Editing Root test

{{Short description|Criterion for the convergence of an infinite series}}
{{Calculus |Series}}

In [[mathematics]], the '''root test''' is a criterion for the [[Convergent series|convergence]] (a [[convergence test]]) of an [[infinite series]].  It depends on the quantity
:<math>\limsup_{n\rightarrow\infty}\sqrt[n]{|a_n|},</math>
where <math>a_n</math> are the terms of the series, and states that the series converges absolutely if this quantity is less than one, but diverges if it is greater than one.  It is particularly useful in connection with [[power series]].

== Root test explanation ==
[[File:Decision diagram for the root test.svg|thumb|Decision diagram for the root test]]
The root test was developed first by [[Augustin-Louis Cauchy]] who published it in his textbook [[Cours d'analyse]] (1821).<ref>{{citation|title=The Higher Calculus: A History of Real and Complex Analysis from Euler to Weierstrass|first=Umberto|last=Bottazzini|publisher=Springer-Verlag|year=1986|isbn=978-0-387-96302-0|pages=[https://archive.org/details/highercalculushi0000bott/page/116 116–117]|url=https://archive.org/details/highercalculushi0000bott/page/116}}. Translated from the Italian by Warren Van Egmond.</ref> Thus, it is sometimes known as the '''Cauchy root test''' or '''Cauchy's radical test'''. For a series

:<math>\sum_{n=1}^\infty a_n</math>

the root test uses the number

:<math>C = \limsup_{n\rightarrow\infty}\sqrt[n]{|a_n|},</math>

where "lim sup" denotes the [[limit superior]], possibly +∞. Note that if

:<math>\lim_{n\rightarrow\infty}\sqrt[n]{|a_n|},</math>

converges then it equals ''C'' and may be used in the root test instead.

The root test states that:
* if ''C'' < 1 then the series [[converges absolutely]],
* if ''C'' > 1 then the series [[divergent series|diverges]],
* if ''C'' = 1 and the limit approaches strictly from above then the series diverges,
* otherwise the test is inconclusive (the series may diverge, converge absolutely or [[converge conditionally]]).

There are some series for which ''C'' = 1 and the series converges, e.g. <math>\textstyle \sum 1/{n^2}</math>, and there are others for which ''C'' = 1 and the series diverges, e.g. <math>\textstyle\sum 1/n</math>.

==Application to power series==

This test can be used with a [[power series]]

:<math>f(z) = \sum_{n=0}^\infty c_n (z-p)^n</math>

where the coefficients ''c''<sub>''n''</sub>, and the center ''p'' are [[complex number]]s and the argument ''z'' is a complex variable.

The terms of this series would then be given by ''a''<sub>''n''</sub> = ''c''<sub>''n''</sub>(''z'' &minus; ''p'')<sup>''n''</sup>. One then applies the root test to the ''a''<sub>''n''</sub> as above. Note that sometimes a series like this is called a power series "around ''p''", because the [[radius of convergence]] is the radius ''R'' of the largest interval or disc centred at ''p'' such that the series will converge for all points ''z'' strictly in the interior (convergence on the boundary of the interval or disc generally has to be checked separately).

A [[corollary]] of the root test applied to a power series is the [[Cauchy–Hadamard theorem]]: the radius of convergence is exactly <math>1/\limsup_{n \rightarrow \infty}{\sqrt[n]{|c_n|}},</math> taking care that we really mean ∞ if the denominator is&nbsp;0.

== Proof ==
The proof of the convergence of a series Σ''a''<sub>''n''</sub> is an application of the [[Direct comparison test|comparison test]].

If for all ''n'' ≥ ''N'' (''N'' some fixed [[natural number]]) we have <math>\sqrt[n]{|a_n|} \le k < 1</math>, then <math>|a_n| \le k^n < 1</math>. Since the [[geometric series]] <math>\sum_{n=N}^\infty k^n</math> converges so does <math>\sum_{n=N}^\infty |a_n|</math> by the comparison test. Hence Σ''a''<sub>''n''</sub> converges absolutely. 

If <math>\sqrt[n]{|a_n|} > 1</math> for infinitely many ''n'', then ''a''<sub>''n''</sub> fails to converge to 0, hence the series is divergent.

'''Proof of corollary''': 
For a power series Σ''a''<sub>''n''</sub> = Σ''c''<sub>''n''</sub>(''z''&nbsp;&minus;&nbsp;''p'')<sup>''n''</sup>, we see by the above that the series converges if there exists an ''N'' such that for all ''n'' ≥ ''N'' we have

:<math>\sqrt[n]{|a_n|} = \sqrt[n]{|c_n(z - p)^n|} < 1,</math>

equivalent to

:<math>\sqrt[n]{|c_n|}\cdot|z - p| < 1</math>

for all ''n'' ≥ ''N'', which implies that in order for the series to converge we must have <math>|z - p| < 1/\sqrt[n]{|c_n|}</math> for all sufficiently large ''n''. This is equivalent to saying

:<math>|z - p| < 1/\limsup_{n \rightarrow \infty}{\sqrt[n]{|c_n|}},</math>

so <math>R \le 1/\limsup_{n \rightarrow \infty}{\sqrt[n]{|c_n|}}.</math> Now the only other place where convergence is possible is when

:<math>\sqrt[n]{|a_n|} = \sqrt[n]{|c_n(z - p)^n|} = 1,</math>

(since points &gt; 1 will diverge) and this will not change the radius of convergence since these are just the points lying on the boundary of the interval or disc, so

:<math>R = 1/\limsup_{n \rightarrow \infty}{\sqrt[n]{|c_n|}}.</math>

==Examples==

''Example 1:'' 
:<math> \sum_{i=1}^\infty \frac{2^i}{i^9} </math>
Applying the root test and using the fact that <math> \lim_{n \to \infty} n^{1/n}=1,</math>
::<math> C = \lim_{n \to \infty}\sqrt[n]{\left|\frac{2^n}{n^9}\right|}= \lim_{n \to \infty}\frac{ \sqrt[n]{2^n} } { \sqrt[n]{n^9} } = \lim_{n \to \infty}\frac{ 2 }   {(n^{1/n})^9 }  = 2 </math> 
Since <math> C=2>1,</math>  the series diverges.<ref>{{cite book
 | first1= William |last1= Briggs|first2= Lyle|last2 = Cochrane
 | title= Calculus: Early Transcendentals
 | url= https://archive.org/details/calculusearlytra0000brig/page/n5/mode/2up
 | url-access= registration
 | publisher= Addison Wesley
 | year=2011
 }}  p. 571. </ref>

''Example 2:'' 
:<math>\sum_{n=0}^\infty \frac{1}{2^{\lfloor n/2 \rfloor}}= 1 + 1 + \frac12 + \frac12 + \frac14 + \frac14 + \frac18 + \frac18  + \ldots </math>
The root test shows convergence because
:: <math>r= \limsup_{n\to\infty}\sqrt[n]{|a_n|} = \limsup_{n\to\infty}\sqrt[2n]{|a_{2n}|} = \limsup_{n\to\infty}\sqrt[2n]{|1/2^n|}=\frac1\sqrt{2}<1.</math>
This example shows how the root test is stronger than the [[ratio test]]. The ratio test is inconclusive for this series as if <math>n</math> is even, <math>a_{n+1}/a_n = 1</math> while if <math>n</math> is odd, <math>a_{n+1}/a_n = 1/2</math>, therefore the limit <math>\lim_{n\to\infty} |a_{n+1}/a_n|</math> does not exist.

== 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]].

==See also==
* [[Ratio test]]
* [[Convergent series]]

== References ==
<references/>
* {{cite book
 | author= Knopp, Konrad
 | title= Infinite Sequences and Series
 | url= https://archive.org/details/infinitesequence0000knop
 | url-access= registration
 | chapter = §&nbsp;3.2
 | publisher=Dover publications, Inc., New York
 | year=1956
 | isbn = 0-486-60153-6}} 
* {{cite book
 |author1=Whittaker, E. T. |author2=Watson, G. N.
  |name-list-style=amp | title= A Course in Modern Analysis
 | chapter = §&nbsp;2.35
 | edition=fourth
 | publisher=Cambridge University Press
 | year=1963
 | isbn = 0-521-58807-3}}

{{PlanetMath attribution|id=3934|title=Proof of Cauchy's root test}}
{{Calculus topics}}

[[Category:Augustin-Louis Cauchy]]
[[Category:Convergence tests]]
[[Category:Articles containing proofs]]

[[pl:Kryteria zbieżności szeregów#Kryterium Cauchy'ego]]