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Subsection 3.3.8 Optional — The Root Test

There is another test that is very similar in spirit to the ratio test. It also comes from a reexamination of the geometric series

\begin{gather*} \sum_{n=0}^\infty a_n = \sum_{n=0}^\infty a r^n \end{gather*}

The ratio test was based on the observation that \(r\text{,}\) which largely determines whether or not the series converges, could be found by computing the ratio \(r = a_{n+1}/a_n\text{.}\) The root test is based on the observation that \(|r|\) can also be determined by looking that the \(n^{\rm th}\) root of the \(n^{\rm th}\) term with \(n\) very large:

\begin{equation*} \lim_{n\to\infty}\root{n}\of{\big|ar^n\big|} =|r|\lim_{n\to\infty}\root{n}\of{\big|a\big|} =|r|\qquad\text{if $a\ne 0$} \end{equation*}

Of course, in general, the \(n^{\rm th}\) term is not exactly \(ar^n\text{.}\) However, if for very large \(n\text{,}\) the \(n^{\rm th}\) term is approximately proportional to \(r^n\text{,}\) with \(|r|\) given by the above limit, we would expect the series to converge when \(|r| \lt 1\) and diverge when \(|r| \gt 1\text{.}\) That is indeed the case.

Warning 3.3.25

Beware that the root test provides absolutely no conclusion about the convergence or divergence of the series \(\sum\limits_{n=1}^\infty a_n\) if \(\lim\limits_{n\rightarrow\infty}\root{n}\of{\big|a_n\big|} = 1\text{.}\)

(a) Pick any number \(R\) obeying \(L \lt R \lt 1\text{.}\) We are assuming that \(\root{n}\of{|a_n|}\) approaches \(L\) as \(n\rightarrow\infty\text{.}\) In particular there must be some natural number \(M\) so that \(\root{n}\of{|a_n|}\le R\) for all \(n\ge M\text{.}\) So \(|a_n|\le R^n\) for all \(n\ge M\) and the series \(\sum\limits_{n=1}^\infty a_n\) converges by comparison to the geometric series \(\sum\limits_{n=1}^\infty R^n\)

(b) We are assuming that \(\root{n}\of{|a_n|}\) approaches \(L \gt 1\) (or grows unboundedly) as \(n\rightarrow\infty\text{.}\) In particular there must be some natural number \(M\) so that \(\root{n}\of{|a_n|}\ge 1\) for all \(n\ge M\text{.}\) So \(|a_n|\ge 1\) for all \(n\ge M\) and the series diverges by the divergence test.

We have already used the ratio test, in Example 3.3.23, to show that this series converges when \(|x| \lt \frac{1}{3}\) and diverges when \(|x| \gt \frac{1}{3}\text{.}\) We'll now use the root test to draw the same conclusions.

  • Write \(a_n= \frac{ (-3)^n \sqrt{n+1}}{2n+3}x^n\text{.}\)
  • We compute
    \begin{align*} \root{n}\of{|a_n|} &= \root{n}\of{ \bigg|\frac{ (-3)^n \sqrt{n+1}}{2n+3}x^n\bigg|}\\ &= 3 |x|\big(n+1\big)^{\frac{1}{2n}} \big(2n+3)^{-\frac{1}{n}} \end{align*}
  • We'll now show that the limit of \(\big(n+1\big)^{\frac{1}{2n}}\) as \(n\to\infty\) is exactly \(1\text{.}\) To do, so we first compute the limit of the logarithm.
    \begin{align*} \lim_{n\to\infty}\log \big(n+1\big)^{\frac{1}{2n}} &=\lim_{n\to\infty}\frac{\log \big(n+1\big)}{2n} \qquad&\text{now apply Theorem }\knowl{./knowl/thm_SRxlimtoanlim.html}{\text{3.1.6}}\\ &=\lim_{t\to\infty}\frac{\log \big(t+1\big)}{2t}\\ &=\lim_{t\to\infty}\frac{\frac{1}{t+1}}{2} \qquad&\text{by l'Hôpital}\\ &=0 \end{align*}
    So
    \begin{gather*} \lim_{n\to\infty}\big(n+1\big)^{\frac{1}{2n}} =\lim_{n\to\infty}\exp\big\{\log \big(n+1\big)^{\frac{1}{2n}}\big\} = e^0=1 \end{gather*}
    An essentially identical computation also gives that \(\lim_{n\to\infty}\big(2n+3)^{-\frac{1}{n}} = e^0=1\text{.}\)
  • So
    \begin{gather*} \lim_{n\to\infty}\root{n}\of{|a_n|} = 3 |x| \end{gather*}

and the root test also tells us that if \(3|x| \gt 1\) the series diverges, while when \(3|x| \lt 1\) the series converges.

We have done the last example once, in Example 3.3.23, using the ratio test and once, in Example 3.3.26, using the root test. It was clearly much easier to use the ratio test. Here is an example that is most easily handled by the root test.

Write \(a_n= \big(\frac{n}{n+1}\big)^{n^2}\text{.}\) Then

\begin{align*} \root{n}\of{|a_n|} &= \root{n}\of{ \Big(\frac{n}{n+1}\Big)^{n^2}} = \Big(\frac{n}{n+1}\Big)^{n} = \Big(1+\frac{1}{n}\Big)^{-n} \end{align*}

Now we take the limit,

\begin{align*} \lim_{n\to\infty}\Big(1+\frac{1}{n}\Big)^{-n} &=\lim_{X\to\infty}\Big(1+\frac{1}{X}\Big)^{-X} \qquad&\text{by Theorem }\knowl{./knowl/thm_SRxlimtoanlim.html}{\text{3.1.6}}\\ &=\lim_{x\to 0}\big(1+x\big)^{-1/x} \qquad&\text{where $x=\frac{1}{X}$}\\ &= e^{-1} \end{align*}

by Example 3.7.20 in the CLP-1 text with \(a=-1\text{.}\) As the limit is strictly smaller than \(1\text{,}\) the series \(\sum_{n=1}^\infty \big(\frac{n}{n+1}\big)^{n^2}\) converges.

To draw the same conclusion using the ratio test, one would have to show that the limit of

\begin{align*} \frac{a_{n+1}}{a_n} &= \Big(\frac{n+1}{n+2}\Big)^{(n+1)^2} \Big(\frac{n+1}{n}\Big)^{n^2} \end{align*}

as \(n\rightarrow\infty\) is strictly smaller than 1. It's clearly better to stick with the root test.

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