Root test

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).[1] Thus, it is sometimes known as the Cauchy root test or Cauchy's radical test. For a series

${\displaystyle \sum _{n=1}^{\infty }a_{n}}$

the root test uses the number

${\displaystyle C=\limsup _{n\rightarrow \infty }{\sqrt[{n}]{|a_{n}|}},}$

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

${\displaystyle \lim _{n\rightarrow \infty }{\sqrt[{n}]{|a_{n}|}},}$

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 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. ${\displaystyle \textstyle \sum 1/{n^{2}}}$, and there are others for which C = 1 and the series diverges, e.g. ${\displaystyle \textstyle \sum 1/n}$.

This test can be used with a power series

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

where the coefficients c_n, and the center p are complex numbers and the argument z is a complex variable.

The terms of this series would then be given by a_n = c_n(z − p)ⁿ. One then applies the root test to the a_n 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 such a power series is the Cauchy–Hadamard theorem: the radius of convergence is exactly ${\displaystyle 1/\limsup _{n\rightarrow \infty }{\sqrt[{n}]{|c_{n}|}},}$ 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 the comparison test. If for all n ≥ N (N some fixed natural number) we have ${\displaystyle {\sqrt[{n}]{|a_{n}|}}\leq k<1}$, then ${\displaystyle |a_{n}|\leq k^{n}<1}$. Since the geometric series ${\displaystyle \sum _{n=N}^{\infty }k^{n}}$ converges so does ${\displaystyle \sum _{n=N}^{\infty }|a_{n}|}$ by the comparison test. Hence Σa_n converges absolutely.

If ${\displaystyle {\sqrt[{n}]{|a_{n}|}}>1}$ for infinitely many n, then a_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 an N such that for all n ≥ N we have

${\displaystyle {\sqrt[{n}]{|a_{n}|}}={\sqrt[{n}]{|c_{n}(z-p)^{n}|}}<1,}$

equivalent to

${\displaystyle {\sqrt[{n}]{|c_{n}|}}\cdot |z-p|<1}$

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

${\displaystyle |z-p|<1/\limsup _{n\rightarrow \infty }{\sqrt[{n}]{|c_{n}|}},}$

so ${\displaystyle R\leq 1/\limsup _{n\rightarrow \infty }{\sqrt[{n}]{|c_{n}|}}.}$ 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

${\displaystyle R=1/\limsup _{n\rightarrow \infty }{\sqrt[{n}]{|c_{n}|}}.}$

Example 1:

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

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

${\displaystyle C={\sqrt[{n}]{|{\frac {2^{n}}{n^{9}}}|}}={\frac {\sqrt[{n}]{2^{n}}}{\sqrt[{n}]{n^{9}}}}={\frac {2}{(n^{1/n})^{9}}}=2}$

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

Example 2:

${\displaystyle 1+1+0.5+0.5+0.25+0.25+0.125+0.125+...}$

The root test shows convergence because

${\displaystyle r=\limsup _{n\to \infty }{\sqrt[{n}]{|a_{n}|}}=\limsup _{n\to \infty }{\sqrt[{n}]{|.5^{n}|}}=.5<1.}$

This example shows how the root test is stronger than the ratio test. The ratio test is inconclusive for this series if ${\displaystyle n}$ is odd so ${\displaystyle a_{n}=a_{n+1}=.5^{n}}$ (though not if ${\displaystyle n}$ is even), because

${\displaystyle r=\lim _{n\to \infty }\left|{\frac {a_{n+1}}{a_{n}}}\right|=\lim _{n\to \infty }\left|{\frac {2\cdot .5^{n}}{2\cdot .5^{n}}}\right|=1.}$

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

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

Let ${\displaystyle K\geq 1}$ be an integer, and let ${\displaystyle \ln _{(K)}(x)}$ denote the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikipedia.org/v1/":): K th iterate of natural logarithm, i.e. ${\displaystyle \ln _{(1)}(x)=\ln(x)}$ and for any ${\displaystyle 2\leq k\leq K}$, ${\displaystyle \ln _{(k)}(x)=\ln _{(k-1)}(\ln(x))}$.

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 ${\displaystyle \liminf _{n\to \infty }\rho _{n}>1}$
• The series diverges, if ${\displaystyle \limsup _{n\to \infty }\rho _{n}<1}$
• 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,

${\displaystyle a_{n}={\begin{cases}\mathrm {e} ^{-1+O(1/n)}{\frac {1}{(n\prod _{k=1}^{K-2}\ln _{(k)}n)\ln _{(K-1)}^{\rho _{n}}n}},&K\geq 2,\\\mathrm {e} ^{-1+O(1/n)}{\frac {1}{n^{\rho _{n}}}},&K=1.\end{cases}}}$

(The empty product is set to 1.)

The final result follows from the integral 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 on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.