_{k=1}^K\ln_{(k)}(n)\frac{a_n}{a_{n+1}}-(n+1)\left[\prod_{k=1}^K\left(\ln_{(k)}(n)+\frac{1}{n\prod_{j=1}^{k-1}\ln_{(j)}(n)}\right)\right]+o(1)
=n\prod_{k=1}^K\ln_{(k)}(n)\left(\frac{a_n}{a_{n+1}}-1\right)-\sum_{j=1}^K\prod_{k=1}^j\ln_{(K-k+1)}(n)-1+o(1).</math>

Hence,

::<math>\rho_\text{Kummer} = \rho_\text{Extended Bertrand}-1.</math>

Note that for these four tests, the higher they are in the De Morgan hierarchy, the more slowly the <math>1/\zeta_n</math> series diverges.

=====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>.