Ratio and Root Tests: Learn It 1

  • Use the ratio test to check if a series converges absolutely
  • Use the root test to check if a series converges absolutely

You’ve already worked through several convergence tests, and you’ve likely noticed that no single test works for every series. The ratio test and root test complete our toolkit by offering particularly useful methods that don’t require finding a comparable series. The ratio test will prove especially valuable when we explore power series in the next module.

Ratio Test

Consider a series [latex]\displaystyle\sum_{n=1}^{\infty}{a}{n}[/latex]. We know that having [latex]\lim{n\to \infty}{a}_{n}=0[/latex] is necessary for convergence, but it’s not sufficient. The terms must approach zero quickly enough.

Think about these two familiar examples:

  • [latex]\displaystyle\sum_{n=1}^{\infty}\frac{1}{n}[/latex] (harmonic series) diverges
  • [latex]\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{2}}[/latex] (p-series with [latex]p=2[/latex]) converges

Both have terms that approach zero, but [latex]\frac{1}{n^{2}}[/latex] approaches zero much faster than [latex]\frac{1}{n}[/latex]. The ratio test provides a systematic way to measure this “speed” of approach to zero.

ratio test

Let [latex]\displaystyle\sum_{n=1}^{\infty}{a}_{n}[/latex] be a series with nonzero terms. Let

[latex]\rho =\lim_{n\to \infty}\left|\frac{a_{n+1}}{a_{n}}\right|[/latex]

Then:

  • If [latex]0\le \rho <1[/latex], the series [latex]\displaystyle\sum_{n=1}^{\infty}{a}_{n}[/latex] converges absolutely
  • If [latex]\rho >1[/latex] or [latex]\rho =\infty[/latex], the series [latex]\displaystyle\sum_{n=1}^{\infty}{a}_{n}[/latex] diverges
  • If [latex]\rho =1[/latex], the test is inconclusive and does not provide any information

Proof

Let [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex] be a series with nonzero terms.

We begin with the proof of part i. In this case, [latex]\rho =\underset{n\to \infty }{\text{lim}}|\frac{{a}_{n+1}}{{a}_{n}}|<1[/latex]. Since [latex]0\le \rho < 1[/latex], there exists [latex]R[/latex] such that [latex]0 \le \rho < R < 1[/latex]. Let [latex]\epsilon =R-\rho >0[/latex]. By the definition of limit of a sequence, there exists some integer [latex]N[/latex] such that

[latex]||\frac{{a}_{n+1}}{{a}_{n}}|-\rho |<\epsilon \text{ for all }n\ge N[/latex].

Therefore,

[latex]|\frac{{a}_{n+1}}{{a}_{n}}|<\rho +\epsilon =R\text{ for all }n\ge N[/latex]

and, thus,

[latex]\begin{array}{l}|{a}_{N+1}| < R|{a}_{N}|\\ |{a}_{N+2}| < R|{a}_{N+1}| < {R}^{2}|{a}_{N}|\\ |{a}_{N+3}| < R|{a}_{N+2}| < {R}^{2}|{a}_{N+1}|<{R}^{3}|{a}_{N}|\\ |{a}_{N+4}| < R|{a}_{N+3}| < {R}^{2}|{a}_{N+2}| < {R}^{3}|{a}_{N+1}| < {R}^{4}|{a}_{N}|\\ \vdots .\end{array}[/latex]

Since [latex]R<1[/latex], the geometric series

[latex]R|{a}_{N}|+{R}^{2}|{a}_{N}|+{R}^{3}|{a}_{N}|+\cdots[/latex]

converges. Given the inequalities above, we can apply the comparison test and conclude that the series

[latex]|{a}_{N+1}|+|{a}_{N+2}|+|{a}_{N+3}|+|{a}_{N+4}|+\cdots[/latex]

converges. Therefore, since

[latex]\displaystyle\sum _{n=1}^{\infty }|{a}_{n}|=\displaystyle\sum _{n=1}^{N}|{a}_{n}|+\displaystyle\sum _{n=N+1}^{\infty }|{a}_{n}|[/latex]

where [latex]\displaystyle\sum _{n=1}^{N}|{a}_{n}|[/latex] is a finite sum and [latex]\displaystyle\sum _{n=N+1}^{\infty }|{a}_{n}|[/latex] converges, we conclude that [latex]\displaystyle\sum _{n=1}^{\infty }|{a}_{n}|[/latex] converges.

For part ii.

[latex]\rho =\underset{n\to \infty }{\text{lim}}|\frac{{a}_{n+1}}{{a}_{n}}|>1[/latex].

Since [latex]\rho >1[/latex], there exists [latex]R[/latex] such that [latex]\rho >R>1[/latex]. Let [latex]\epsilon =\rho -R>0[/latex]. By the definition of the limit of a sequence, there exists an integer [latex]N[/latex] such that

[latex]||\frac{{a}_{n+1}}{{a}_{n}}|-\rho |<\epsilon \text{for all}n\ge N[/latex].

Therefore,

[latex]R=\rho -\epsilon <|\frac{{a}_{n+1}}{{a}_{n}}|\text{for all}n\ge N[/latex],

and, thus,

[latex]\begin{array}{l}|{a}_{N+1}|>R|{a}_{N}|\\ |{a}_{N+2}|>R|{a}_{N+1}|>{R}^{2}|{a}_{N}|\\ |{a}_{N+3}|>R|{a}_{N+2}|>{R}^{2}|{a}_{N+1}|>{R}^{3}|{a}_{N}|\\ |{a}_{N+4}|>R|{a}_{N+3}|>{R}^{2}|{a}_{N+2}|>{R}^{3}|{a}_{N+1}|>{R}^{4}|{a}_{N}|.\end{array}[/latex]

Since [latex]R>1[/latex], the geometric series

[latex]R|{a}_{N}|+{R}^{2}|{a}_{N}|+{R}^{3}|{a}_{N}|+\cdots[/latex]

diverges. Applying the comparison test, we conclude that the series

[latex]|{a}_{N+1}|+|{a}_{N+2}|+|{a}_{N+3}|+\cdots[/latex]

diverges, and therefore the series [latex]\displaystyle\sum _{n=1}^{\infty }|{a}_{n}|[/latex] diverges.

For part iii. we show that the test does not provide any information if [latex]\rho =1[/latex] by considering the [latex]p-\text{series}[/latex] [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{p}}[/latex]. For any real number [latex]p[/latex],

[latex]\rho =\underset{n\to \infty }{\text{lim}}\frac{\frac{1}{{\left(n+1\right)}^{p}}}{\frac{1}{{n}^{p}}}=\underset{n\to \infty }{\text{lim}}\frac{{n}^{p}}{{\left(n+1\right)}^{p}}=1[/latex].

However, we know that if [latex]p\le 1[/latex], the [latex]p-\text{series}[/latex] [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{p}}[/latex] diverges, whereas [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{p}}[/latex] converges if [latex]p>1[/latex].

[latex]_\blacksquare[/latex]

When to Use the Ratio Test

The ratio test works particularly well for series containing factorials or exponentials, where the ratio [latex]\frac{a_{n+1}}{a_n}[/latex] often simplifies nicely. This makes it convenient since you don’t need to hunt for a comparison series. However, the test has limitations—sometimes it provides no useful information, especially when [latex]\rho = 1[/latex].

Before applying the ratio test, let’s review some essential algebra that will make your calculations much smoother.

Simplifying Exponential Expressions

For any [latex]b > 0[/latex] and real numbers [latex]m[/latex] and [latex]n[/latex]: [latex]\frac{b^m}{b^n} = b^{m-n} = \frac{1}{b^{n-m}}[/latex]

Examples:

  • [latex]\frac{e^{n+1}}{e^n} = e[/latex]
  • [latex]\frac{3^{2n+1}}{3^{2n+3}} = \frac{1}{3^2} = \frac{1}{9}[/latex]

Factorial Notation

The expression [latex]n![/latex], called “[latex]n[/latex] factorial,” is defined as: [latex]n! = n(n-1)(n-2)\ldots 3 \cdot 2 \cdot 1[/latex]

Key facts:

  • [latex]5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120[/latex]
  • [latex]0! = 1[/latex] (by definition)

Simplifying factorial ratios:

  • [latex]\frac{10!}{8!} = \frac{10 \cdot 9 \cdot 8!}{8!} = 10 \cdot 9 = 90[/latex]
  • [latex]\frac{n!}{(n+2)!} = \frac{n!}{(n+2)(n+1)n!} = \frac{1}{(n+2)(n+1)}[/latex]

For each of the following series, use the ratio test to determine whether the series converges or diverges.

  1. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{{2}^{n}}{n\text{!}}[/latex]
  2. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{{n}^{n}}{n\text{!}}[/latex]
  3. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{{\left(-1\right)}^{n}{\left(n\text{!}\right)}^{2}}{\left(2n\right)\text{!}}[/latex]