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.

Using the Ratio Test

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

1. \(\sum\limits_{k\,=\,1}^{\infty }\frac{k}{4^{k}}\)
2. \(\sum\limits_{k\,=\,1}^{\infty }\frac{2^{k}}{k}\)
3. \(\sum\limits_{k\,=\,1}^{\infty }\frac{3k+1}{k^{2}}\)
4. \(\sum\limits_{k\,=\,1}^{\infty }\frac{1}{k!}\)

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.

Using 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.

Using 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.

Using 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).