Root criterion

The root criterion is a mathematical convergence criterion for infinite series . Like the quotient criterion , it is based on a comparison with a geometric series .

The basic idea is as follows: A geometric series with positive, real terms converges if and only if the quotient of successive terms is less than 1. The -th root of the -th summand of this geometric series tends towards . If another series behaves in the same way, it is also convergent. Since it is even an absolute convergence , the rule can be generalized by looking at the amounts. ${\displaystyle q}$ ${\displaystyle n}$ ${\displaystyle n}$ ${\displaystyle q}$

The root criterion was first published in 1821 by the French mathematician Augustin Louis Cauchy in his textbook "Cours d'analysis". This is why it is also called the “Cauchy root criterion”.

Formulations

Decision tree for the root criterion

Given an infinite series with real or complex summands . If you now ${\displaystyle S=\sum _{n=0}^{\infty }a_{n}}$ ${\displaystyle a_{n}}$

${\displaystyle \limsup _{n\to \infty }{\sqrt[{n}]{|a_{n}|}}<1}$( stands for the Limes superior ) or ${\displaystyle \limsup }$

${\displaystyle {\sqrt[{n}]{|a_{n}|}}\leq C}$for one and almost all indices ${\displaystyle C<1}$ ${\displaystyle n}$

can prove, the series is absolutely convergent . I.e. the series itself and also the series converges. ${\displaystyle S}$ ${\displaystyle S}$ ${\displaystyle \sum _{n=0}^{\infty }|a_{n}|}$

But it is

${\displaystyle \limsup _{n\to \infty }{\sqrt[{n}]{|a_{n}|}}>1}$ or

${\displaystyle {\sqrt[{n}]{|a_{n}|}}\geq 1}$for an infinite number of indices , ${\displaystyle n}$

so the series diverges because the series members do not form a zero sequence .

In the case

${\displaystyle \limsup _{n\to \infty }{\sqrt[{n}]{|a_{n}|}}=1}$ and

${\displaystyle {\sqrt[{n}]{|a_{n}|}}<1}$for almost all indices ${\displaystyle n}$

nothing can be said about the convergence of the series. For example, with the root criterion, no statement can be made about the convergence of the general harmonic series for because ${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{\alpha }}}}$ ${\displaystyle \alpha \geq 1}$

${\displaystyle \limsup _{n\to \infty }{\sqrt[{n}]{|a_{n}|}}=\lim _{n\to \infty }{\sqrt[{n}]{1/n^{\alpha }}}=\left(\lim _{n\to \infty }{\sqrt[{n}]{1/n}}\right)^{\alpha }=1}$.

For is the general harmonic series divergent, for convergent; however, the root criterion cannot distinguish the two cases. ${\displaystyle \alpha =1}$ ${\displaystyle \alpha >1}$

Examples

Example 1. We examine the series

${\displaystyle \sum _{n=1}^{\infty }\left(1-{\frac {1}{n}}\right)^{n^{2}}}$

on convergence. Using the root criterion we get:

${\displaystyle \lim _{n\to \infty }{\sqrt[{n}]{|a_{n}|}}=\lim _{n\to \infty }{\sqrt[{n}]{\left(1-{\frac {1}{n}}\right)^{n^{2}}}}=\lim _{n\to \infty }\left(1-{\frac {1}{n}}\right)^{n}={\frac {1}{e}}<1}$

with Euler's number . Thus this series is convergent. ${\displaystyle e}$

Example 2. We now check the series

${\displaystyle \sum _{n=1}^{\infty }{\frac {n^{n}}{2^{n}n!}}}$

on convergence. We obtain:

${\displaystyle \lim _{n\to \infty }{\sqrt[{n}]{|a_{n}|}}=\lim _{n\to \infty }{\sqrt[{n}]{\frac {n^{n}}{2^{n}n!}}}=\lim _{n\to \infty }{\frac {1}{2}}\cdot {\frac {n}{\sqrt[{n}]{n!}}}={\frac {e}{2}}>1.}$

So this series is divergent.

Evidence sketch

The root criterion was first proven by Augustin Louis Cauchy. It follows with the majorant criterion from properties of the geometric series :

• Because if it applies to all , then the majorante criterion is fulfilled with a convergent geometric series as majorante. ${\displaystyle n\in \mathbb {N} :\;{\sqrt[{n}]{|a_{n}|}}\leq C<1}$ ${\displaystyle \forall n\in \mathbb {N} :\;|a_{n}|\leq C^{n}}$ ${\displaystyle \sum _{n=0}^{\infty }C^{n}={\frac {1}{1-C}}}$

• This also does not change anything if this criterion is not met for the first N terms of the series.

• If it holds , then it is fulfilled for almost all n , according to the definition of the largest accumulation point , with which a majorante can again be constructed. ${\displaystyle \limsup _{n\to \infty }{\sqrt[{n}]{|a_{n}|}}=C<1}$ ${\displaystyle {\sqrt[{n}]{|a_{n}|}}\leq {\frac {1+C}{2}}<1}$

Residual link estimation

If the row according to the root criterion converges, one receives an error estimate, i.e. H. an estimate of the remainder of the sum after N summands:

${\displaystyle S-S_{N}=\sum _{n=N+1}^{\infty }a_{n}\leq C^{N+1}{\frac {1}{1-C}}}$.

The root criterion is more stringent than the quotient criterion

Be a positive consequence and be ${\displaystyle (a_{n})\,}$

${\displaystyle \alpha =\liminf {\frac {a_{n+1}}{a_{n}}}\quad ,\quad \alpha '=\liminf {\sqrt[{n}]{a_{n}}}\quad ,\quad \beta '=\limsup {\sqrt[{n}]{a_{n}}}\quad ,\quad \beta =\limsup {\frac {a_{n+1}}{a_{n}}}}$.

Provides that for a range

Quotient criterion a decision (i.e. in the case of convergence or in the case of divergence), ${\displaystyle \beta <1}$ ${\displaystyle \alpha >1}$

so the root criterion also provides a decision (that is, in the case of convergence or in the case of divergence). ${\displaystyle \beta '<1}$ ${\displaystyle \alpha '>1}$

This is induced by the chain of inequalities

${\displaystyle 0\leq \alpha \leq \alpha '\leq \beta '\leq \beta \leq \infty .}$

If there is no restriction and , there is an index limit for every small but positive ( ) , from which the following applies: ${\displaystyle \alpha >0\,}$ ${\displaystyle \beta <\infty }$ ${\displaystyle \varepsilon }$ ${\displaystyle <\alpha \,}$ ${\displaystyle m}$

${\displaystyle \alpha -\varepsilon <{\frac {a_{k+1}}{a_{k}}}<\beta +\varepsilon \qquad \forall k\geq m.}$

If you multiply the inequality from to by, you get a telescope product in the middle : ${\displaystyle k=m}$ ${\displaystyle n-1}$

${\displaystyle (\alpha -\varepsilon )^{n-m}<{\frac {a_{n}}{a_{m}}}<(\beta +\varepsilon )^{n-m}.}$

If you then multiply by and pull the -th root, then is ${\displaystyle a_{m}}$ ${\displaystyle n}$

${\displaystyle {\sqrt[{n}]{a_{m}}}\,(\alpha -\varepsilon )^{1-{\frac {m}{n}}}<{\sqrt[{n}]{a_{n}}}<{\sqrt[{n}]{a_{m}}}\,(\beta +\varepsilon )^{1-{\frac {m}{n}}}.}$

For the left side converges against and the right side against . thats why ${\displaystyle n\to \infty }$ ${\displaystyle \alpha -\varepsilon }$ ${\displaystyle \beta +\varepsilon }$

${\displaystyle \alpha -\varepsilon \leq \liminf {\sqrt[{n}]{a_{n}}}}$     and     ${\displaystyle \limsup {\sqrt[{n}]{a_{n}}}\leq \beta +\varepsilon .}$

Since arbitrarily small can be chosen, it follows ${\displaystyle \varepsilon \,}$

${\displaystyle \alpha \leq \alpha '}$     and     ${\displaystyle \beta '\leq \beta .}$

For example, if the series members are and , then is and . ${\displaystyle a_{2n}={\frac {1}{2^{2n}}}}$ ${\displaystyle a_{2n+1}={\frac {4}{2^{2n+1}}}}$ ${\displaystyle {\frac {a_{2n+1}}{a_{2n}}}=2}$ ${\displaystyle {\frac {a_{(2n+1)+1}}{a_{2n+1}}}={\frac {1}{8}}}$

Here is and , according to which the quotient criterion does not provide a decision. ${\displaystyle \alpha ={\frac {1}{8}}\leq 1}$ ${\displaystyle \beta =2\geq 1}$

The root criterion provides a decision here because is. ${\displaystyle \alpha '=\beta '=\lim {\sqrt[{n}]{a_{n}}}={\frac {1}{2}}}$

From follows the convergence of . So the root criterion is really more stringent than the quotient criterion. ${\displaystyle \beta '={\frac {1}{2}}<1}$ ${\displaystyle \sum a_{n}}$

Web links

Wikibooks: Math for Non-Freaks: Root Criterion  - Learning and Teaching Materials

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1. ↑ See the answer to the question "Where is the root test first proved" on the Q&A website "History of Science and Mathematics"
2. ↑ Konrad Knopp: Theory and application of the infinite series. 5th edition Springer-Verlag, Berlin / Heidelberg 1964, ISBN 3-540-03138-3 . P. 286, sentence 161