Calculus II 09.06 The Ratio and Root Tests

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9.6 The Ratio and Root Tests

• Use the Ratio Test to determine whether a series converges or diverges.
• Use the Root Test to determine whether a series converges or diverges.
• Apply The Ratio and Root Tests to an infinite series.

The Ratio Test

Apply the Ratio Test to a series to find absolute convergence.

Theorem 9.6.1 Ratio Test

Let \( \sum a_{n} \) be a series with nonzero terms. $$1. \:\: \text{ The series }\: \sum a_{n} \: \text{ converges absolutely when } \lim_{n \to \infty} \left | \frac{a_{n+1}}{a_{n}} \right | < 1. $$ $$2. \:\: \text{ The series }\: \sum a_{n} \: \text{ diverges when } \lim_{n \to \infty} \left | \frac{a_{n+1}}{a_{n}} \right | > 1 \text{ or } \lim_{n \to \infty} \left | \frac{a_{n+1}}{a_{n}} \right | = \infty. $$ $$3. \:\: \text{ The Ratio Test is inconclusive when } \lim_{n \to \infty} \left | \frac{a_{n+1}}{a_{n}} \right | = 1. $$ Proof To prove Property 1, assume that

$$\lim_{n \to \infty} \left | \frac{a_{n+1}}{a_{n}} \right | = r< 1$$

and choose \(R\) such that \( 0 \leqslant r < R < 1\). By the Sequence Limit, there exists some \(N > 0\) such that \( |a_{n+1}/a_{n}| < R \) for all \(n > N\). Therefore the following inequalities hold.

\(|a_{n+1}|\)	\( < |a_{N}|R\)
\(|a_{n+2}|\)	\( < |a_{N+1}|R < |a_{N}|R^{2} \)
\(|a_{n+3}|\)	\( < |a_{N+2}|R < |a_{N+1}|R^{2} < |a_{N}|R^{3} \)
	\(\vdots\)

The geometric series

$$ \lim_{n \to \infty} |a_{N}|R^{n} = |a_{N}|R + |a_{N}|R^{2} + \cdots + |a_{N}|R^{n} + \cdots $$

converges. By the Direct Comparison Test, the series

$$ \lim_{n \to \infty}|a_{N+n}|=|a_{N=1}|+|a_{N=2}|+\cdots+|a_{N+n}|+\cdots $$

also converges. This in turn implies that the series \(\sum|a_{n}|\) also converges, because discarding a finite number of terms \((n=N-1\) does not affect convergence. Consequently, the series converges absolutely by the Ratio Test Theorem. The proof for Property 2 is similar.

Ratio Test is particularly useful for series that converge rapidly. Factorial and exponential series frequently converge rapidly and thus yield to the Ratio Test.

Example 9.6.1 Using the Ratio Test with Factorial

Determine the convergence or divergence for

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

Solution Because

$$ a_{n} = \frac{2^{n}}{n!}$$

the series can be written as

$$ \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_{n}}\right|$$	$$=\lim_{n \to \infty}\left[\frac{2^{n+1}}{(n+1)!} \div \frac{2^{n}}{n!}\right] $$
	$$= \lim_{n \to \infty}\left[\frac{2^{n+1}}{(n+1)!} \cdot \frac{n!}{2^{n}}\right] $$
	$$= \lim_{n \to \infty} \frac{2}{n+1} $$
	\(= 0 < 1\)

The series converges because the limit is less than 1.

Example 9.6.2 Using the Ratio Test with Factorial with \(n\)-Powers

Determine whether each series converges or diverges. $$a.\:\: \sum_{n=0}^{\infty} \frac{n^{2}2^{n+1}}{3^{n}}\:\:\:\:b.\:\: \sum_{n=1}^{\infty} \frac{n^{n}}{n!}$$ Solution
a. The series converges because the limit is less than 1.

$$ \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_{n}}\right|$$	$$=\lim_{n \to \infty}\left[(n+1)^{2} \frac{2^{n+2}}{3^{n+1}} \left ( \frac{2^{n+2}}{3^{n+1}}\right) \left( \frac{3^{n}}{n^{2}2^{n+1}}\right)\right] $$
	$$= \lim_{n \to \infty}\left[(n+1)^{2} \frac{2^{n+2}}{3^{n+1}} \left ( \frac{2^{n+2}}{3^{n+1}}\right) \left( \frac{3^{n}}{n^{2}2^{n+1}}\right)\right] $$
	$$= \frac{2}{3} < 1$$

b. The series diverges because the limit is greater than 1.

$$ \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_{n}}\right|$$	$$=\lim_{n \to \infty}\left[\frac{(n+1)^{n+1}}{(n+1)!} \left ( \frac{n!}{n^{n}}\right) \right] $$
	$$= \lim_{n \to \infty}\left[\frac{(n+1)^{n+1}}{(n+1)!} \left ( \frac{1}{n^{n}}\right) \right]$$
	$$=\lim_{n \to \infty} \frac{(n+1)^{n}}{n^{n}}$$
	$$=\lim_{n \to \infty} \left (1+ \frac{1}{n} \right)^{n}$$
	$$= e > 1$$

Example 9.6.3 Ratio Test Failure

Determine the convergence or divergence for

$$\sum_{n=1}^{\infty} (- 1 )^{n} \frac{\sqrt{n}}{n+1}. $$

Solution The limit is equal to 1.

$$ \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_{n}}\right|$$	$$=\lim_{n \to \infty}\left[ \left( \frac{\sqrt{n+1}}{n+2} \right ) \left ( \frac{n+1}{\sqrt{n}}\right) \right] $$
	$$= \lim_{n \to \infty}\left[ \sqrt{ \frac{n+1}{n}} \left ( \frac{n+1}{n+2}\right) \right] $$
	$$= \sqrt{1}(1) = 1$$

Therefore the Ratio Test is inconclusive. To determine convergence a different test is needed. Using the Alternating Series Test to show that \(a_{n+1} \leqslant a_{n}\), let

$$f(x)=\frac{\sqrt{x}}{x+1}. $$

Then the derivative is

$$f(x)=\frac{-x+1}{2\sqrt{x}(x+1)^{2}}. $$

The function \(f\) is decreasing because the derivative is negative for \(x > 1\). By L’Hôpital’s Rule

$$ \lim_{n \to \infty} \frac{\sqrt{x}}{x+1} $$	$$= \lim_{n \to \infty} \frac{1/(2\sqrt{x})}{1}$$
	$$= \lim_{n \to \infty} \frac{1}{2\sqrt{x}}$$
	$$= 0. $$

The series conditionally converges by the Alternating Series Test because the absolute version diverges, but the regular version does not.

The Root Test

The Root Test works well with series involving \(n\)th powers. The proof is similar for the Ratio Test.

Theorem 9.6.2 Root Test

$$1.\:\:\text{The series } \sum a_{n}\:\text{ converges absolutely when }\:\lim_{n \to \infty}\sqrt[n]{|a_{n}|}<1.$$ $$2.\:\:\text{The series } \sum a_{n}\:\text{ diverges when }\:\lim_{n \to \infty}\sqrt[n]{|a_{n}|}>1 \text{ or }\lim_{n \to \infty}\sqrt[n]{|a_{n}|}= \infty.$$ $$3.\:\:\text{The Root Test is inconclusive when }\:\lim_{n \to \infty}\sqrt[n]{|a_{n}|}=1.$$

Example 9.6.4 Using the Root Test

Determine whether the series converges or diverges.

$$\sum_{n=1}^{\infty} \frac{e^{2n}}{n^{n}}.$$

Solution Apply the Root Test

$$ \lim_{n \to \infty}\sqrt[n]{\left|a_{n}\right|}$$	$$=\lim_{n \to \infty}\sqrt[n]{ \frac{e^{2n}}{n^{n}}}$$
	$$= \lim_{n \to \infty} \frac{e^{2n/n}}{n^{n/n}} $$
	$$= 0 < 1$$

The series converges absolutely because the limit is less than 1. When choosing between the Root or Ratio test pick the one that makes finding the limit easier.

Strategies for Testing Series

• Does the \(n\)th term approach zero? If not, the series diverges.
• What type is it? Geometric, \(p\)-series, telescoping, or alternating.
• Can the Integral Test, the Root Test, or the Ratio Test be applied?
• Can the series be compared favorably to one special type?

Example 9.6.5 Applying the Strategies for Testing Series

Determine the convergence or divergence for each series.

$$a. \:\: \sum_{n=1}^{\infty} \frac{n+1}{3n+1}$$	$$b. \:\: \sum_{n=1}^{\infty} \left ( \frac{\pi}{6} \right )^{n}$$	$$c. \:\: \sum_{n=1}^{\infty} ne^{-n^{2}}$$
$$d. \:\: \sum_{n=1}^{\infty} \frac{1}{3n+1}$$	$$e. \:\: \sum_{n=1}^{\infty} (-1)^{n} \frac{3}{4n+1}$$	$$f. \:\: \sum_{n=1}^{\infty} \frac{n!}{10^{n}}$$
$$g. \:\: \sum_{n=1}^{\infty} \left ( \frac{n+1}{2n+1} \right )^{n}$$

Solution
a. The limit for the \(n\)th term is not 0, \(a_{n} \to 1/3 \text{ as } n \to \infty \). The series diverges by the \(n\)th-Term Test.
b. This series is geometric. Moreover, because the terms have the ratio

$$ r=\frac{\pi}{6} $$

which is less than 1 in absolute value, the series converges.
c. Because the function

$$ f(x)= xe^{-x^{2}} $$

is easily integrated. By the Integral Test the series converges.
d. The \(n\)th term for this series can be compared to the \(n\)th term for the harmonic series. The series diverges by The Limit Comparison Test.
e. This is an alternating series whose \(n\)th term approaches 0. Because \(a_{n+1} \leqslant a_{n}\), the series converges by the Alternating Series Test.
f. The \(n\)th term involves a factorial, which indicates that the Ratio Test may work well. After applying the Ratio Test, you can conclude that the series diverges.
g. The \(n\)th term for this series involves a variable that is raised to the \(n\)th power, which indicates that the Root Test may work well. After applying the Root Test, you can conclude that the series converges.

Tests for Series.
Figure 9.6.1

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Parent Article: Calculus II 09 Infinite Series