OBJECTIVES

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

1. Determine whether a series has a sum (p. 554)
2. Analyze a geometric series (p. 557)
3. Analyze the harmonic series (p. 561)
   Using a geometric series in a biology application (p. 562)

spanDEFINITIONspan Infinite Series

If \(a_{1}\), \(a_{2}\), \(\ldots ,\) \(a_{n},\) \(\ldots\) is an infinite collection of numbers, the expression \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{\sum\limits_{k\,=\,1}^{\infty}a_{k}=a_{1}+~a_{2}+ \cdots +~a_{n}+ \cdots }} \]

is called an infinite series or, simply, a series.

NEED TO REVIEW?

Sums and summation notation are discussed in Appendix A.5, pp. A-40 to A-42.

spanDEFINITIONspan Convergence, Divergence of an Infinite Series

If the sequence \(\{S_{n}\}\) of partial sums of an infinite series \(\sum\limits_{k\,=\,1}^{\infty }a_{k}\) has a limit \(S\), then the series converges and is said to have the sum \(S\). That is, if \(\lim\limits_{n\,\rightarrow \,\infty }S_{n}=S\), then \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{\sum\limits_{k\,=\,1}^{\infty}a_{k}=a_{1}+ a_{2}+\cdots + a_{n}+\cdots =S }} \]

An infinite series diverges if the sequence of partial sums diverges.

Finding the Sum of a Series

Show that \[ \sum_{k\,=\,1}^{\infty }\frac{1}{k(k+1)}=\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3 }+\frac{1}{3\cdot 4}+\cdots =\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\cdots =1 \]

Solution We begin with the sequence \(\{S_{n}\}\) of partial sums, \[ \begin{array}{cl} S_{1} &= \dfrac{1}{1\cdot 2} \\[10pt] S_{2} &= \dfrac{1}{1\cdot 2}+\dfrac{1}{2\cdot 3} \\[10pt] S_{3} &= \dfrac{1}{1\cdot 2}+\dfrac{1}{2\cdot 3}+\dfrac{1}{3\cdot 4} \\ \vdots & \\ S_{n} &= \dfrac{1}{1\cdot 2}+\dfrac{1}{2\cdot 3}+\dfrac{1}{3\cdot 4}+\cdots +\dfrac{1}{n(n+1)} \\ \vdots & \end{array} \]

Since \[ \begin{array}{l@{\hspace{14.5pt}}l} \dfrac{1}{n(n+1)}=\dfrac{1}{n}-\dfrac{1}{n+1} & {\color{#0066A7}{{\hbox{Use partial fractions.}}}} \end{array} \]

\(S_{n}\) can be written as \[ S_{n}=\left( {\frac{1}{1}-\frac{1}{2}}\right) +\left( {\frac{1}{2}-\frac{1}{3 }}\right) +\cdots +\left( {\frac{1}{n-1}-\frac{1}{n}}\right) +\left( {\frac{1 }{n}-\frac{1}{n+1}}\right) \]

After removing parentheses notice that all the terms except the first and last cancel, so that \[ S_{n}=1-\frac{1}{n+1} \]

Then \[ \lim_{n\rightarrow \infty }S_{n}=\lim_{n\rightarrow \infty }\left( {1-\frac{1 }{n+1}}\right) =1 \]

The series \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{k(k+1)}\) converges, and its sum is \(1\).

NOTE

Sums for which the middle terms cancel, as in Example 1, are called telescopingsums.

NOW WORK

Problem 11.

Showing a Series Diverges

Show that the series \(\sum\limits_{k\,=\,1}^{\infty }(-1)^{k}=-1+1-1+\cdots\) diverges.

Solution The sequence \(\{S_{n}\}\) of partial sums for this series is \[ \begin{array}{c@{}l} S_{1} &= -1 \\ S_{2} &= -1+1=0 \\ S_{3} &= -1+1-1=-1 \\ S_{4} &= -1+1-1+1=0 \\ \vdots & \\ S_{n} &= \left\{ \begin{array}{r@{\qquad}l@{\quad}l} -1 & \hbox{if} & n \hbox{ is odd} \\ 0  & \hbox{if} & n \hbox{ is even} \end{array} \right.\end{array} \]

Since \(\lim\limits_{n\rightarrow \infty }S_{n}\) does not exist, the sequence \(\{S_{n}\}\) of partial sums diverges. Therefore, the series diverges.

NOW WORK

Problem 49.

Determining Whether a Series Converges or Diverges

Determine whether the series \(\sum\limits_{k\,=\,1}^{\infty }k=1+2+3+\cdots\) converges or diverges.

Solution The sequence \(\{S_{n}\}\) of partial sums is \[ \begin{array}{c@{}l} S_{1} & =1 \\ S_{2} & =1+2 \\ S_{3} & =1+2+3 \\ \vdots & \\ S_{n} & =1+2+3+\cdots +n \end{array} \]

To express \(S_{n}\) in a way that will make it easy to find \(\lim\limits_{n\,\rightarrow \,\infty }S_{n},\) we use the formula for the sum of the first \(n\) integers: \[ S_{n}=\sum\limits_{k=1}^{n}k=1+2+3+ \cdots +n=\dfrac{n(n+1) }{2} \]

Since \(\lim\limits_{n\,\rightarrow \,\infty }S_{n}=\lim\limits_{n\,\rightarrow \,\infty }\dfrac{n(n+1) }{2} =\infty ,\) the sequence \(\{S_{n}\}\) of partial sums diverges. So, the series \(\sum\limits_{k\,=\,1}^{\infty }k\) diverges.

RECALL

\[ \sum\limits_{k=1}^{n}{ k=1+2+\cdots +n} =\dfrac{n(n+1)}{2}. \]

(See Appendix A.5, p. A-41.)

NOW WORK

Problem 43.

spanDEFINITIONspan Geometric Series

A series of the form \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \sum\limits_{k\,=\,0}^{\infty}ar^{k}=\sum\limits_{k\,=\,1}^{\infty }ar^{k-1}=a+ar+ar^{2}+\cdots +ar^{n-1}+\cdots }} \]

where \(a\neq 0\) is called a geometric series.

THEOREM Convergence of a Geometric Series

• If \(\vert r\vert <1\), the geometric series \(\sum\limits_{k\,=\,1}^{\infty }ar^{k-1}\) converges, and its sum is \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{\sum\limits_{k\,=\,1}^{\infty }ar^{k-1}=\dfrac{a}{1-r}}} \]
• If \(\vert r\vert \geq 1\), the geometric series \(\sum\limits_{k\,=\,1}^{\infty }ar^{k-1}\) diverges.

Determining Whether a Geometric Series Converges

Determine whether each geometric series converges or diverges. If it converges, find its sum.

1. \(\sum\limits_{k\,=\,1}^{\infty }8 \left( \frac{2}{5}\right)^{k-1}\)
2. \(\sum\limits_{k\,=\,1}^{\infty }\left(-\frac{5}{9}\right) ^{k-1}\)
3. \(\sum\limits_{k\,=\,1}^{\infty }3 \left( \frac{3}{2}\right) ^{k-1}\)
4. \(\sum\limits_{k\,=\,1}^{\infty }\frac{1}{2^{k}}\)
5. \(\sum\limits_{k\,=\,0}^{\infty }\left( \frac{1}{3}\right) ^{k-1}\)

Solution We compare each series to \(\sum\limits_{k\,=\,1}^{\infty }ar^{k-1}\).

(a) In this series \(a=8\) and \(r=\dfrac{2}{5}\). Since \(\vert r\vert=\dfrac{2}{5}<1\), the series converges and \[ \sum\limits_{k\,=\,1}^{\infty }8 \left( \frac{2}{5}\right) ^{k-1}=\frac{8}{1-\dfrac{2}{5}}=8 \left( \frac{5}{3}\right) =\frac{40}{3} \]

(b) Here, \(a=1\) and \(r=-\dfrac{5}{9}\). Since \(\vert r\vert =\dfrac{5}{9}<1\), the series converges and \[ \sum\limits_{k\,=\,1}^{\infty }\left( -\frac{5}{9}\right) ^{k-1}=\frac{1}{ 1-\left( -\dfrac{5}{9}\right) }=\frac{9}{14} \]

(c) Here, \(a=3\) and \(r=\dfrac{3}{2}\). Since \( \vert r\vert =\dfrac{3}{2}>1\), the series \(\sum\limits_{k\,=\,1}^{\infty}3 \left( \dfrac{3}{2}\right) ^{k-1}\) diverges.

(d) \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{2^{k}}\) is not in the form \(\sum\limits_{k\,=\,1}^{\infty }ar^{k-1}\). To place it in this form, we proceed as follows: \[ \begin{eqnarray*} &\sum\limits_{k\,=\,1}^{\infty }~\frac{1}{2^{k}}=\sum\limits_{k\,=\,1}^{ \infty }\left( \frac{1}{2}\right) ^{k} = \sum\limits_{k\,=\,1}^{\infty }\left[ \dfrac{1}{2}\cdot \left( \dfrac{1}{2} \right) ^{k-1}\right] \\[-13pt] &\hspace{2pc} \underset{\color{#0066A7}{{{\hbox{Write in the form}}}} {\color{#0066A7}{\hbox{\(\sum\limits_{k=1}^{\infty}ar^{k-1}\)}}}}{\color{#0066A7}{{{\kern1pt\left\uparrow{\vphantom{\vrule width0pc height8pt depth0pt}}\right.}}}} \end{eqnarray*} \]

So, \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{2^{k}}\) is a geometric series with \(a=\dfrac{1}{2}\) and \(r=\dfrac{1}{2}\). Since \(\vert r\vert <1,\) the series converges, and its sum is \[ \sum\limits_{k\,=\,1}^{\infty }\frac{1}{2^{k}}=\frac{\dfrac{1}{2}}{1-\dfrac{ 1}{2}}=1 \]

which agrees with the sum we found earlier.

(e) \(\sum\limits_{k\,=\,0}^{\infty }\left( \dfrac{1}{3}\right) ^{k-1}\) starts at \(0,\) so it is not in the form, \(\sum\limits_{k\,=1}^{ \infty }ar^{k-1}\). To place it in this form, change the index to \(l\), where \(l=k+1.\) Then when \(k=0,\) \(l=1\) and \[ \sum\limits_{k\,=\,0}^{\infty }\left( \dfrac{1}{3}\right) ^{k-1} = \sum\limits_{l=1}^{\infty }\left(\dfrac{1}{3}\right) ^{l-2}=\sum\limits_{l=1}^{\infty }\left(\dfrac{1}{3}\right) ^{-1}\left( \dfrac{1}{3}\right) ^{l-1}=\sum\limits_{l=1}^{\infty }3 \left( \dfrac{1}{3} \right)^{l-1} \]

That is, \(\sum\limits_{k\,=\,0}^{\infty }\left( \dfrac{1}{3}\right) ^{k-1}=\sum\limits_{l=1}^{\infty }3 \left( \dfrac{1}{3}\right) ^{l-1}\) is a geometric series with \(a=3\) and \(r=\dfrac{1}{3}.\) Since \(\vert r\vert <1,\) the series converges, and its sum is \[ \sum\limits_{k\,=\,0}^{\infty }\left( \dfrac{1}{3}\right) ^{k-1}=\dfrac{3}{1-\dfrac{1}{3}}=\dfrac{9}{2} \]

NOW WORK

Problems 21 and 29.

Writing a Repeating Decimal as a Fraction

Express the repeating decimal \(0.090909\ldots\) as a quotient of two integers.

Solution We write the infinite decimal \(0.090909\ldots\) as an infinite series: \[ \begin{eqnarray*} 0.090909\ldots &=& 0.09 + 0.0009+0.000009+0.00000009+\cdots \\ &=& \dfrac{9}{100}+\dfrac{9}{10000} +\dfrac{9}{1000000}+\cdots \\ &=& \dfrac{9}{100}\left( 1+\dfrac{1}{100}+\dfrac{1}{10000}+\dfrac{1}{1000000} +\cdots\right) \\ &=& \sum\limits_{k\,=\,1}^{\infty }\dfrac{9}{100}\left( \frac{1}{ 100}\right) ^{k-1} \end{eqnarray*} \]

This is a geometric series with \(a=\dfrac{9}{100}\) and \(r=\dfrac{1}{100}\). Since \(\left\vert  r\right\vert <1\), the series converges and its sum is \[ \sum\limits_{k\,=\,1}^{\infty }\dfrac{9}{100}\left( \frac{1}{100}\right) ^{k-1}=\frac{\dfrac{9}{100}}{1-\dfrac{1}{100}}=\frac{9}{99}=\frac{1}{11} \]

So, \(0.090909\ldots \,=\dfrac{1}{11}\).

NOW WORK

Problem 59.

Using a Geometric Series with a Bouncing Ball

A ball is dropped from a height of 12m. Each time it strikes the ground, it bounces back to a height three-fourths the distance from which it fell. Find the total distance traveled by the ball. See Figure 18.

Solution Let \(h_{n}\) denote the height of the ball on the \(n\)th bounce. Then \[ \begin{array}{cl} h_{0} & =12 \\ h_{1} & =\dfrac{3}{4}(12) \\ h_{2} & =\dfrac{3}{4}\left[ \dfrac{3}{4}(12)\right] =\left( \dfrac{3}{4}\right) ^{2}(12) \\ \vdots & \\ h_{n} & =\left( \dfrac{3}{4}\right) ^{n}(12) \end{array} \]

After the first bounce, the ball travels up a distance \(h_{1}=\dfrac{3}{4}(12)\) and then the same distance back down. Between the first and the second bounce, the total distance traveled is therefore \(h_{1}+h_{1}=2h_{1}\). The total distance \(H\) traveled by the ball is \[ \begin{eqnarray*} H=h_{0}+2h_{1}+2h_{2}+2h_{3}+\cdots &=& h_{0}+\sum\limits_{k\,=\,1}^{\infty }\left( 2h_{k}\right) =12+\sum\limits_{k\,=\,1}^{\infty }2 \left[ 12 \left( { \frac{3}{4}}\right) ^{k}\right]\\[8pt] &=& 12+\sum\limits_{k\,=\,1}^{\infty }24\left[ \frac{3}{4}\left( {\frac{3}{4}}\right) ^{k-1}\right] \\ &=& 12+\sum\limits_{k\,=\,1}^{\infty }18\left( {\frac{3}{4}}\right) ^{k-1} \end{eqnarray*} \]

The sum is a geometric series with \(a=18\) and \(r=\dfrac{3}{4}\). The series converges and \[ H=12+\sum\limits_{k\,=\,1}^{\infty }18\left( \frac{3}{4}\right) ^{k-1}=12 + \frac{18}{1-\dfrac{3}{4}}=84 \]

The ball travels a total distance of \(84\) m.

NOW WORK

Problem 63.

spanDEFINITIONspan Harmonic Series

The infinite series \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{k}=1+ \dfrac{1}{2}+\dfrac{1}{3}+\cdots }} \]

is called the harmonic series.

THEOREM

The harmonic series \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{k}\) diverges.

RECALL

An unbounded sequence diverges (p. 547).

Proof

To show that the harmonic series diverges, we look at the partial sums whose indexes of summation are powers of \(2\). That is, we investigate the sequence \(S_{1}\), \(S_{2}\), \(S_{4}\), \(S_{8}\), and so on. We can show that this sequence is not bounded as follows: \[ \begin{array}{l} S_{1}=1>\dfrac{1}{2}=1\left( \dfrac{1}{2}\right)\\[10pt] S_{2}=1+\dfrac{1}{2}>\dfrac{1}{2}+\dfrac{1}{2}=2\left( \dfrac{1}{2}\right)\\[10pt] S_{4}=1+\dfrac{1}{2}+\left( \dfrac{1}{3}+\dfrac{1}{4}\right) >2\left( \dfrac{ 1}{2}\right) +\left( \dfrac{1}{4}+\dfrac{1}{4}\right) =3\left( \dfrac{1}{2}\right)\\ S_{8}=1+\dfrac{1}{2}+\left( \dfrac{1}{3}+\dfrac{1}{4}\right) +\left( \dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}\right)\\ \quad \, >3\left( \dfrac{1}{2}\right) +\left( \dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}\right) =4\left( \dfrac{1}{2}\right)\\ \vdots \\ \hspace{-10pt} S_{2^{n-1}}>n\left( \dfrac{1}{2}\right) \end{array} \]

We conclude that the sequence \(\left\{ S_{2^{n-1}}\right\}\) is not bounded, so the sequence \(\{S_{n}\}\) of partial sums is not bounded. It follows that the sequence \(\{S_{n}\}\) of partial sums diverges. Therefore, the harmonic series \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{k}\) diverges.