Root test

Application to power series

This test can be used with a power series

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

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 a power series is the Cauchy–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.

Proof

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 ${\displaystyle \sqrt[n]{|a_n|}\le k<1}$, then ${\displaystyle |a_n|\le k^n<1}$. Since the geometric 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 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

${\displaystyle |z-p|<1/\underset{n\to \mathrm{\infty }}{lim sup}\sqrt[n]{|c_n|},}$

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

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

Examples

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.

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

${\displaystyle r=\underset{n\to \mathrm{\infty }}{lim sup}\sqrt[n]{|a_n|}=\underset{n\to \mathrm{\infty }}{lim sup}\sqrt[2n]{|a_{2n}|}=\underset{n\to \mathrm{\infty }}{lim sup}\sqrt[2n]{|1/2^n|}=\frac{1}{\sqrt{2}}<1.}$

This example shows how the root test is stronger than the ratio test. The ratio test is inconclusive for this series as if ${\displaystyle n}$ 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

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 ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{a}_n}$ 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 ${\displaystyle K}$th iterate of natural 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 ${\displaystyle \underset{n\to \mathrm{\infty }}{lim sup}{\rho }_n<1}$
• Otherwise, the test is inconclusive.

Proof

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,

${\displaystyle a_n=\{\begin{array}{ll}{\mathrm{e}}^{-1+O(1/n)}\frac{1}{(n\prod _{k=1}^{K-2}{\ln }_{(k)}n){\ln }_{(K-1)}^{{\rho }_n}n},& K\ge 2,\\ {\mathrm{e}}^{-1+O(1/n)}\frac{1}{n^{{\rho }_n}},& K=1.\end{array}}$

(The empty product is set to 1.)

The final result follows from the integral test for convergence.