Root test

In mathematics, the root test is a criterion for the convergence (a convergence test) of an infinite series. It depends on the quantity
 * $$\limsup_{n\rightarrow\infty}\sqrt[n]{|a_n|},$$

where $$a_n$$ 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 with power series.

Root test explanation
The root test was developed first by Augustin-Louis Cauchy who published it in his textbook Cours d'analyse (1821). Thus, it is sometimes known as the Cauchy root test or Cauchy's radical test. For a series


 * $$\sum_{n=1}^\infty a_n$$

the root test uses the number


 * $$C = \limsup_{n\rightarrow\infty}\sqrt[n]{|a_n|},$$

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


 * $$\lim_{n\rightarrow\infty}\sqrt[n]{|a_n|},$$

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

Application to power series
This test can be used with a power series


 * $$f(z) = \sum_{n=0}^\infty c_n (z-p)^n$$

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 &minus; 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 $$1/\limsup_{n \rightarrow \infty}{\sqrt[n]{|c_n|}},$$ 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 $$\sqrt[n]{|a_n|} \le k < 1$$, then $$|a_n| \le k^n < 1$$. Since the geometric series $$\sum_{n=N}^\infty k^n$$ converges so does $$\sum_{n=N}^\infty |a_n|$$ by the comparison test. Hence Σan converges absolutely.

If $$\sqrt[n]{|a_n|} > 1$$ 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 &minus; p)n, we see by the above that the series converges if there exists an N such that for all n ≥ N we have


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

equivalent to


 * $$\sqrt[n]{|c_n|}\cdot|z - p| < 1$$

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


 * $$|z - p| < 1/\limsup_{n \rightarrow \infty}{\sqrt[n]{|c_n|}},$$

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


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

(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


 * $$R = 1/\limsup_{n \rightarrow \infty}{\sqrt[n]{|c_n|}}.$$

Examples
Example 1:
 * $$ \sum_{i=1}^\infty \frac{2^i}{i^9} $$

Applying the root test and using the fact that $$ \lim_{n \to \infty} n^{1/n}=1,$$
 * $$ C = \lim_{n \to \infty}\sqrt[n]{\left|\frac{2^n}{n^9}\right|}= \lim_{n \to \infty}\frac{ \sqrt[n]{2^n} } { \sqrt[n]{n^9} } = \lim_{n \to \infty}\frac{ 2 }  {(n^{1/n})^9 }  = 2 $$

Since $$ C=2>1,$$ the series diverges.

Example 2:
 * $$1 + 1 + 0.5 + 0.5 + 0.25 + 0.25 + 0.125 + 0.125 + ...  $$

The root test shows convergence because
 * $$r=\limsup_{n\to\infty}\sqrt[2n]{|a_{2n}|} = \limsup_{n\to\infty}\sqrt[2n]{|.5^n|}=\sqrt{.5}<1.$$

This example shows how the root test is stronger than the ratio test. The ratio test is inconclusive for this series if $$n$$ is odd so $$a_n=a_{n+1} = .5^n$$ (though not if $$n$$ is even), because
 * $$r=\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| = \lim_{n\to\infty}\left|\frac{ 2 \cdot.5^{n}}{2 \cdot.5^{n}}\right| =1. $$

Root tests hierarchy
Root tests hierarchy 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 $$\sum_{n=1}^\infty a_n$$ with positive terms we have the following tests for convergence/divergence.

Let $$K\geq1$$ be an integer, and let $$\ln_{(K)}(x)$$ denote the $$K$$th iterate of natural logarithm, i.e. $$\ln_{(1)}(x)=\ln (x)$$ and for any $$2\leq k\leq K$$, $$\ln_{(k)}(x)=\ln_{(k-1)}(\ln (x))$$.

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


 * $$\sqrt[-n]{a_n}=1+\frac{1}{n}+\frac{1}{n}\sum_{i=1}^{K-1}\frac{1}{\prod_{k=1}^i\ln_{(k)}(n)}+\frac{\rho_n}{n\prod_{k=1}^K\ln_{(k)}(n)}.$$

(The empty sum is assumed to be 0.)


 * The series converges, if $$\liminf_{n\to\infty}\rho_n>1$$
 * The series diverges, if $$\limsup_{n\to\infty}\rho_n<1$$
 * Otherwise, the test is inconclusive.

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


 * $$\mathrm{e}^{-\frac{1}{n}\ln a_n}=1+\frac{1}{n}+\frac{1}{n}\sum_{i=1}^{K-1}\frac{1}{\prod_{k=1}^i\ln_{(k)}(n)}+\frac{\rho_n}{n\prod_{k=1}^K\ln_{(k)}(n)}.$$

From this,


 * $$ \ln a_n=-n\ln\left(1+\frac{1}{n}+\frac{1}{n}\sum_{i=1}^{K-1}\frac{1}{\prod_{k=1}^i\ln_{(k)}(n)}+\frac{\rho_n}{n\prod_{k=1}^K\ln_{(k)}(n)}\right).$$

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


 * $$ \ln a_n=-1-\sum_{i=1}^{K-1}\frac{1}{\prod_{k=1}^i\ln_{(k)}(n)}-\frac{\rho_n}{\prod_{k=1}^K\ln_{(k)}(n)}+O\left(\frac{1}{n}\right).$$

Hence,


 * $$a_n=\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} $$ (The empty product is set to 1.)

The final result follows from the integral test for convergence.