Ratio and Root Tests

Learning Outcomes

Use the ratio test to determine absolute convergence of a series
Use the root test to determine absolute convergence of a series

Ratio Test

Consider a series $\displaystyle\sum _{n=1}^{\infty }{a}_{n}$ . From our earlier discussion and examples, we know that $\underset{n\to \infty }{\text{lim}}{a}_{n}=0$ is not a sufficient condition for the series to converge. Not only do we need ${a}_{n}\to 0$ , but we need ${a}_{n}\to 0$ quickly enough. For example, consider the series $\displaystyle\sum _{n=1}^{\infty }\frac{1}{n}$ and the series $\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{2}}$ . We know that $\frac{1}{n}\to 0$ and $\frac{1}{{n}^{2}}\to 0$ . However, only the series $\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{2}}$ converges. The series $\displaystyle\sum _{n=1}^{\infty }\frac{1}{n}$ diverges because the terms in the sequence $\left\{\frac{1}{n}\right\}$ do not approach zero fast enough as $n\to \infty$ . Here we introduce the ratio test, which provides a way of measuring how fast the terms of a series approach zero.

theorem: Ratio Test

Let $\displaystyle\sum _{n=1}^{\infty }{a}_{n}$ be a series with nonzero terms. Let

ρ = lim_{n \to \infty} | \frac{a_{n + 1}}{a_{n}} |

$\rho =\underset{n\to \infty }{\text{lim}}|\frac{{a}_{n+1}}{{a}_{n}}|$ .

If $0\le \rho <1$ , then $\displaystyle\sum _{n=1}^{\infty }{a}_{n}$ converges absolutely.
If $\rho >1$ or $\rho =\infty$ , then $\displaystyle\sum _{n=1}^{\infty }{a}_{n}$ diverges.
If $\rho =1$ , the test does not provide any information.

Proof

Let $\displaystyle\sum _{n=1}^{\infty }{a}_{n}$ be a series with nonzero terms.

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

| | \frac{a_{n + 1}}{a_{n}} | - ρ | < ϵ for all n \geq N

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

Therefore,

| \frac{a_{n + 1}}{a_{n}} | < ρ + ϵ = R for all n \geq N

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

and, thus,

[latex]\begin{array}{l}|{a}_{N+1}|

Since $R<1$ , the geometric series

R | a_{N} | + R^{2} | a_{N} | + R^{3} | a_{N} | + \dots

$R|{a}_{N}|+{R}^{2}|{a}_{N}|+{R}^{3}|{a}_{N}|+\cdots$

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

| a_{N + 1} | + | a_{N + 2} | + | a_{N + 3} | + | a_{N + 4} | + \dots

$|{a}_{N+1}|+|{a}_{N+2}|+|{a}_{N+3}|+|{a}_{N+4}|+\cdots$

converges. Therefore, since

\sum_{n = 1}^{\infty} | a_{n} | = \sum_{n = 1}^{N} | a_{n} | + \sum_{n = N + 1}^{\infty} | a_{n} |

$\displaystyle\sum _{n=1}^{\infty }|{a}_{n}|=\displaystyle\sum _{n=1}^{N}|{a}_{n}|+\displaystyle\sum _{n=N+1}^{\infty }|{a}_{n}|$

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

For part ii.

ρ = lim_{n \to \infty} | \frac{a_{n + 1}}{a_{n}} | > 1

$\rho =\underset{n\to \infty }{\text{lim}}|\frac{{a}_{n+1}}{{a}_{n}}|>1$ .

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

| | \frac{a_{n + 1}}{a_{n}} | - ρ | < ϵ for all n \geq N

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

Therefore,

R = ρ - ϵ < | \frac{a_{n + 1}}{a_{n}} | for all n \geq N

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

and, thus,

\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}

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

Since $R>1$ , the geometric series

R | a_{N} | + R^{2} | a_{N} | + R^{3} | a_{N} | + \dots

$R|{a}_{N}|+{R}^{2}|{a}_{N}|+{R}^{3}|{a}_{N}|+\cdots$

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

| a_{N + 1} | + | a_{N + 2} | + | a_{N + 3} | + \dots

$|{a}_{N+1}|+|{a}_{N+2}|+|{a}_{N+3}|+\cdots$

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

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

ρ = lim_{n \to \infty} \frac{\frac{1}{{(n + 1)}^{p}}}{\frac{1}{n^{p}}} = lim_{n \to \infty} \frac{n^{p}}{{(n + 1)}^{p}} = 1

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

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

$_\blacksquare$

The ratio test is particularly useful for series whose terms contain factorials or exponentials, where the ratio of terms simplifies the expression. The ratio test is convenient because it does not require us to find a comparative series. The drawback is that the test sometimes does not provide any information regarding convergence.

Before we apply the Ratio Test, we briefly review some laws of exponents and factorial notation, both of which will be of critical importance for using the Ratio Test.

Recall: Simplifying exponential expressions

For any $b > 0$ and real numbers $m$ and $n$ : $\frac{b^m}{b^n} = b^{m-n} = \frac{1}{b^{n-m}}$ .

Consider the examples below:

$\frac{e^{n+1}}{e^n} = e$
$\frac{3^{2n+1}}{3^{2n+3}} = \frac{1}{3^2} = \frac{1}{9}$

Recall: Factorial Notation

The expression $n!$ , called $n$ factorial, is defined as the product of the integer $n$ with all positive integers less than $n$ .

$n! = n (n-1) (n-2) \ldots 3 \cdot 2 \cdot 1$

For example, $5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120$

By definition, $0! = 1$

Note that a ratio of factorial expressions can be simplified. Consider the examples below:

$\frac{10!}{8!} = \frac{10 \cdot 9 \cdot 8 \cdot 7 \ldots 3 \cdot 2 \cdot 1 }{8 \cdot 7 \ldots 3 \cdot 2 \cdot 1} = 10\cdot 9 = 90$
$\frac {n!}{(n+2)!} = \frac{n!}{(n+2)(n+1)n!} = \frac{1}{(n+2)(n+1)}$

Example: Using the Ratio Test

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

$\displaystyle\sum _{n=1}^{\infty }\frac{{2}^{n}}{n\text{!}}$
$\displaystyle\sum _{n=1}^{\infty }\frac{{n}^{n}}{n\text{!}}$
$\displaystyle\sum _{n=1}^{\infty }\frac{{\left(-1\right)}^{n}{\left(n\text{!}\right)}^{2}}{\left(2n\right)\text{!}}$

Show Solution

From the ratio test, we can see that

$\rho =\underset{n\to \infty }{\text{lim}}\frac{\frac{{2}^{n+1}}{\left(n+1\right)\text{!}}}{\frac{{2}^{n}}{n\text{!}}}=\underset{n\to \infty }{\text{lim}}\frac{{2}^{n+1}}{\left(n+1\right)\text{!}}\cdot \frac{n\text{!}}{{2}^{n}}$ .

Since $\left(n+1\right)\text{!}=\left(n+1\right)\cdot n\text{!}$ ,

$\rho =\underset{n\to \infty }{\text{lim}}\frac{2}{n+1}=0$ .

Since $\rho <1$ , the series converges.
We can see that

$\begin{array}{cc}\rho \hfill & =\underset{n\to \infty }{\text{lim}}\frac{\frac{{\left(n+1\right)}^{n+1}}{\left(n+1\right)\text{!}}}{\frac{{n}^{n}}{n\text{!}}}\hfill \\ & =\underset{n\to \infty }{\text{lim}}\frac{{\left(n+1\right)}^{n+1}}{\left(n+1\right)\text{!}}\cdot \frac{n\text{!}}{{n}^{n}}\hfill \\ & =\underset{n\to \infty }{\text{lim}}{\left(\frac{n+1}{n}\right)}^{n}=\underset{n\to \infty }{\text{lim}}{\left(1+\frac{1}{n}\right)}^{n}=e.\hfill \end{array}$

Since $\rho >1$ , the series diverges.
Since

$\begin{array}{cc}\hfill |\frac{\frac{{\left(-1\right)}^{n+1}{\left(\left(n+1\right)\text{!}\right)}^{2}}{\left(2\left(n+1\right)\right)\text{!}}}{\frac{{\left(-1\right)}^{n}{\left(n\text{!}\right)}^{2}}{\left(2n\right)\text{!}}}|& =\frac{\left(n+1\right)\text{!}\left(n+1\right)\text{!}}{\left(2n+2\right)\text{!}}\cdot \frac{\left(2n\right)\text{!}}{n\text{!}n\text{!}}\hfill \\ & =\frac{\left(n+1\right)\left(n+1\right)}{\left(2n+2\right)\left(2n+1\right)}\hfill \end{array}$

we see that

$\rho =\underset{n\to \infty }{\text{lim}}\frac{\left(n+1\right)\left(n+1\right)}{\left(2n+2\right)\left(2n+1\right)}=\frac{1}{4}$ .

Since $\rho <1$ , the series converges.

try it

Use the ratio test to determine whether the series $\displaystyle\sum _{n=1}^{\infty }\frac{{n}^{3}}{{3}^{n}}$ converges or diverges.

Hint

Evaluate $\underset{n\to \infty }{\text{lim}}\frac{{\left(n+1\right)}^{3}}{{3}^{n+1}}\cdot \frac{{3}^{n}}{{n}^{3}}$ .

Show Solution

Watch the following video to see the worked solution to the above Try IT.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “5.6.1” here (opens in new window).

Try It

Root Test

The approach of the is similar to that of the ratio test. Consider a series $\displaystyle\sum _{n=1}^{\infty }{a}_{n}$ such that $\underset{n\to \infty }{\text{lim}}\sqrt[n]{|{a}_{n}|}=\rho$ for some real number $\rho$ . Then for $N$ sufficiently large, $|{a}_{N}|\approx {\rho }^{N}$ . Therefore, we can approximate $\displaystyle\sum _{n=N}^{\infty }|{a}_{n}|$ by writing

| a_{N} | + | a_{N + 1} | + | a_{N + 2} | + \dots \approx ρ^{N} + ρ^{N + 1} + ρ^{N + 2} + \dots

$|{a}_{N}|+|{a}_{N+1}|+|{a}_{N+2}|+\cdots \approx {\rho }^{N}+{\rho }^{N+1}+{\rho }^{N+2}+\cdots$ .

The expression on the right-hand side is a geometric series. As in the ratio test, the series $\displaystyle\sum _{n=1}^{\infty }{a}_{n}$ converges absolutely if $0\le \rho <1$ and the series diverges if $\rho \ge 1$ . If $\rho =1$ , the test does not provide any information. For example, for any p-series, $\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{p}}$ , we see that

ρ = lim_{n \to \infty} \sqrt[n]{| \frac{1}{n^{p}} |} = lim_{n \to \infty} \frac{1}{n^{p n}}

$\rho =\underset{n\to \infty }{\text{lim}}\sqrt[n]{|\frac{1}{{n}^{p}}|}=\underset{n\to \infty }{\text{lim}}\frac{1}{{n}^{{p}{n}}}$ .

To evaluate this limit, we use the natural logarithm function. Doing so, we see that

ln ρ = ln (lim_{n \to \infty} \frac{1}{n^{\frac{p}{n}}}) = lim_{n \to \infty} ln {(\frac{1}{n})}^{\frac{p}{n}} = lim_{n \to \infty} \frac{p}{n} \cdot ln (\frac{1}{n}) = lim_{n \to \infty} \frac{p ln (\frac{1}{n})}{n}

$\text{ln}\rho =\text{ln}\left(\underset{n\to \infty }{\text{lim}}\frac{1}{{n}^{\frac{p}{n}}}\right)=\underset{n\to \infty }{\text{lim}}\text{ln}{\left(\frac{1}{n}\right)}^{\frac{p}{n}}=\underset{n\to \infty }{\text{lim}}\frac{p}{n}\cdot \text{ln}\left(\frac{1}{n}\right)=\underset{n\to \infty }{\text{lim}}\frac{p\text{ln}\left(\frac{1}{n}\right)}{n}$ .

Using L’Hôpital’s rule, it follows that $\text{ln}\rho =0$ , and therefore $\rho =1$ for all $p$ . However, we know that the p-series only converges if $p>1$ and diverges if $p<1$ .

theorem: Root Test

Consider the series $\displaystyle\sum _{n=1}^{\infty }{a}_{n}$ . Let

ρ = lim_{n \to \infty} \sqrt[n]{| a_{n} |}

$\rho =\underset{n\to \infty }{\text{lim}}\sqrt[n]{|{a}_{n}|}$ .

If $0\le \rho <1$ , then $\displaystyle\sum _{n=1}^{\infty }{a}_{n}$ converges absolutely.
If $\rho >1$ or $\rho =\infty$ , then $\displaystyle\sum _{n=1}^{\infty }{a}_{n}$ diverges.
If $\rho =1$ , the test does not provide any information.

The root test is useful for series whose terms involve exponentials. In particular, for a series whose terms ${a}_{n}$ satisfy $|{a}_{n}|={b}_{n}^{n}$ , then $\sqrt[n]{|{a}_{n}|}={b}_{n}$ and we need only evaluate $\underset{n\to \infty }{\text{lim}}{b}_{n}$ .

Example: Using the Root Test

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

$\displaystyle\sum _{n=1}^{\infty }\frac{{\left({n}^{2}+3n\right)}^{n}}{{\left(4{n}^{2}+5\right)}^{n}}$
$\displaystyle\sum _{n=1}^{\infty }\frac{{n}^{n}}{{\left(\text{ln}\left(n\right)\right)}^{n}}$

Show Solution

To apply the root test, we compute

$\rho =\underset{n\to \infty }{\text{lim}}\sqrt[n]{\frac{{\left({n}^{2}+3n\right)}^{n}}{{\left(4{n}^{2}+5\right)}^{n}}}=\underset{n\to \infty }{\text{lim}}\frac{{n}^{2}+3n}{4{n}^{2}+5}=\frac{1}{4}$ .

Since $\rho <1$ , the series converges absolutely.
We have

$\rho =\underset{n\to \infty }{\text{lim}}\sqrt[n]{\frac{{n}^{n}}{{\left(\text{ln}n\right)}^{n}}}=\underset{n\to \infty }{\text{lim}}\frac{n}{\text{ln}n}=\infty \text{by L'Hôpital's rule}$ .

Since $\rho =\infty$ , the series diverges.

try it

Use the root test to determine whether the series $\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{n}}$ converges or diverges.

Hint

Evaluate $\underset{n\to \infty }{\text{lim}}\sqrt[n]{\frac{1}{{n}^{n}}}$ using L’Hôpital’s rule.

Show Solution

Watch the following video to see the worked solution to the above Try IT.

You can view the transcript for this segmented clip of “5.6.2” here (opens in new window).

Module 5: Sequences and Series

Learning Outcomes

Ratio Test

theorem: Ratio Test

Proof

Recall: Simplifying exponential expressions

Recall: Factorial Notation

Example: Using the Ratio Test

try it

Try It

Root Test

theorem: Root Test

Example: Using the Root Test

try it

Try It

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