Editing Ratio test (section)

=== Ali's ''m''th ratio test ===

This test is a direct extension of the second ratio test.<ref name="Ali2008"/><ref name="Blackburn2012"/> For <math>0\leq k\leq m-1,</math> and positive <math>a_n</math> define:
{|
|<math>L_k\equiv\lim_{n\rightarrow \infty} \frac{a_{mn+k}}{a_n}</math>
|-
|<math>L\equiv\max(L_0,L_1,\ldots,L_{m-1})</math>
|}
By the <math>m</math>th ratio test, the series will:
* Converge if <math>L<\frac{1}{m}</math>
* Diverge if <math>L>\frac{1}{m}</math>
* If <math>L=\frac{1}{m}</math> then the test is inconclusive.

If the above limits do not exist, it may be possible to use the limits superior and inferior. For <math>0\leq k\leq m-1</math> define:

{|
|<math>L_k\equiv\limsup_{n\rightarrow \infty} \frac{a_{mn+k}}{a_n}</math>
|-
|<math>\ell_k\equiv\liminf_{n\rightarrow \infty} \frac{a_{mn+k}}{a_n}</math>
|-
|<math>L\equiv\max(L_0,L_1,\ldots,L_{m-1})</math>
|
|<math>\ell\equiv\min(\ell_0,\ell_1,\ldots,\ell_{m-1})</math>
|}

Then the series will:
* Converge if <math>L<\frac{1}{m}</math>
* Diverge if <math>\ell>\frac{1}{m}</math>
* If <math>\ell \leq \frac{1}{m} \leq L</math>, then the test is inconclusive.