9 Sequences and Series 9.5 Alternating Series and Absolute Convergence 9.7 Strategy for Testing Series

9.6 Ratio and Root Tests

Theorem 9.2.4 states that if a series $\displaystyle\sum_{n=1}^{\infty}a_{n}$ converges, then $\displaystyle\lim_{n\to\infty}a_{n}=0$ . 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 $\displaystyle\lim_{n\to\infty}1/n=0$ , the Harmonic Series $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n}$ diverges as the terms of $\{1/n\}$ do not approach 0 “fast enough.”

The comparison tests of Section 9.4 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 9.6.1 Ratio Test

Let $\{a_{n}\}$ be a sequence where $\displaystyle\lim_{n\to\infty}\left\lvert\frac{a_{n+1}}{a_{n}}\right\rvert=L$ .

(a)

If $L<1$ , then $\displaystyle\sum_{n=1}^{\infty}a_{n}$ converges.
(b)

If $L>1$ or $L=\infty$ , then $\displaystyle\sum_{n=1}^{\infty}a_{n}$ diverges.
(c)

If $L=1$ , the Ratio Test is inconclusive.

The principle of the Ratio Test is this: if $\displaystyle\lim_{n\to\infty}\left\lvert\frac{a_{n+1}}{a_{n}}\right\rvert=L<1$ , then for large $n$ , each term of $\{a_{n}\}$ is significantly smaller than its previous term which is enough to ensure convergence. A full proof can be found at http://tutorial.math.lamar.edu/Classes/CalcII/RatioTest.aspx.

Watch the video:
Using the Ratio Test to Determine if a Series Converges #1 from https://youtu.be/iy8mhbZTY7g

Example 9.6.1 Applying the Ratio Test

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

\text{1. }\sum_{n=1}^{\infty}\frac{2^{n}}{n!}\qquad\qquad\text{2. }\sum_{n=1}^% {\infty}\frac{3^{n}}{n^{3}}\qquad\qquad\text{3. }\sum_{n=1}^{\infty}\frac{1}{n% ^{2}+1}.

Solution

(a)

$\displaystyle\sum_{n=1}^{\infty}\frac{2^{n}}{n!}$ :
$\displaystyle\lim_{n\to\infty}\frac{2^{n+1}/(n+1)!}{2^{n}/n!}$ $\displaystyle=\lim_{n\to\infty}\frac{2^{n+1}n!}{2^{n}(n+1)!}$ $\displaystyle=\lim_{n\to\infty}\frac{2}{n+1}$ $\displaystyle=0.$ Since the limit is $0<1$ , by the Ratio Test $\displaystyle\sum_{n=1}^{\infty}\frac{2^{n}}{n!}$ converges.
(b)

$\displaystyle\sum_{n=1}^{\infty}\frac{3^{n}}{n^{3}}$ :
$\displaystyle\lim_{n\to\infty}\frac{3^{n+1}/(n+1)^{3}}{3^{n}/n^{3}}$ $\displaystyle=\lim_{n\to\infty}\frac{3^{n+1}n^{3}}{3^{n}(n+1)^{3}}$ $\displaystyle=\lim_{n\to\infty}\frac{3n^{3}}{(n+1)^{3}}$ $\displaystyle=3.$ Since the limit is $3>1$ , by the Ratio Test $\displaystyle\sum_{n=1}^{\infty}\frac{3^{n}}{n^{3}}$ diverges.
(c)

$\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{2}+1}$ :
$\displaystyle\qquad\lim_{n\to\infty}\frac{1/\bigl{(}(n+1)^{2}+1\bigr{)}}{1/(n^% {2}+1)}$ $\displaystyle=\lim_{n\to\infty}\frac{n^{2}+1}{(n+1)^{2}+1}$ $\displaystyle=1.$ 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 $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{2}}$ .

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 9.6.2 Applying the Ratio Test

Determine the convergence of $\displaystyle\sum_{n=1}^{\infty}\frac{n!n!}{(2n)!}$ .

SolutionBefore we begin, be sure to note the difference between $(2n)!$ and $2n!$ . When $n=4$ , the former is $8!=8\cdot 7\cdot\dotsm\cdot 2\cdot 1=40{,}320$ , whereas the latter is $2(4\cdot 3\cdot 2\cdot 1)=48$ .

Applying the Ratio Test:

	$\displaystyle\lim_{n\to\infty}\frac{(n+1)!(n+1)!/\bigl{(}2(n+1)\bigr{)}!}{n!n!% /(2n)!}$	$\displaystyle=\lim_{n\to\infty}\frac{(n+1)!(n+1)!(2n)!}{n!n!(2n+2)!}$
Noting that $(2n+2)!=(2n+2)\cdot(2n+1)\cdot(2n)!$ , we have
		$\displaystyle=\lim_{n\to\infty}\frac{(n+1)(n+1)}{(2n+2)(2n+1)}$
		$\displaystyle=1/4.$

Since the limit is $1/4<1$ , by the Ratio Test we conclude $\displaystyle\sum_{n=1}^{\infty}\frac{n!n!}{(2n)!}$ 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 9.6.2 Root Test

Let $\{a_{n}\}$ be a sequence where $\displaystyle\lim_{n\to\infty}\left\lvert a_{n}\right\rvert^{1/n}=L$ .

(a)

If $L<1$ , then $\displaystyle\sum_{n=1}^{\infty}a_{n}$ converges.
(b)

If $L>1$ or $L=\infty$ , then $\displaystyle\sum_{n=1}^{\infty}a_{n}$ diverges.
(c)

If $L=1$ , the Root Test is inconclusive.

Example 9.6.3 Applying the Root Test

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

\text{1.}\quad\sum_{n=1}^{\infty}\left(\frac{3n+1}{5n-2}\right)^{n}\qquad% \qquad\text{2.}\quad\sum_{n=2}^{\infty}\frac{n^{4}}{(\ln n)^{n}}\qquad\qquad% \text{3.}\quad\sum_{n=1}^{\infty}\frac{2^{n}}{n^{2}}.

Solution

(a)

$\displaystyle\lim_{n\to\infty}\left(\left(\frac{3n+1}{5n-2}\right)^{n}\right)^% {1/n}=\lim_{n\to\infty}\frac{3n+1}{5n-2}=\frac{3}{5}.$ Since the limit is less than 1, we conclude the series converges. Note: it is difficult to apply the Ratio Test to this series.
(b)

$\displaystyle\lim_{n\to\infty}\left(\frac{n^{4}}{(\ln n)^{n}}\right)^{1/n}=% \lim_{n\to\infty}\frac{\bigl{(}n^{1/n}\bigr{)}^{4}}{\ln n}$ . As $n$ grows, the numerator approaches 1 (apply L’Hôpital’s Rule) and the denominator grows to infinity. Thus

$\lim_{n\to\infty}\frac{\bigl{(}n^{1/n}\bigr{)}^{4}}{\ln n}=0.$

Since the limit is less than 1, we conclude the series converges.
(c)

$\displaystyle\lim_{n\to\infty}\left(\frac{2^{n}}{n^{2}}\right)^{1/n}=\lim_{n% \to\infty}\frac{2}{\bigl{(}n^{1/n}\bigr{)}^{2}}=2$ . Since this is greater than 1, we conclude the series diverges.

We end here our study of tests to determine convergence. The next section of this text provides strategies for testing series, while the back of the book contains a table summarizing the tests that one may find useful.

While series are worthy of study in and of themselves, our ultimate goal within calculus is the study of Power Series, which we will consider in Section 9.8. We will use power series to create functions where the output is the result of an infinite summation.

Exercises 9.6

Terms and Concepts

1.

The Ratio Test is not effective when the terms of a sequence only contain xxxxxx functions.
2.

The Ratio Test is most effective when the terms of a sequence contains xxxxxx and/or xxxxxx functions.
3.

What three convergence tests do not work well with terms containing factorials?
4.

The Root Test works particularly well on series where each term is xxxxxx to a xxxxxx.

Problems

In Exercises 5–16., determine the convergence of the given series using the Ratio Test. If the Ratio Test is inconclusive, state so and determine convergence with another test.

5.

$\displaystyle\sum_{n=0}^{\infty}\frac{2n}{n!}$
6.

$\displaystyle\sum_{n=0}^{\infty}\frac{5^{n}-3n}{4^{n}}$
7.

$\displaystyle\sum_{n=0}^{\infty}\frac{n!10^{n}}{(2n)!}$
8.

$\displaystyle\sum_{n=1}^{\infty}\frac{5^{n}+n^{4}}{7^{n}+n^{2}}$
9.

$\displaystyle\sum_{n=1}^{\infty}\frac{1}{n}$
10.

$\displaystyle\sum_{n=1}^{\infty}\frac{1}{3n^{3}+7}$
11.

$\displaystyle\sum_{n=1}^{\infty}\frac{10\cdot 5^{n}}{7^{n}-3}$
12.

$\displaystyle\sum_{n=1}^{\infty}n\cdot\left(\frac{3}{5}\right)^{n}$
13.

$\displaystyle\sum_{n=1}^{\infty}\frac{2\cdot 4\cdot 6\cdot 8\dotsm 2n}{3\cdot 6% \cdot 9\cdot 12\dotsm 3n}$
14.

$\displaystyle\sum_{n=1}^{\infty}\frac{n!}{5\cdot 10\cdot 15\dotsm(5n)}$
15.

$\displaystyle\sum_{n=1}^{\infty}e^{-n}n!$
16.

$\displaystyle\sum_{n=1}^{\infty}\frac{e^{1/n}}{n^{3}}$

In Exercises 17–26., determine the convergence of the given series using the Root Test. If the Root Test is inconclusive, state so and determine convergence with another test.

17.

$\displaystyle\sum_{n=1}^{\infty}\left(\frac{2n+5}{3n+11}\right)^{n}$
18.

$\displaystyle\sum_{n=1}^{\infty}\left(\frac{.9n^{2}-n-3}{n^{2}+n+3}\right)^{n}$
19.

$\displaystyle\sum_{n=1}^{\infty}\frac{2^{n}n^{2}}{3^{n}}$
20.

$\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{n}}$
21.

$\displaystyle\sum_{n=1}^{\infty}\frac{3^{n}}{n^{2}2^{n+1}}$
22.

$\displaystyle\sum_{n=1}^{\infty}\frac{4^{n+7}}{7^{n}}$
23.

$\displaystyle\sum_{n=1}^{\infty}\left(\frac{n^{2}-n}{n^{2}+n}\right)^{n}$
24.

$\displaystyle\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n^{2}}\right)^{n}$
25.

$\displaystyle\sum_{n=2}^{\infty}\frac{1}{\bigl{(}\ln n\bigr{)}^{n}}$
26.

$\displaystyle\sum_{n=2}^{\infty}\frac{n^{2}}{\bigl{(}\ln n\bigr{)}^{n}}$

27.

We know that the harmonic series $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots$ diverges. Suppose we remove some terms by considering the series $\sum_{n=1}^{\infty}\frac{1}{F_{n}}$ where $F_{n}$ is the $n^{\text{th}}$ Fibonacci number (so $F_{1}=1$ , $F_{2}=1$ , $F_{3}=2$ , $F_{4}=3$ , $F_{5}=5$ , …, and in general $F_{n}=F_{n-1}+F_{n-2}$ for $n\geq 3$ ). Determine if this series converges or diverges, using the fact that $\lim_{n\to\infty}\frac{F_{n+1}}{F_{n}}=\phi$ where $\phi=\frac{1}{2}(1+\sqrt{5})$ is known as the Golden Ratio.