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Inmathematics, theroot test is a criterion for theconvergence (aconvergence test) of aninfinite series. It depends on the quantity

${\displaystyle \underset{n\to \mathrm{\infty }}{lim sup}\sqrt[n]{|a_n|},}$

where${\displaystyle a_n}$ 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 withpower series.

The root test was developed first byAugustin-Louis Cauchy who published it in his textbookCours d'analyse (1821).^[1] Thus, it is sometimes known as theCauchy root test orCauchy's radical test. For a series

${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{a}_n}$

the root test uses the number

${\displaystyle C=\underset{n\to \mathrm{\infty }}{lim sup}\sqrt[n]{|a_n|},}$

where "lim sup" denotes thelimit superior, possibly +∞. Note that if

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\sqrt[n]{|a_n|},}$

converges then it equalsC and may be used in the root test instead.

The root test states that:

• ifC < 1 then the seriesconverges absolutely,
• ifC > 1 then the seriesdiverges,
• ifC = 1 and the limit approaches strictly from above then the series diverges,
• otherwise the test is inconclusive (the series may diverge, converge absolutely orconverge conditionally).

There are some series for whichC = 1 and the series converges, e.g.${\displaystyle \sum 1/n^2}$, and there are others for whichC = 1 and the series diverges, e.g.${\displaystyle \sum 1/n}$.

This test can be used with apower series

${\displaystyle f(z)=\sum _{n=0}^{\mathrm{\infty }}{c}_n(z-p{)}^n}$

where the coefficientsc_n, and the centerp arecomplex numbers and the argumentz is a complex variable.

The terms of this series would then be given bya_n =c_n(z −p)ⁿ. One then applies the root test to thea_n as above. Note that sometimes a series like this is called a power series "aroundp", because theradius of convergence is the radiusR of the largest interval or disc centred atp such that the series will converge for all pointsz strictly in the interior (convergence on the boundary of the interval or disc generally has to be checked separately).

Acorollary of the root test applied to a power series is theCauchy–Hadamard theorem: the radius of convergence is exactly${\displaystyle 1/\underset{n\to \mathrm{\infty }}{lim sup}\sqrt[n]{|c_n|},}$ taking care that we really mean ∞ if the denominator is 0.

The proof of the convergence of a series Σa_n is an application of thecomparison test.

If for alln ≥N (N some fixednatural number) we have, then. Since thegeometric series${\displaystyle \sum _{n=N}^{\mathrm{\infty }}{k}^n}$ converges so does${\displaystyle \sum _{n=N}^{\mathrm{\infty }}|a_n|}$ by the comparison test. Hence Σa_n converges absolutely.

If${\displaystyle \sqrt[n]{|a_n|}>1}$ for infinitely manyn, thena_n fails to converge to 0, hence the series is divergent.

Proof of corollary: For a power series Σa_n = Σc_n(z − p)ⁿ, we see by the above that the series converges if there exists anN such that for alln ≥N we have

equivalent to

for alln ≥N, which implies that in order for the series to converge we must have for all sufficiently largen. This is equivalent to saying

so${\displaystyle R\le 1/\underset{n\to \mathrm{\infty }}{lim sup}\sqrt[n]{|c_n|}.}$ Now the only other place where convergence is possible is when

${\displaystyle \sqrt[n]{|a_n|}=\sqrt[n]{|c_n(z-p{)}^n|}=1,}$

(since points > 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

${\displaystyle R=1/\underset{n\to \mathrm{\infty }}{lim sup}\sqrt[n]{|c_n|}.}$

Example 1:

${\displaystyle \sum _{i=1}^{\mathrm{\infty }}\frac{2^i}{i^9}}$

Applying the root test and using the fact that${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{n}^{1/n}=1,}$

${\displaystyle C=\underset{n\to \mathrm{\infty }}{lim}\sqrt[n]{\left|\frac{2^n}{n^9}\right|}=\underset{n\to \mathrm{\infty }}{lim}\frac{\sqrt[n]{2^n}}{\sqrt[n]{n^9}}=\underset{n\to \mathrm{\infty }}{lim}\frac{2}{(n^{1/n}{)}^9}=2}$

Since${\displaystyle C=2>1,}$ the series diverges.^[2]

Example 2:

${\displaystyle \sum _{n=0}^{\mathrm{\infty }}\frac{1}{2^{\lfloor n/2\rfloor }}=1+1+\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\dots }$

The root test shows convergence because

This example shows how the root test is stronger than theratio test. The ratio test is inconclusive for this series as if${\displaystyle n}$ is even,${\displaystyle a_{n+1}/a_n=1}$ while if${\displaystyle n}$ is odd,${\displaystyle a_{n+1}/a_n=1/2}$, therefore the limit${\displaystyle \underset{n\to \mathrm{\infty }}{lim}|a_{n+1}/a_n|}$ does not exist.

Root tests hierarchy^[3][4] is built similarly to the ratio tests hierarchy (see Section 4.1 ofratio test, and more specifically Subsection 4.1.4 there).

For a series${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{a}_n}$ with positive terms we have the following tests for convergence/divergence.

Let${\displaystyle K\ge 1}$ be an integer, and let${\displaystyle {\ln }_{(K)}(x)}$ denote the${\displaystyle K}$thiterate ofnatural logarithm, i.e.${\displaystyle {\ln }_{(1)}(x)=\ln (x)}$ and for any${\displaystyle 2\le k\le K}$,${\displaystyle {\ln }_{(k)}(x)={\ln }_{(k-1)}(\ln (x))}$.

Suppose that${\displaystyle \sqrt[-n]{a_n}}$, when${\displaystyle n}$ is large, can be presented in the form

${\displaystyle \sqrt[-n]{a_n}=1+\frac{1}{n}+\frac{1}{n}\sum _{i=1}^{K-1}\frac{1}{\prod _{k=1}^i{\ln }_{(k)}(n)}+\frac{{\rho }_n}{n\prod _{k=1}^K{\ln }_{(k)}(n)}.}$

(The empty sum is assumed to be 0.)

• The series converges, if${\displaystyle \underset{n\to \mathrm{\infty }}{lim inf}{\rho }_n>1}$
• The series diverges, if
• Otherwise, the test is inconclusive.

Since${\displaystyle \sqrt[-n]{a_n}={\mathrm{e}}^{-\frac{1}{n}\ln a_n}}$, then we have

${\displaystyle {\mathrm{e}}^{-\frac{1}{n}\ln a_n}=1+\frac{1}{n}+\frac{1}{n}\sum _{i=1}^{K-1}\frac{1}{\prod _{k=1}^i{\ln }_{(k)}(n)}+\frac{{\rho }_n}{n\prod _{k=1}^K{\ln }_{(k)}(n)}.}$

From this,

${\displaystyle \ln a_n=-n\ln \left(1+\frac{1}{n}+\frac{1}{n}\sum _{i=1}^{K-1}\frac{1}{\prod _{k=1}^i{\ln }_{(k)}(n)}+\frac{{\rho }_n}{n\prod _{k=1}^K{\ln }_{(k)}(n)}\right).}$

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

${\displaystyle \ln a_n=-1-\sum _{i=1}^{K-1}\frac{1}{\prod _{k=1}^i{\ln }_{(k)}(n)}-\frac{{\rho }_n}{\prod _{k=1}^K{\ln }_{(k)}(n)}+O\left(\frac{1}{n}\right).}$

Hence,

(The empty product is set to 1.)

The final result follows from theintegral test for convergence.

• Ratio test
• Convergent series

• Knopp, Konrad (1956). "§ 3.2".Infinite Sequences and Series. Dover publications, Inc., New York.ISBN 0-486-60153-6.
• Whittaker, E. T. & Watson, G. N. (1963). "§ 2.35".A Course in Modern Analysis (fourth ed.). Cambridge University Press.ISBN 0-521-58807-3.

This article incorporates material from Proof of Cauchy's root test onPlanetMath, which is licensed under theCreative Commons Attribution/Share-Alike License.

• Augustin-Louis Cauchy
• Convergence tests

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