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Root test explanation

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

∑ n = 1 ∞ a n {\displaystyle \sum _{n=1}^{\infty }a_{n}}

the root test uses the number

C = lim sup n → ∞ | a n | n , {\displaystyle C=\limsup _{n\rightarrow \infty }{\sqrt[{n}]{|a_{n}|}},}

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

lim n → ∞ | a n | n , {\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. ∑ 1 / n 2 {\displaystyle \textstyle \sum 1/{n^{2}}} , and there are others for which C = 1 and the series diverges, e.g. ∑ 1 / n {\displaystyle \textstyle \sum 1/n} .

Application to power series

This test can be used with a power series

f ( z ) = ∑ n = 0 ∞ c n ( z − p ) n {\displaystyle f(z)=\sum _{n=0}^{\infty }c_{n}(z-p)^{n}}

where the coefficients cn, 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 an = cn(z − p)n. One then applies the root test to the an 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 1 / lim sup n → ∞ | c n | n , {\displaystyle 1/\limsup _{n\rightarrow \infty }{\sqrt[{n}]{|c_{n}|}},} taking care that we really mean ∞ if the denominator is 0.

Proof

The proof of the convergence of a series Σan is an application of the comparison test.

If for all n ≥ N (N some fixed natural number) we have | a n | n ≤ k < 1 {\displaystyle {\sqrt[{n}]{|a_{n}|}}\leq k<1} , then | a n | ≤ k n < 1 {\displaystyle |a_{n}|\leq k^{n}<1} . Since the geometric series ∑ n = N ∞ k n {\displaystyle \sum _{n=N}^{\infty }k^{n}} converges so does ∑ n = N ∞ | a n | {\displaystyle \sum _{n=N}^{\infty }|a_{n}|} by the comparison test. Hence Σan converges absolutely.

If | a n | n > 1 {\displaystyle {\sqrt[{n}]{|a_{n}|}}>1} for infinitely many n, then an fails to converge to 0, hence the series is divergent.

Proof of corollary: For a power series Σan = Σcn(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

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

equivalent to

| c n | n ⋅ | z − p | < 1 {\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 | z − p | < 1 / | c n | n {\displaystyle |z-p|<1/{\sqrt[{n}]{|c_{n}|}}} for all sufficiently large n. This is equivalent to saying

| z − p | < 1 / lim sup n → ∞ | c n | n , {\displaystyle |z-p|<1/\limsup _{n\rightarrow \infty }{\sqrt[{n}]{|c_{n}|}},}

so R ≤ 1 / lim sup n → ∞ | c n | n . {\displaystyle R\leq 1/\limsup _{n\rightarrow \infty }{\sqrt[{n}]{|c_{n}|}}.} Now the only other place where convergence is possible is when

| a n | n = | c n ( z − p ) n | n = 1 , {\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

R = 1 / lim sup n → ∞ | c n | n . {\displaystyle R=1/\limsup _{n\rightarrow \infty }{\sqrt[{n}]{|c_{n}|}}.}

Examples

Example 1:

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

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

C = lim n → ∞ | 2 n n 9 | n = lim n → ∞ 2 n n n 9 n = lim n → ∞ 2 ( n 1 / n ) 9 = 2 {\displaystyle 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}

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

Example 2:

∑ n = 0 ∞ 1 2 ⌊ n / 2 ⌋ = 1 + 1 + 1 2 + 1 2 + 1 4 + 1 4 + 1 8 + 1 8 + … {\displaystyle \sum _{n=0}^{\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}}+\ldots }

The root test shows convergence because

r = lim sup n → ∞ | a n | n = lim sup n → ∞ | a 2 n | 2 n = lim sup n → ∞ | 1 / 2 n | 2 n = 1 2 < 1. {\displaystyle 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}|}}={\frac {1}{\sqrt {2}}}<1.}

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

Root tests hierarchy

Root tests hierarchy³⁴ 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 ∑ n = 1 ∞ a n {\displaystyle \sum _{n=1}^{\infty }a_{n}} with positive terms we have the following tests for convergence/divergence.

Let K ≥ 1 {\displaystyle K\geq 1} be an integer, and let ln ( K ) ⁡ ( x ) {\displaystyle \ln _{(K)}(x)} denote the K {\displaystyle K} th iterate of natural logarithm, i.e. ln ( 1 ) ⁡ ( x ) = ln ⁡ ( x ) {\displaystyle \ln _{(1)}(x)=\ln(x)} and for any 2 ≤ k ≤ K {\displaystyle 2\leq k\leq K} , ln ( k ) ⁡ ( x ) = ln ( k − 1 ) ⁡ ( ln ⁡ ( x ) ) {\displaystyle \ln _{(k)}(x)=\ln _{(k-1)}(\ln(x))} .

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

a n − n = 1 + 1 n + 1 n ∑ i = 1 K − 1 1 ∏ k = 1 i ln ( k ) ⁡ ( n ) + ρ n n ∏ k = 1 K ln ( k ) ⁡ ( n ) . {\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 lim inf n → ∞ ρ n > 1 {\displaystyle \liminf _{n\to \infty }\rho _{n}>1}
• The series diverges, if lim sup n → ∞ ρ n < 1 {\displaystyle \limsup _{n\to \infty }\rho _{n}<1}
• Otherwise, the test is inconclusive.

Proof

Since a n − n = e − 1 n ln ⁡ a n {\displaystyle {\sqrt[{-n}]{a_{n}}}=\mathrm {e} ^{-{\frac {1}{n}}\ln a_{n}}} , then we have

e − 1 n ln ⁡ a n = 1 + 1 n + 1 n ∑ i = 1 K − 1 1 ∏ k = 1 i ln ( k ) ⁡ ( n ) + ρ n n ∏ k = 1 K ln ( k ) ⁡ ( n ) . {\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,

ln ⁡ a n = − n ln ⁡ ( 1 + 1 n + 1 n ∑ i = 1 K − 1 1 ∏ k = 1 i ln ( k ) ⁡ ( n ) + ρ n n ∏ k = 1 K ln ( k ) ⁡ ( n ) ) . {\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:

ln ⁡ a n = − 1 − ∑ i = 1 K − 1 1 ∏ k = 1 i ln ( k ) ⁡ ( n ) − ρ n ∏ k = 1 K ln ( k ) ⁡ ( n ) + O ( 1 n ) . {\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,

a n = { e − 1 + O ( 1 / n ) 1 ( n ∏ k = 1 K − 2 ln ( k ) ⁡ n ) ln ( K − 1 ) ρ n ⁡ n , K ≥ 2 , e − 1 + O ( 1 / n ) 1 n ρ n , K = 1. {\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.

See also

• 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.

References

1. Bottazzini, Umberto (1986), The Higher Calculus: A History of Real and Complex Analysis from Euler to Weierstrass, Springer-Verlag, pp. 116–117, ISBN 978-0-387-96302-0. Translated from the Italian by Warren Van Egmond. 978-0-387-96302-0 ↩

2. Briggs, William; Cochrane, Lyle (2011). Calculus: Early Transcendentals. Addison Wesley. p. 571. https://archive.org/details/calculusearlytra0000brig/page/n5/mode/2up ↩

3. Abramov, Vyacheslav M. (2022). "Necessary and sufficient conditions for the convergence of positive series" (PDF). Journal of Classical Analysis. 19 (2): 117--125. arXiv:2104.01702. doi:10.7153/jca-2022-19-09. http://files.ele-math.com/articles/jca-19-09.pdf ↩

4. Bourchtein, Ludmila; Bourchtein, Andrei; Nornberg, Gabrielle; Venzke, Cristiane (2012). "A hierarchy of convergence tests related to Cauchy's test" (PDF). International Journal of Mathematical Analysis. 6 (37--40): 1847--1869. http://www.m-hikari.com/ijma/ijma-2012/ijma-37-40-2012/bourchteinIJMA37-40-2012.pdf ↩