Editing Root test (section)

== Root tests hierarchy ==

Root tests hierarchy<ref>{{cite journal|url=http://files.ele-math.com/articles/jca-19-09.pdf |last1=Abramov |first1=Vyacheslav M. |date=2022 |title=Necessary and sufficient conditions for the convergence of positive series |journal=Journal of Classical Analysis |volume=19 |issue=2 |pages=117--125 |doi=10.7153/jca-2022-19-09 |arxiv=2104.01702 }}</ref><ref>{{cite journal|url=http://www.m-hikari.com/ijma/ijma-2012/ijma-37-40-2012/bourchteinIJMA37-40-2012.pdf |last1=Bourchtein |first1=Ludmila |last2=Bourchtein |first2=Andrei |last3=Nornberg |first3=Gabrielle |last4=Venzke |first4=Cristiane |date=2012 |title=A hierarchy of convergence tests related to Cauchy's test |journal=International Journal of Mathematical Analysis |volume=6 |issue=37--40 |pages=1847--1869 }}</ref> is built similarly to the [[ ratio test | ratio tests ]] hierarchy (see Section 4.1 of [[ ratio test]], and more specifically Subsection 4.1.4 there).

For a series <math>\sum_{n=1}^\infty a_n</math> with positive terms we have the following tests for convergence/divergence.

Let <math>K\geq1</math> be an integer, and let <math>\ln_{(K)}(x)</math> denote the <math>K</math>th [[iteration|iterate]] of [[natural logarithm]], i.e. <math>\ln_{(1)}(x)=\ln (x)</math> and for any <math>2\leq k\leq K</math>, 
<math>\ln_{(k)}(x)=\ln_{(k-1)}(\ln (x))</math>.

Suppose that <math>\sqrt[-n]{a_n}</math>, when <math>n</math> is large, can be presented in the form

:<math>\sqrt[-n]{a_n}=1+\frac{1}{n}+\frac{1}{n}\sum_{i=1}^{K-1}\frac{1}{\prod_{k=1}^i\ln_{(k)}(n)}+\frac{\rho_n}{n\prod_{k=1}^K\ln_{(k)}(n)}.</math>
(The empty sum is assumed to be 0.)

*	The series converges, if <math>\liminf_{n\to\infty}\rho_n>1</math>
*	The series diverges, if <math>\limsup_{n\to\infty}\rho_n<1</math>
*	Otherwise, the test is inconclusive.

=== Proof ===
Since <math>\sqrt[-n]{a_n}=\mathrm{e}^{-\frac{1}{n}\ln a_n}</math>, then we have

:<math>\mathrm{e}^{-\frac{1}{n}\ln a_n}=1+\frac{1}{n}+\frac{1}{n}\sum_{i=1}^{K-1}\frac{1}{\prod_{k=1}^i\ln_{(k)}(n)}+\frac{\rho_n}{n\prod_{k=1}^K\ln_{(k)}(n)}.</math>

From this,

:<math> \ln a_n=-n\ln\left(1+\frac{1}{n}+\frac{1}{n}\sum_{i=1}^{K-1}\frac{1}{\prod_{k=1}^i\ln_{(k)}(n)}+\frac{\rho_n}{n\prod_{k=1}^K\ln_{(k)}(n)}\right).</math>

From [[Taylor Series| Taylor's expansion]] applied to the right-hand side, we obtain:

:<math> \ln a_n=-1-\sum_{i=1}^{K-1}\frac{1}{\prod_{k=1}^i\ln_{(k)}(n)}-\frac{\rho_n}{\prod_{k=1}^K\ln_{(k)}(n)}+O\left(\frac{1}{n}\right).</math>

Hence,

:<math>a_n=\begin{cases}\mathrm{e}^{-1+O(1/n)}\frac{1}{(n\prod_{k=1}^{K-2}\ln_{(k)}n)\ln^{\rho_n}_{(K-1)}n}, &K\geq2,\\
\mathrm{e}^{-1+O(1/n)}\frac{1}{n^{\rho_n}}, &K=1.
\end{cases}
</math>
(The empty product is set to 1.)

The final result follows from the [[integral test for convergence]].