OBJECTIVES

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

1. Write the terms of a sequence (p. 539)
2. Find the \(n\)th term of a sequence (p. 539)
3. Use properties of convergent sequences (p. 542)
4. Use a related function or the Squeeze Theorem to show a sequence converges (p. 543)
5. Determine whether a sequence converges or diverges (p. 545)

NEED TO REVIEW?

Sequences are discussed in Appendix A.5, pp. A-38 to A-39.

spanDEFINITIONspan

A sequence is a function whose domain is the set of positive integers and whose range is a subset of the real numbers.

IN WORDS

\(\{s_{n}\}_{n=1}^{\infty}=\left\{ \dfrac{1}{n}\right\} _{n=1}^{\infty }\) or \(\{s_{n}\}=\left\{\dfrac{1}{n}\right\}\) is the name of the sequence; \(s_{n}=\dfrac{1}{n}\) is the \(n\)th term of the sequence.

Writing the Terms of a Sequence

Write the first three terms of each sequence:

1. \(\{b_{n}\} _{n=1}^{\infty }=\left\{ \dfrac{1}{3n-2}\right\} _{n=1}^{\infty }\)
2. \(\left\{ c_{n}\right\}=\left\{ \dfrac{2n-1}{n^{3}}\right\}\)

Solution (a) The \(n\)th term of this sequence is \(b_{n}=\dfrac{1}{3n-2}.\) The first three terms are \[ b_{1}=\dfrac{1}{3\cdot 1-2}=1\qquad b_{2}=\dfrac{1}{3\cdot 2-2}=\dfrac{1}{4}\qquad b_{3}=\dfrac{1}{3\cdot 3-2}=\dfrac{1}{7} \]

(b) The \(n\)th term of this sequence is \(c_{n}=\dfrac{2n-1}{n^{3}}\). Then \[ c_{1}=\dfrac{2(1)-1}{1^{3}}=1\qquad c_{2}=\dfrac{2(2) -1}{2^{3}}=\dfrac{3}{8}\qquad c_{3}=\dfrac{2 (3) -1}{3^{3}}=\dfrac{5}{27} \]

NOW WORK

Problem 15.

Writing the Terms of a Sequence

Write the first five terms of the sequence \[ \{a_{n}\} =\left\{ (-1)^{n}\left( \dfrac{1}{2}\right) ^{n}\right\} \]

Solution The \(n\)th term of this sequence is \(a_{n}=(-1)^{n}\left(\dfrac{1}{2}\right) ^{n}\). So, \[ \begin{eqnarray*} a_{1}&=&(-1)^{1}\left(\dfrac{1}{2}\right) ^{1}=-\dfrac{1}{2}\quad a_{2}=(-1)^{2}\left(\dfrac{1}{2}\right) ^{2}=\dfrac{1}{4}\quad a_{3}=(-1)^{3}\left(\dfrac{1}{2}\right) ^{3}=-\dfrac{1}{8}\\[6pt] a_{4}&=&\dfrac{1}{16}\quad a_{5}=-\dfrac{1}{32} \end{eqnarray*} \]

The first five terms of the sequence \(\{a_{n}\}\) are \(-\dfrac{1}{2}\), \(\dfrac{1}{4}\), \(-\dfrac{1}{8}\), \(\dfrac{1}{16}\), \(-\dfrac{1}{32}\).

NOW WORK

Problem 17.

Finding the \(n\)th Term of a Sequence

Find the \(n\)th term of each of the following sequences. Assume that the indicated patterns continue.

Solution

NOW WORK

Problem 25.

spanDEFINITIONspan Convergent Sequence

Let \(L\) be a real number and let \(\{ s_{n}\}\) be a sequence. The sequence \(\{ s_{n}\}\) converges to a real number \(L\) if, for any number \(\varepsilon >0,\) there is a positive integer \(N\) so that \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{\left\vert s_{n}-L\right\vert <\varepsilon \qquad \hbox{for all integers}\, n>N}} \]

If \(\{ s_{n}\} \) converges to \(L\), then \(L\) is called the limit of the sequence and we write \(\lim\limits_{n\rightarrow \infty}s_{n}=L.\) If a sequence converges, it is said to be convergent; otherwise, it is said to be divergent.

Showing a Sequence Converges

Show that:

1. \(\lim\limits_{n\rightarrow \infty }c=c\)
2. \(\lim\limits_{n\rightarrow \infty }\dfrac{1}{n}=0\)

Solution (a) The graph of the sequence \(\{ s_{n}\} =\{c\}\) suggests that \(\{ s_{n}\} \) converges to \(c.\) See Figure 4.

To show the sequence converges to \(c,\) we look at \(\vert s_{n}-c\vert .\) Then for any \(\varepsilon >0,\) \[ \vert s_{n}-c\vert =\left\vert c-c\right\vert =0<\varepsilon \qquad \hbox{ for all }n \]

The sequence \(\{ s_{n}\} =\{c\} \) converges to \(c.\)

(b) The graph of the sequence \(\{s_{n}\} =\left\{\dfrac{1}{n}\right\} \) shown in Figure 5 suggests that \(\{ s_{n}\} \) converges to \(0.\)

To show the sequence \(\{ s_{n}\} =\left\{ \dfrac{1}{n}\right\}\) converges to \(0,\) we look at \[ \left\vert s_{n}-0\right\vert =\left\vert \dfrac{1}{n}-0\right\vert =\dfrac{1}{n} \]

For any \(\varepsilon > 0\), choose any integer \(N>\dfrac{1}{\varepsilon }\). Then for all \(n>N>\dfrac{1}{\varepsilon },\) we have \(\left\vert s_{n}-0\right\vert =\dfrac{1}{n}<\dfrac{1}{N}<\varepsilon\), so the sequence \(\{ s_{n}\} =\left\{ \dfrac{1}{n}\right\}\) converges to \(0\).

THEOREM Properties of Convergent Sequences

If \(\{ s_{n}\}\) and \(\{ t_{n}\} \) are convergent sequences and if \(c\) is a number, then

• Constant multiple property: \[ \begin{equation} \bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{\lim\limits_{n\,\rightarrow \,\infty}(cs_{n}) =c\lim\limits_{n\,\rightarrow \,\infty }s_{n}}} \tag{1} \end{equation} \]
• Sum and difference properties: \[ \begin{equation} \bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{\lim\limits_{n\,\rightarrow \,\infty }\left( s_{n}\pm t_{n}\right) =\lim\limits_{n\,\rightarrow \,\infty }s_{n}\pm\lim\limits_{n\,\rightarrow \,\infty }t_{n}}} \tag{2} \end{equation} \]
• Product property: \[ \begin{equation} \bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{\lim\limits_{n\,\rightarrow \,\infty }(s_{n}\cdot t_{n}) =\left( \lim\limits_{n\,\rightarrow \,\infty }s_{n}\right) \left(\lim\limits_{n\,\rightarrow \,\infty }t_{n}\right) \hspace{-1pt}}} \tag{3} \end{equation} \]
• Quotient property: \[ \begin{equation} \bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{\lim\limits_{n\,\rightarrow \,\infty }\dfrac{s_{n}}{t_{n}}=\dfrac{\lim\limits_{n\,\rightarrow \,\infty }s_{n}}{\lim\limits_{n\,\rightarrow \,\infty }t_{n}}\quad \hbox{ provided } \quad \lim\limits_{n\,\rightarrow \,\infty }t_{n}\neq 0}} \tag{4} \end{equation} \]
• Power property: \[ \begin{equation} \bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{\lim\limits_{n\rightarrow \infty }s_{n}^{p}=\left[ \lim\limits_{n\rightarrow \infty }s_{n}\right]^{p}\qquad p\geq 2 \hbox{ is an integer }\hspace{-1.5pt}}} \tag{5} \end{equation} \]
• Root property: \[ \begin{equation} \bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{\lim\limits_{n\rightarrow \infty }\sqrt[p]{s_{n}}=\sqrt[p]{\lim\limits_{n\rightarrow \infty }s_{n}}\qquad p\geq 2 \hbox{ and }s_{n}\geq 0 \hbox{ if } p \hbox{ is even}}} \tag{6} \end{equation} \]

Using Properties of Convergent Sequences

Use properties of convergent sequences to find \(\lim\limits_{n\rightarrow \infty }s_{n}.\)

1. \(\{ s_{n}\} =\left\{ \dfrac{2}{n}+3\right\}\)
2. \(\{ s_{n}\} =\left\{\dfrac{4}{n^{2}}\right\}\)
3. \(\{ s_{n}\} =\left\{ \sqrt[3]{\dfrac{16n^{2}+3n}{2n^{2}}}\right\}\)

Solution
(a) \[ \begin{eqnarray*} \lim\limits_{n\rightarrow \infty}s_{n}&=&\lim\limits_{n\rightarrow \infty }\left( \dfrac{2}{n}+3\right)=\lim\limits_{n\rightarrow \infty }\dfrac{2}{n}+\lim\limits_{n\rightarrow\infty }3\\ &=&2\lim\limits_{n\rightarrow \infty }\dfrac{1}{n}+\lim\limits_{n\rightarrow \infty }3 \underset{\underset{\color{#0066A7}{\lim\limits_{n\rightarrow \infty}\dfrac{1}{n}=0;\,\,\lim\limits_{n\rightarrow\infty} 3=3}}{\color{#0066A7}{\uparrow}}}{=} 2\cdot 0+3=3\\ \end{eqnarray*} \]

(b) \[ \begin{eqnarray*} \lim\limits_{n\rightarrow \infty }s_{n}=\lim\limits_{n\rightarrow \infty }\dfrac{4}{n^{2}}=4\lim\limits_{n\rightarrow \infty }\dfrac{1}{n^{2}}=4\lim\limits_{n\rightarrow \infty }\dfrac{1}{n}\cdot \lim\limits_{n\rightarrow \infty }\dfrac{1}{n} \underset{\underset{\color{#0066A7}{\lim\limits_{n\rightarrow \infty }\dfrac{1}{n}=0}}{\color{#0066A7}{\uparrow}}}{=} 4\cdot 0\cdot 0=0\\ \hspace{-7pc} \end{eqnarray*} \]

(c) \[ \begin{eqnarray*} \lim\limits_{n\rightarrow \infty}s_{n} &=&\lim\limits_{n\rightarrow \infty }\sqrt[3]{\dfrac{16n^{2}+3n}{2n^{2}}}=\sqrt[3]{\lim\limits_{n\rightarrow \infty }\left( \dfrac{16n^{2}+3n}{2n^{2}} \right) }=\sqrt[3]{\lim\limits_{n\rightarrow \infty }\left( 8+\dfrac{3}{2n}\right) }\\ &=&\sqrt[3]{\lim\limits_{n\rightarrow \infty }8+\lim\limits_{n\rightarrow \infty }\dfrac{ 3}{2n}} \underset{\underset{\color{#0066A7}{\lim\limits_{n\rightarrow \infty }8=8}}{\color{#0066A7}{\uparrow}}}{=} \sqrt[3]{8+\dfrac{3}{2}\lim\limits_{n\rightarrow \infty } \dfrac{1}{n}} \underset{\underset{\color{#0066A7}{\lim\limits_{n\rightarrow \infty }\dfrac{1}{n}=0}}{\color{#0066A7}{\uparrow}}}{=} \sqrt[3]{8+\dfrac{3}{2}\cdot 0}=\sqrt[3]{8}=2 \\ \end{eqnarray*} \]

NOW WORK

Problem 37.

THEOREM

Let \(\{ s_{n}\} \) be a sequence of real numbers. If \(\lim\limits_{n\rightarrow \infty }s_{n}=L\) and if \(f\) is a function that is continuous at \(L\) and is defined for all numbers \(s_{n},\) then \(\lim\limits_{n\rightarrow \infty }f( s_{n}) =f(L).\)

Showing a Sequence Converges

Show \(\left\{\ln \left(\dfrac{2}{n}+3\right) \right\}\) converges and find its limit.

Solution Since \(\lim\limits_{n\rightarrow \infty }\left( \dfrac{2}{n}+3\right) =3\) [from Example 5(a)], the sequence \(\{ s_{n}\}=\left\{ \dfrac{2}{n}+3\right\} \) converges to \(3.\) The function \(f(x) =\ln x\) is continuous on its domain, so it is continuous at \(3\). Then \[ \lim\limits_{n\rightarrow \infty }f( s_{n}) =\lim\limits_{n\rightarrow \infty }f \left( \dfrac{2}{n}+3\right) =\lim\limits_{n\rightarrow \infty }\ln \left( \dfrac{2}{n}+3\right) =\ln\left[\lim\limits_{n\rightarrow \infty } \left(\dfrac{2}{n}+3\right) \right] =\ln 3 \]

So, the sequence \(\left\{\ln \left( \dfrac{2}{n}+3\right) \right\}\) converges to \(\ln 3\).

NOW WORK

Problem 45.

spanDEFINITIONspan Related Function of a Sequence

A related function \(f\) of the sequence \(\{s_{n}\}\) has the following two properties:

• \(f\) is defined on the open interval \((0,\,\infty)\); that is, the domain of \(f\) is the set of positive real numbers.
• \(f(n) =s_{n}\) for all integers \(n\geq 1\).

Identifying a Related Function of a Sequence

If \(\{ s_{n}\} =\left\{\dfrac{n}{e^{n}}\right\}\), then a related function is given by \(f(x)=\dfrac{x}{e^{x}},\) where \(x>0,\) as shown in Figure 6.

THEOREM

Let \(\{ s_{n}\} \) be a sequence of real numbers and let \(f\) be a related function of \(\{ s_{n}\} \).Suppose \(L\) is a real number. \[ \begin{equation} \bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{\hbox{If } \lim\limits_{x\,\rightarrow \,\infty } f(x)=L\qquad \hbox{ then}\quad \lim\limits_{n\,\rightarrow \,\infty }s_{n}=L} }\tag{7} \end{equation} \]

NEED TO REVIEW?

Limits at infinity are discussed in Section 1.5, pp. 117-125.

Using a Related Function to Show a Sequence Converges

Show that \(\left\{{\dfrac{3n^{2}+5n-2}{6n^{2}-6n+5}}\right\}\) converges and find its limit.

Solution The function \[ f(x)= \dfrac{3x^{2}+5x-2}{6x^{2}-6x+5} \quad x>0 \]

is a related function of the sequence \(\left\{ \dfrac{3n^{2}+5n-2}{6n^{2}-6n+5}\right\}\). Since \[ \begin{eqnarray*} \lim\limits_{x\,\rightarrow \,\infty }\,f(x)&=&\lim_{ x \rightarrow \,\infty }\frac{3x^{2}+5x-2}{6x^{2}-6x+5} =\lim_{ x \rightarrow \,\infty }\frac{\dfrac{3x^{2}}{6x^{2}}+\dfrac{5x}{6x^{2}}-\dfrac{2}{6x^{2}}}{1-\dfrac{6x}{6x^{2}}+\dfrac{5}{6x^{2}}}\\[5pt] &=&\lim_{x\rightarrow \,\infty }\frac{\dfrac{1}{2}+\dfrac{5}{6x}-\dfrac{1}{3x^{2}}}{1-\dfrac{1}{x}+\dfrac{5}{6x^{2}}}=\dfrac{\dfrac{1}{2}+0-0}{1-0+0}=\dfrac{1}{2} \end{eqnarray*} \]

the sequence \(\left\{ \dfrac{3n^{2}+5n-2}{6n^{2}-6n+5}\right\}\) converges and \(\lim\limits_{n\,\rightarrow \,\infty } \dfrac{3n^{2}+5 n-2}{6n^{2}-6n+5}=\dfrac{1}{2}\).

NOW WORK

Problem 51.

NEED TO REVIEW?

L’Hôpital’s Rule is discussed in Section 4.5, pp. 299-302.

Using L’Hôpital’s Rule to Show a Sequence Converges

Show that \(\left\{ \dfrac{n}{e^{n}}\right\} \) converges and find its limit.

Solution We begin with the related function \(f(x)=\dfrac{x}{e^{x}},\) \(x>0\). To find \(\lim\limits_{x\rightarrow \infty }f(x)\), we use L’Hôpital’s Rule. \[ \begin{eqnarray*} \lim_{x\,\rightarrow \,\infty }\,f(x) &=& \lim_{x\,\rightarrow \,\infty }\frac{x}{e^{x}} \underset{\underset{\color{#0066A7}{\rm Use L’Hôpital’s Rule}}{\color{#0066A7}{\uparrow}}}{=} \lim_{x\,\rightarrow\,\infty }\frac{1}{e^{x}}=0 \\[-8pt] \end{eqnarray*} \]

Since \(\lim\limits_{x\,\rightarrow \,\infty }\,f(x)=0\), the sequence \(\left\{ \dfrac{n}{e^{n}}\right\}\) converges and \(\lim\limits_{n\,\rightarrow \,\infty}\dfrac{n}{e^{n}}=0\).

NOW WORK

Problem 57.

NEED TO REVIEW?

The Squeeze Theorem for functions is discussed in Section 1.4, pp. 106-107.

THEOREM The Squeeze Theorem for Sequences

Suppose \(\{a_{n}\}\), \(\{b_{n}\}\), and \(\{ s_{n}\}\) are sequences and \(N\) is a positive integer. If \(a_{n}\leq s_{n}\leq b_{n}\) for every integer \(n>N,\) and if \(\lim\limits_{n\rightarrow \infty }a_{n}=\lim\limits_{n\rightarrow \infty }b_{n}=L\), then \(\lim\limits_{n\rightarrow \infty }s_{n}=L.\)

Using the Squeeze Theorem for Sequences

Show that \(\{ s_{n}\} =\left\{(-1) ^{n}\dfrac{1}{n}\right\}\) converges and find its limit.

Solution We seek two sequences that "squeeze" \(\{ s_{n}\} = \left\{ (-1) ^{n}\dfrac{1}{n}\right\} \) as \(n\) becomes large. We begin with \(\vert s_{n}\vert\): \[ \begin{eqnarray*} \vert s_{n}\vert &=&\left\vert (-1) ^{n}\dfrac{1}{n}\right\vert \leq \dfrac{1}{n} \\[5pt] -\dfrac{1}{n} &\leq &(-1) ^{n}\dfrac{1}{n}\leq \dfrac{1}{n} \end{eqnarray*} \]

Notice that \(s_{n}\) is bounded by \(\{a_{n}\} =\left\{ -\dfrac{1}{n}\right\} \) and \(\{b_{n}\} =\left\{ \dfrac{1}{n}\right\}\). Since \(a_{n}\leq s_{n}\leq b_{n}\) for all \(n\) and \(\lim\limits_{n\rightarrow \infty }a_{n}=\lim\limits_{n\rightarrow \infty }\left( -\dfrac{1}{n}\right) =0\), and \(\lim\limits_{n\rightarrow \infty }b_{n}=\lim\limits_{n\rightarrow \infty }\dfrac{1}{n}=0\), then by the Squeeze Theorem, the sequence \(\{ s_{n}\} =\left\{ (-1) ^{n}\dfrac{1}{n}\right\}\) converges and \(\lim\limits_{n\rightarrow \infty } s_n=0 \).

NOW WORK

Problem 59.

spanDEFINITIONspan Divergence of a Sequence to Infinity

The sequence \(\{ s_{n}\} \) diverges to infinity, that is, \[ \lim_{n\rightarrow \infty }s_{n}=\infty \]

if, given any positive number \(M,\) there is a positive integer \(N\) so that whenever \(n>N\), then \(s_{n}>M.\)

Showing a Sequence Diverges

Show that the following sequences diverge:

1. \(\{1+(-1)^{n}\}\)
2. \(\{n\} \)

Solution (a) The terms of the sequence are \(0, 2, 0, 2, 0, 2, \ldots \,\). See Figure 8. Since the terms alternate between \(0\) and \(2\), the terms of the sequence \(\{1+(-1)^{n}\}\) do not approach a single number \(L\). So, the sequence \(\{1+(-1)^{n}\}\) is divergent.

(b) The terms of the sequence \(\{ s_{n}\}=\{n\} \) are \(1,\,2,\,3,\,4,\ldots.\) Given any positive number \(M\), we choose a positive integer \(N>M.\) Then whenever \(n>N,\) we have \(s_{n}=n>N>M.\) That is, the sequence \(\{n\} \) diverges to infinity. See Figure 9.

NOW WORK

Problem 63.

The sequence \(\{s_n \} = \{ r^{n}\}\), where \(r\) is a real number,

• converges to \(0,\) if \(-1<r<1\).
• converges to \(1,\) if \(r=1\).
• diverges for all other numbers.

Determine Whether \({\{r^{n}\}}\) Converges or Diverges

Determine whether the following sequences converge or diverge:

1. \(\{ s_{n}\} =\left\{ \left( \dfrac{3}{4}\right) ^{n}\right\}\)
2. \(\{ t_{n}\} =\left\{\left( \dfrac{4}{3}\right) ^{n}\right\} \)

Solution (a) The sequence \(\{ s_{n}\} =\left\{ \left(\dfrac{3}{4}\right) ^{n}\right\}\) converges to \(0\) because \(-1<\dfrac{3}{4}<1.\)

(b) The sequence \(\{ t_{n}\} =\left\{ \left( \dfrac{4}{3}\right) ^{n}\right\} \) diverges because \(\dfrac{4}{3}>1\).

NOW WORK

Problem 67.

Determining Whether a Sequence Is Bounded from Above or Bounded from Below

1. The sequence \(\{ s_{n}\} =\left\{ \dfrac{3n}{n+2}\right\} \) is bounded both from above and below because \[ \dfrac{3n}{n+2}=\dfrac{3}{1+\dfrac{2}{n}}<3 \qquad \hbox{ and }\qquad \dfrac{3n}{n+2}>0 \quad \hbox{ for all } n\geq 1 \] See Figure 12(a).

   547

2. The sequence \(\{a_{n}\} =\left\{ \dfrac{4n}{3}\right\}\) is bounded from below because \(\dfrac{4n}{3}>1\) for all \(n\geq 1.\) It is not bounded from above because \(\lim\limits_{n\rightarrow \infty }\dfrac{4n}{3}=\dfrac{4}{3}\lim\limits_{n\rightarrow \infty }n=\infty.\) See Figure 12(b).
3. The sequence \(\{b_{n}\} =\left\{ (-1)^{n+1}n\right\}\) is neither bounded from above nor bounded from below.If \(n\) is odd, \(\lim\limits_{n\rightarrow \infty }b_{n} = \lim\limits_{n\rightarrow \infty } n =\infty\), and if \(n\) is even, \(\lim\limits_{n\rightarrow \infty }b_{n}= \lim\limits_{n\rightarrow \infty } (-n) = -\infty.\) See Figure 12(c).

NOW WORK

Problem 73.

THEOREM Boundedness Theorem

A convergent sequence is bounded.

NEED TO REVIEW?

The Triangle Inequality is discussed in Appendix A.1, p. A-7

Proof

If \(\{ s_{n}\} \) is a convergent sequence, there is a number \(L\) for which \(\lim\limits_{n\,\rightarrow \,\infty}s_{n}=L\). We use the definition of the limit of a sequence with \(\varepsilon =1\). Then there is a positive integer \(N\) so that \[ \left\vert s_{n}-L\right\vert <1 \qquad \hbox{for all }n>N \]

Then, for all \(n>N,\) \[ \begin{eqnarray*} &&\vert s_{n}\vert =\vert s_{n}-L+L \vert \leq \vert s_{n}-L \vert + \vert L \vert <1+ \vert L \vert \\ &&\hspace{0pc} \qquad\qquad \, \underset{\color{#0066A7}{\rm Triangle \, Inequality}}{\color{#0066A7}{\uparrow}} \end{eqnarray*} \]

If we choose \(K\) to be the largest number in the finite collection \[ |s_{1}|,\quad |s_{2}|,\quad |s_{3}|,\quad \ldots ,\quad |s_{N}|,\quad 1+|L| \]

it follows that \(\vert s_{n}\vert \leq K\) for all integers \(n\geq 1\). That is, the sequence \(\{s_{n}\} \) is bounded.

THEOREM Test for Divergence of a Sequence

If a sequence is not bounded from above or if it is not bounded from below, then it diverges.

CAUTION

The converse of the boundedness theorem is not true. Bounded sequences may converge or they may diverge. For example, the sequence \(\{1+(-1)^{n}\}\) in Example 11(a) is bounded, but it diverges.

spanDEFINITIONspan

A sequence \(\{ s_{n}\} \) is said to be:

1. Increasing if \(s_{n}<s_{n+1}\) for \(n\geq 1\).
2. Nondecreasing if \(s_{n} \leq s_{n+1}\) for \(n\geq 1\).
3. Decreasing if \(s_{n} >s_{n+1}\) for \(n\geq 1\).
4. Nonincreasing if \(s_{n} \geq s_{n+1}\) for \(n\geq 1\).

If any of the above conditions hold, the sequence is called monotonic.

Showing a Sequence Is Monotonic

Show that each of the following sequences is monotonic by determining whether it is increasing, nondecreasing, decreasing, or nonincreasing:

1. \(\{ s_{n}\} =\left\{ {\dfrac{n}{n+1}}\right\} \)
2. \(\{ s_{n}\} =\left\{ {\dfrac{e^{n}}{n!}}\right\}\)
3. \(\{ s_{n}\} =\{\ln n\}\)

Solution
(a) We use the algebraic difference test. \[ \begin{eqnarray*} s_{n+1}-s_{n} &=& \frac{n+1}{n+2}-\frac{n}{n+1}=\frac{n^{2}+2n+1-n^{2}-2n}{(n+2)(n+1)}\\[5pt] &=&\frac{1}{(n+2)(n+1)}>0 \qquad \hbox{ for all }n\geq 1 \end{eqnarray*} \]

So, \(\{ s_{n}\} \) is an increasing sequence.

(b) When the sequence contains a factorial, the algebraic ratio test is usually easiest to use. \[ \frac{s_{n+1}}{s_{n}}=\frac{\dfrac{e^{n+1}}{(n+1)!}}{\dfrac{e^{n}}{n!}}=\left( {\frac{e^{n+1}}{e^{n}}}\right) \frac{n!}{(n+1)!}=\frac{e}{n+1}<1 \qquad \hbox{ for all }n\geq 2 \]

After the first term, \(\{ s_{n}\} =\left\{ \dfrac{e^{n}}{n!}\right\} \) is a decreasing sequence.

(c) Here, we use the derivative of the related function \(f(x) =\ln x\) of the sequence \(\{ s_{n}\} =\{\ln n\}.\) Since \(\dfrac{d}{dx}\ln x=\dfrac{1}{x}>0\) for all \(x>0\), it follows that \(f\) is an increasing function and so the sequence \(\{\ln n\}\) is an increasing sequence.

RECALL

\(n!=n(n-1) (n-2)\cdots 2\cdot 1\) So \(\dfrac{n!}{(n+1)!} =\dfrac{1}{n+1}\).

NOW WORK

Problem 81.

IN WORDS

A bounded, monotonic sequence converges.

THEOREM

An increasing (or nondecreasing) sequence \(\{ s_{n}\} \) that is bounded from above converges. A decreasing (or nonincreasing) sequence \(\{ s_{n}\} \) that is bounded from below converges.

Determining if a Sequence Converges or Diverges

Determine if the sequence \(\{ s_{n}\} =\left\{ {\dfrac{2^{n}}{n!}}\right\}\) converges or diverges.

Solution To see if \(\left\{ \dfrac{2^{n}}{n!}\right\} \) is monotonic, find the algebraic ratio \(\dfrac{s_{n+1}}{s_{n}}\): \[ \frac{s_{n+1}}{s_{n}}=\frac{\dfrac{2^{n+1}}{(n+1)!}}{\dfrac{2^{n}}{n!}}=\frac{2^{n+1}\,n!}{(n+1)!\,2^{n}}=\frac{2}{n+1}\leq 1 \qquad \hbox{ for all }n \geq 1 \]

Since \(s_{n+1}\leq s_{n}\) for \(n\geq 1,\) the sequence \(\{ s_{n}\}\) is nonincreasing.

Next, since each term of the sequence is positive, \(s_{n}>0\) for \(n\geq 1,\) the sequence \(\{ s_{n}\} \) is bounded from below.

Since \(\{ s_{n}\}\) is nonincreasing and bounded from below, it converges.

NOW WORK

Problem 89.

NOTE

Although the sequence \(\{ s_{n}\} \) converges, the theorem does not tell us what \(\lim\limits_{n\,\rightarrow \,\infty }s_{n}\) equals.