OBJECTIVES

When you finish this section, you should be able to:

1. Use the Ratio Test (p. 591)
2. Use the Root Test (p. 593)

THEOREM Ratio Test

Let $\sum\limits_{k\,=\,1}^{\infty }a_{k}$ be a series of nonzero terms.

1. Let $\lim\limits_{n\rightarrow \infty }\left\vert \dfrac{a_{n+1}}{a_{n}}\right\vert =L,$ a number.
   • If $\ L<1$, then the series $\sum\limits_{k\,=\,1}^{\infty }a_{k}$ converges absolutely and so $\sum\limits_{k\,=\,1}^{\infty }a_{k}$ is convergent.
   • If $L=1$, the test fails to indicate whether the series converges or diverges.
   • If $L>1,$ then the series $\sum\limits_{k\,=\,1}^{\infty }a_{k}$ diverges.
2. If $\lim\limits_{n\,\rightarrow \,\infty }\left\vert \dfrac{a_{n+1}}{ a_{n}}\right\vert =\infty$, then the series $\sum\limits_{k\,=\,1}^{ \infty }a_{k}$ diverges.

EXAMPLE 1Using the Ratio Test

Use the Ratio Test to determine whether each series converges or diverges.

Solution (a) $\sum\limits_{k\,=\,1}^{\infty }\dfrac{k}{4^{k}}$ is a series of nonzero terms; $a_{n+1}=\dfrac{n+1}{4^{n+1}}$ and $a_{n}= \dfrac{n}{4^{n}}$. The absolute value of their ratio is $$\left\vert \frac{a_{n+1}}{a_{n}}\right\vert =\dfrac{\dfrac{n+1}{4^{n+1}}}{ \dfrac{n}{4^{n}}}=\dfrac{n+1}{4^{n+1}}\cdot \dfrac{4^{n}}{n}=\frac{n+1}{4n}$$

Then $$\lim\limits_{n\,\rightarrow \,\infty }\,\left\vert \frac{a_{n+1}}{a_{n}} \right\vert =\lim\limits_{n\,\rightarrow \,\infty }\frac{n+1}{4n}=\frac{1}{4} <1$$

Since the limit is less than $1$, the series converges.

(b) $\sum\limits_{k\,=\,1}^{\infty }\dfrac{2^{k}}{k}$ is a series of nonzero terms; $a_{n+1}=\dfrac{2^{n+1}}{n+1}$ and $a_{n}=\dfrac{ 2^{n}}{n}.$ The absolute value of their ratio is $$\left\vert \frac{a_{n+1}}{a_{n}}\right\vert =\frac{2^{n+1}}{n+1} \cdot \frac{ n}{2^{n}} =\frac{2n}{n+1}$$

Then $$\lim\limits_{n\,\rightarrow \,\infty }\left\vert \frac{a_{n+1}}{a_{n}} \right\vert =\lim\limits_{n\,\rightarrow \,\infty }\frac{2n}{n+1}=2>1$$

Since the limit is greater than $1$, the series diverges.

(c) $\sum\limits_{k\,=\,1}^{\infty }\dfrac{3k+1}{k^{2}}$ is a series of nonzero terms; $a_{n+1}=\dfrac{3(n+1) +1}{(n+1) ^{2}}=\dfrac{3n+4}{(n+1) ^{2}}$ and $a_{n}=\dfrac{3n+1 }{n^{2}}$. The absolute value of their ratio is $$\left\vert \frac{a_{n+1}}{a_{n}}\right\vert {=}\left[ \frac{3n+4}{(n+1)^{2}} \right] \left( \frac{n^{2}}{3n+1}\right) {=}\left( \frac{3n+4}{3n+1}\right) \left( \frac{n^{2}}{n^{2}+2n+1}\right) =\dfrac{3n^{3}+4n^{2}}{3n^{3}+7n^{2}+5n+1}$$

Then $$\lim\limits_{n\,\rightarrow \,\infty }\left\vert \frac{a_{n+1}}{a_{n}} \right\vert =\lim\limits_{n\,\rightarrow \,\infty }\dfrac{3n^{3}+4n^{2}}{3n^{3}+7n^{2}+5n+1}=1$$

The Ratio Test gives no information about this series. Another test must be used. $\bigg({\rm{You}}\, {\rm{can}}\, {\rm{show}}\, {\rm{that}}\, {\rm{the}}\, {\rm{series}}\, {\rm{diverges}}\, {\rm{by}}\, {\rm{comparing}}\, {\rm{it}}\, {\rm{to}}\, {\rm{the}}\, {\rm{harmonic}}\, {\rm{series}}\, \sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{k}.\bigg)$

(d) $\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{k!}$ is a series of nonzero terms; $a_{n+1}=\dfrac{1}{(n+1) !}$ and $a_{n}=\dfrac{ 1}{n!}$. The absolute value of their ratio is $$\left\vert \frac{a_{n+1}}{a_{n}}\right\vert =\frac{n!}{(n+1)!} = \frac{1}{n+1}$$

Then $$\lim_{n\rightarrow \infty }\left\vert \frac{a_{n+1}}{a_{n}}\right\vert =\lim_{n\rightarrow \infty }\frac{1}{n+1}=0$$

Since the limit is less than $1,$ the series $\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{k!}$ converges.

NOW WORK

Problem 7.

EXAMPLE 2Using the Ratio Test

Use the Ratio Test to determine whether the series $\sum\limits_{k\,=\,1}^{\infty }\dfrac{k!}{k^{k}}$ converges or diverges.

Solution $\sum\limits_{k\,=\,1}^{\infty }\dfrac{k!}{k^{k}}$ is a series of nonzero terms. Since $a_{n+1}=\dfrac{(n+1)!}{(n+1) ^{n+1}}$ and $a_{n}=\dfrac{n!}{n^{n}}$, the absolute value of their ratio is $$\begin{eqnarray*} \begin{array}{@{\hspace*{-0pc}}rcl} \displaystyle\left\vert \dfrac{a_{n+1}}{a_{n}}\right\vert &=& \dfrac{\dfrac{(n+1)!}{(n+1) ^{n+1}}}{\dfrac{n!}{n^{n}}}=\dfrac{(n+1)!}{(n+1)^{n+1}}\cdot \dfrac{n^{n}}{n!}=\dfrac{n^{n}}{(n+1)^{n}}\\ &=&\left( \dfrac{n}{n+1}\right)^{n} = \left( \dfrac{1}{1+\dfrac{1}{n}}\right)^{n} = \dfrac{1}{\left( 1+\dfrac{1}{n}\right) ^{n}}\qquad \end{array} \end{eqnarray*}$$

So, $$\lim\limits_{n\,\rightarrow \,\infty }\left\vert \frac{a_{n+1}}{a_{n}} \right\vert =\lim\limits_{n\,\rightarrow \,\infty }\frac{1}{\left( 1+\dfrac{1 }{n}\right) ^{n}}=\frac{\lim\limits_{n\,\rightarrow \,\infty }1}{ \lim\limits_{n\,\rightarrow \,\infty }\,\left( 1+\dfrac{1}{n}\right) ^{n}}=\frac{1}{e}$$

Since $\dfrac{1}{e}<1$, the series converges.

NOW WORK

Problem 15.

CAUTION

In Example 2, the ratio $\left\vert \dfrac{a_{n+1}}{a_{n}} \right\vert$ converges to $\dfrac{1}{e}$. This does not mean that $\sum\limits_{k\,=\,1}^{\infty }\dfrac{k!}{k^{k}}$ converges to $\dfrac{1}{e}$. In fact, the sum of the series $\sum\limits_{k\,=\,1}^{\infty }\dfrac{ k!}{k^{k}}$ is not known; all that is known is that the series converges.

THEOREM Root Test

Let $\sum\limits_{k\,=\,1}^{\infty }a_{k}$ be a series of nonzero terms. Suppose $\ \lim\limits_{n\rightarrow \infty }\sqrt[{n}]{\,\vert a_{n}\vert }=L$, a number.

• If $L<1$, then $\sum\limits_{k\,=\,1}^{\infty }a_{k}$ is absolutely convergent, so the series $\sum\limits_{k\,=\,1}^{\infty }a_{k}$ converges.
• If $L>1$, then $\sum\limits_{k\,=\,1}^{\infty }a_{k}$ diverges.
• If $L=1$, the test is inconclusive.

EXAMPLE 3Using the Root Test

Use the Root Test to determine whether the series $\sum\limits_{k\,=\,1}^{ \infty }\dfrac{e^{k}}{k^{k}}$ converges or diverges.

Solution $\sum\limits_{k\,=\,1}^{\infty }\dfrac{e^{k}}{k^{k}}$ is a series of nonzero terms. The $n$th term is $a_{n}=$ $\dfrac{e^{n}}{n^{n}}=\left(\dfrac{e}{n}\right)^{n}$. Since $a_{n}$ involves an $n$th power, we use the Root Test. $$\lim\limits_{n\,\rightarrow \,\infty }\sqrt[n]{\,\vert a_{n}\vert }=\lim\limits_{n\,\rightarrow \,\infty }\sqrt[n]{\left( \frac{e}{n}\right) ^{n}}=\lim\limits_{n\,\rightarrow \,\infty }\frac{e}{n}=0<1$$

The series $\sum\limits_{k\,=\,1}^{\infty }\dfrac{e^{k}}{k^{k}}$ converges.

NOW WORK

Problem 27.

EXAMPLE 4Using the Root Test

Use the Root Test to determine whether the series $\sum\limits_{k\,=\,1}^{ \infty }\left( {\dfrac{8k+3}{5k-2}}\right) ^{\!\!k}$ converges or diverges.

Solution $\sum\limits_{k\,=\,1}^{\infty }\left( {\dfrac{8k+3}{ 5k-2}}\right) ^{\!\!k}$ is a series of nonzero terms. The $n$th term is $a_{n}=$ $\left( {\dfrac{8n+3}{5n-2}}\right) ^{\!\!n}\!$. Since $a_{n}$ involves an $n$th power, we use the Root Test. $$\lim\limits_{n\,\rightarrow \,\infty }\sqrt[n]{\,\vert a_{n}\vert }=\lim\limits_{n\,\rightarrow \,\infty }\sqrt[n]{\left( {\dfrac{8n+3}{5n-2}} \right) ^{\!\!n}}=\lim\limits_{n\,\rightarrow \,\infty }\frac{8n+3}{5n-2}=\frac{8}{5}>1$$

The series diverges.

NOW WORK

Problem 23.

Proof of the Ratio Test

Let $\sum\limits_{k\,=\,1}^{\infty }a_{k}$ be a series of nonzero terms.

Case 1: $\lim\limits_{n\,\rightarrow \,\infty }\,\left\vert \dfrac{a_{n+1}}{a_{n}}\right\vert =L,$ a number.

• $0\leq L<1$ Let $r$ be any number for which $L<r<1$. Since $\lim\limits_{n\,\rightarrow \,\infty }\,\left\vert \dfrac{ a_{n+1}}{a_{n}}\right\vert =L$ and $L<r$, then by the definition of the limit of a sequence, we can find a number $N$ so that for any number $n>N$, the ratio $\left\vert \dfrac{a_{n+1}}{a_{n}}\right\vert$ can be made as close as we please to $L$ and be less than $r$. Then $$\begin{array}{ll@{\qquad}l@{\qquad}l} \left\vert \dfrac{a_{N+1}}{~a_{N}}\right\vert &< r & \hbox{or} & \left\vert a_{N\,+\,1}\right\vert <r\cdot \left\vert a_{N}\right\vert \\[16pt] \left\vert \dfrac{a_{N+2}}{~a_{N+1}}\right\vert &< r & \hbox{or} & \left\vert a_{N\,+\,2}\right\vert <r\cdot \left\vert a_{N\,+\,1}\right\vert <r^{2}\cdot \left\vert a_{N}\right\vert \\[16pt] \left\vert \dfrac{a_{N+3}}{~a_{N+2}}\right\vert &< r~ & \text{ or} & \left\vert a_{N\,+\,3}\right\vert <r\cdot \left\vert a_{N\,+\,2}\right\vert <r^{3}\cdot \left\vert a_{N}\right\vert \end{array}$$ Each term of the series $\vert a_{N\,+\,1}\vert +\vert a_{N\,+\,2} \vert +\cdots$ is less than the corresponding term of the geometric series $\vert a_{N}\vert \,r+ \vert a_{N} \vert \,r^{2}+\vert a_{N} \vert \,r^{3}+\cdots$. Since $\vert r\vert <1$, the geometric series converges. By the Comparison Test for Convergence, the series $\vert a_{N\,+\,1}\vert +\vert a_{N\,+\,2}\vert +\cdots$ also converges. So, the series $\sum\limits_{k\,=\,1}^{\infty }\vert a_{k}\vert$ converges. By the Absolute Convergence Test, the series $\sum\limits_{k\,=\,1}^{\infty }a_{k}$ converges.
• $L=1$ To show that the test fails for $L=1$, we exhibit two series, one that diverges and another that converges, to show that no conclusion can be drawn. Consider $\sum\limits_{k\,=\,1}^{\infty } \dfrac{1}{k}$ and $\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{k^{2}}$. The first is the harmonic series, which diverges. The second is a $p$-series with $p>1,$ which converges. It is left to you to show that $\lim\limits_{n\rightarrow \infty }\left\vert \dfrac{a_{n+1}}{a_{n}} \right\vert =1$ in each case. (See Problems 45 and 46.)
• $L>1$ Let $r$ be any number for which $1<r<L$. Since $\lim\limits_{n\,\rightarrow \,\infty }\,\left\vert \dfrac{a_{n+1}}{a_{n}} \right\vert =L,$ there is a number $N$ so that for any number $n>N$, the ratio $\left\vert \dfrac{a_{n+1}}{a_{n}}\right\vert$ can be made as close as we please to $L$ and will be greater than $r$. That is, for all numbers $n>N$, the ratio $\left\vert \dfrac{a_{n+1}}{a_{n}}\right\vert >r>1$ so that $\left\vert a_{n+1}\right\vert >\vert a_{n}\vert$. After the $N$th term, the absolute value of the terms are positive and increasing. So, $\lim\limits_{n\,\rightarrow \,\infty }\,\vert a_{n}\vert \neq 0$ and, therefore, $\lim\limits_{n\rightarrow \infty }a_{n}\neq 0.$ By the Test for Divergence, the series diverges.

Case 2: $\lim\limits_{n\,\rightarrow \,\infty }\,\left\vert \dfrac{a_{n+1}}{a_{n}}\right\vert =\infty$ The proof that this series diverges is left as an exercise (Problem 54).