Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

8.4: Ratio and Root Tests

  • Page ID
    4202
  •  

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    The \(n^\text{th}\)--Term Test of Theorem 63 states that in order for a series \(\sum\limits_{n=1}^\infty a_n\) to converge, \(\lim\limits_{n\to\infty}a_n = 0\). That is, the terms of \(\{a_n\}\) must get very small. Not only must the terms approach 0, they must approach 0 "fast enough'': while \(\lim\limits_{n\to\infty}1/n=0\), the Harmonic Series \(\sum\limits_{n=1}^\infty \dfrac1n\) diverges as the terms of \(\{1/n\}\) do not approach 0 "fast enough.''

    The comparison tests of the previous section determine convergence by comparing terms of a series to terms of another series whose convergence is known. This section introduces the Ratio and Root Tests, which determine convergence by analyzing the terms of a series to see if they approach 0 "fast enough.''

    Ratio Test

    theorem 68: ratio test

    Let \(\{a_n\}\) be a positive sequence where \(\lim\limits_{n\to\infty}\dfrac{a_{n+1}}{a_n} = L\).

    1. If \(L<1\), then \(\sum\limits_{n=1}^\infty a_n\) converges.
    2. If \(L>1\) or \(L=\infty\), then \(\sum\limits_{n=1}^\infty a_n\) diverges.
    3. If \(L=1\), the Ratio Test is inconclusive.

    Theorem 64 allows us to apply the Ratio Test to series where \(\{a_n\}\) is positive for all but a finite number of terms.

    The principle of the Ratio Test is this: if \(\lim\limits_{n\to\infty}\dfrac{a_{n+1}}{a_n} = L<1\), then for large \(n\), each term of \(\{a_n\}\) is significantly smaller than its previous term which is enough to ensure convergence.

    Example \(\PageIndex{1}\): Applying the Ratio Test

    Use the Ratio Test to determine the convergence of the following series:

    1. \(\sum\limits_{n=1}^\infty \dfrac{2^n}{n!}\).
    2. \(\sum\limits_{n=1}^\infty \dfrac{3^n}{n^3} \)
    3. \(\sum\limits_{n=1}^\infty \dfrac{1}{n^2+1}.\)

    SOLUTION

    1. \(\sum\limits_{n=1}^\infty \dfrac{2^n}{n!}\): \[\begin{align*}\lim\limits_{n\to\infty}\dfrac{2^{n+1}/(n+1)!}{2^n/n!} &= \lim\limits_{n\to\infty} \dfrac{2^{n+1}n!}{2^n(n+1)!}\\&= \lim\limits_{n\to\infty} \dfrac{2}{n+1}\\&=0.\end{align*}\] Since the limit is \(0<1\), by the Ratio Test \(\sum\limits_{n=1}^\infty \dfrac{2^n}{n!}\) converges.
    2. \(\sum\limits_{n=1}^\infty \dfrac{3^n}{n^3}\): \[\begin{align*}\lim\limits_{n\to\infty} \dfrac{3^{n+1}/(n+1)^3}{3^n/n^3} &= \lim\limits_{n\to\infty}\dfrac{3^{n+1}n^3}{3^n(n+1)^3}\\&= \lim\limits_{n\to\infty} \dfrac{3n^3}{(n+1)^3}\\&= 3.\end{align*}\] Since the limit is \(3>1\), by the Ratio Test \(\sum\limits_{n=1}^\infty \dfrac{3^n}{n^3}\) diverges.
    3. \(\sum\limits_{n=1}^\infty \dfrac{1}{n^2+1}\): \[\begin{align*}\lim\limits_{n\to\infty} \dfrac{1/\big((n+1)^2+1\big)}{1/(n^2+1)} &= \lim\limits_{n\to\infty} \dfrac{n^2+1}{(n+1)^2+1}\\&= 1.\end{align*}\] Since the limit is 1, the Ratio Test is inconclusive. We can easily show this series converges using the Direct or Limit Comparison Tests, with each comparing to the series \(\sum\limits_{n=1}^\infty \dfrac{1}{n^2}\).

    The Ratio Test is not effective when the terms of a series only contain algebraic functions (e.g., polynomials). It is most effective when the terms contain some factorials or exponentials. The previous example also reinforces our developing intuition: factorials dominate exponentials, which dominate algebraic functions, which dominate logarithmic functions. In Part 1 of the example, the factorial in the denominator dominated the exponential in the numerator, causing the series to converge. In Part 2, the exponential in the numerator dominated the algebraic function in the denominator, causing the series to diverge.

    While we have used factorials in previous sections, we have not explored them closely and one is likely to not yet have a strong intuitive sense for how they behave. The following example gives more practice with factorials.

    Example \(\PageIndex{2}\): Applying the Ratio Test

    Determine the convergence of \(\sum\limits_{n=1}^\infty \dfrac{n!n!}{(2n)!}\).

    SOLUTION

    Before we begin, be sure to note the difference between \((2n)!\) and \(2n!\). When \(n=4\), the former is \(8!=8\cdot7\cdot\ldots\cdot 2\cdot1=40,320\), whereas the latter is \(2(4\cdot3\cdot2\cdot1) = 48\).

    Applying the Ratio Test:

    \[\begin{align*}
    \lim\limits_{n\to\infty} \dfrac{(n+1)!(n+1)!/\big(2(n+1)\big)!}{n!n!/(2n)!} &= \lim\limits_{n\to\infty}\dfrac{(n+1)!(n+1)!(2n)!}{n!n!(2n+2)!} \\ 
    \text{Noting that \((2n+2)! = (2n+2)\cdot(2n+1)\cdot(2n)!\), we have}& \\
    &= \lim\limits_{n\to\infty}\dfrac{(n+1)(n+1)}{(2n+2)(2n+1)}\\
    &= 1/4.
    \end{align*}\]

    Since the limit is \(1/4<1\), by the Ratio Test we conclude \(\sum\limits_{n=1}^\infty \dfrac{n!n!}{(2n)!}\) converges.

    Root Test

    The final test we introduce is the Root Test, which works particularly well on series where each term is raised to a power, and does not work well with terms containing factorials. 

    theorem 69: root test

    Let \(\{a_n\}\) be a positive sequence and let \(\lim\limits_{n\to \infty} (a_n)^{1/n} = L\).

    1. If \(L<1\), then \(\sum\limits_{n=1}^\infty a_n\) converges.
    2. If \(L>1\) or \(L=\infty\), then \(\sum\limits_{n=1}^\infty a_n\) diverges.
    3. If \(L=1\), the Root Test is inconclusive.

    Example \(\PageIndex{3}\): Applying the Root Test

    Determine the convergence of the following series using the Root Test:

    1. \(\sum\limits_{n=1}^\infty \left(\dfrac{3n+1}{5n-2}\right)^n\qquad\qquad 2. \sum\limits_{n=1}^\infty\dfrac{n^4}{(\ln n)^n}\qquad\qquad 3. \sum\limits_{n=1}^\infty \dfrac{2^n}{n^2}.\)

    SOLUTION

    1. \(\lim\limits_{n\to\infty} \left(\left(\dfrac{3n+1}{5n-2}\right)^n\right)^{1/n} = \lim\limits_{n\to\infty} \dfrac{3n+1}{5n-2} = \dfrac 35.\)
      Since the limit is less than 1, we conclude the series converges. Note: it is difficult to apply the Ratio Test to this series.
    2. \(\lim\limits_{n\to\infty} \left(\dfrac{n^4}{(\ln n)^n}\right)^{1/n} = \lim\limits_{n\to\infty} \dfrac {\big(n^{1/n}\big)^4}{\ln n} \). 
      As \(n\) grows, the numerator approaches 1 (apply L'H\^opital's Rule) and the denominator grows to infinity. Thus \[ \lim\limits_{n\to\infty} \dfrac{\big(n^{1/n}\big)^4}{\ln n} = 0.\] Since the limit is less than 1, we conclude the series converges.
    3. \(\lim\limits_{n\to\infty} \left(\dfrac{2^n}{n^2}\right)^{1/n} = \lim\limits_{n\to\infty} \dfrac{2}{\big(n^{1/n}\big)^2} = 2\).
      Since this is greater than 1, we conclude the series diverges.

    Each of the tests we have encountered so far has required that we analyze series from positive sequences. The next section relaxes this restriction by considering alternating series, where the underlying sequence has terms that alternate between being positive and negative.

    Contributors

    • Gregory Hartman (Virginia Military Institute). Contributions were made by Troy Siemers and Dimplekumar Chalishajar of VMI and Brian Heinold of Mount Saint Mary's University. This content is copyrighted by a Creative Commons Attribution - Noncommercial (BY-NC) License. http://www.apexcalculus.com/