# Modeling, Functions, and Graphs

## Section9.4Infinite Geometric Series

### SubsectionSummation Notation

There is a convenient notation for representing a series. For example, consider the sum of the first fifteen terms of the sequence
\begin{equation*} 4, 7, 10, \cdots, 3n+1, \cdots \end{equation*}
To find the terms of the series we replace $$n$$ in the general term $$3n+1$$ by the numbers 1 through 15. So, instead of writing out all the terms, we might express the sum as
\begin{equation*} \text{The sum, as}~ n~ \text{runs from 1 to 15, of}~ 3n+1 \end{equation*}
We use the Greek letter $$\Sigma$$ (called "sigma") to stand for "the sum," and we show the first and last values of $$n$$ below and above the summation symbol $$\Sigma\text{,}$$ like this:
\begin{equation*} \sum_{n=1}^{15} 3n+1 \end{equation*}
Writing such an expression is sometimes called "using sigma notation." The letter $$n$$ is called the index of summation; it is like a variable because it represents numerical values.

#### Note9.53.

Any letter can be used for the index of summation; $$i,~j,$$ and $$k$$ are other common choices. Of course, the letter used for the index of summation does not affect the sum.

#### Example9.54.

Use sigma notation to represent the sum of the first 20 terms of the sequence
\begin{equation*} -1, 2, 7, \cdots, k^2-2, \cdots \end{equation*}
Solution.
The general term of the sequence is $$k^2-2\text{,}$$ and the first term is $$-1\text{,}$$ which we find by letting $$k=1$$ in the formula for the general term. Thus,
\begin{equation*} \sum_{k=1}^{20} (k^2-2) \end{equation*}

#### Checkpoint9.55.Practice 1.

Use sigma notation to represent the sum of the first 20 terms of the sequence
\begin{equation*} \begin{gathered} 5, 8, 11, \ldots, 3k+2, \ldots \end{gathered} \end{equation*}
\begin{equation*} \sum_{k=1}^{b} a_k =5+ 8+11+\cdots+ (3k+2)+\cdots \end{equation*}
where $$b=$$ and $$a_k=$$
$$20$$
$$3k+2$$
Solution.
$$\displaystyle\sum_{k=1}^{20} (3k+2)$$

#### Checkpoint9.56.QuickCheck 1.

What does the notation $$\displaystyle{\sum_{j=1}^{8} \frac{1}{j}}$$ mean?
• List 8 terms of the sequence $$a_j=\displaystyle\frac{1}{j}\text{.}$$
• Add the numbers from 1 to 8 and take the reciprocal of the sum.
• Add the reciprocals of the numbers from 1 to 8.
• Add the first and eighth term of the sequence described.
$$\text{Choice 3}$$
Solution.
Add the reciprocals of the numbers from 1 to 8.
The expanded form of a series written in sigma notation is obtained by writing out all the terms of the series. For example,
\begin{equation*} \sum_{m=4}^{8} \dfrac{3}{m} = \dfrac{3}{4} + \dfrac{3}{5} + \dfrac{3}{6} + \dfrac{3}{7} + \dfrac{3}{8} \end{equation*}

#### Note9.57.

Notice that the series above has five terms, which is one more than the difference of the upper and lower limits of summation, $$8 - 4\text{.}$$
Recall that the general term of an arithmetic series is a linear function of the index, and the general term of a geometric series is an exponential function. If we recognize a given series as one of these two types, we can use the formulas developed in the last section to evaluate the sum.

#### Example9.58.

Compute the value of each series.
1. $$\displaystyle \displaystyle{\sum_{i=1}^{13} (90-5i)}$$
2. $$\displaystyle \displaystyle{\sum_{k=0}^{9} 2^k}$$
Solution.
1. Because the general term $$90-5i$$ is linear, this is an arithmetic series. By writing out the first few terms of the series,
\begin{equation*} 85+80+75+70+\cdots \end{equation*}
we can verify that the first term of the series is $$a_1=85$$ and the common difference is $$d=-5\text{.}$$ We also need to know the last term of the series, so we substitute $$n=13$$ in the general term to find $$a_{13} = 90-5(13) = 25$$ Thus,
\begin{equation*} \sum_{i=1}^{13} (90-5i) = \dfrac{13}{2}(85+25) = 715 \end{equation*}
2. The general term of this series is exponential, so the series is geometric. By writing out a few terms of the series,
\begin{equation*} 1+2+4+8+\cdots \end{equation*}
we confirm that the first term of the series is $$a_1=1$$ and the common ratio is $$r=2\text{.}$$ The series has 10 terms, from $$k=0$$ to $$k=9\text{,}$$ so $$n=10\text{.}$$ Finally, we substitute these values into the formula for geometric series to obtain
\begin{equation*} \sum_{k=0}^{9} 2^k = \dfrac{1(1-2^{10})}{1-2} = 1023 \end{equation*}

#### Checkpoint9.59.Practice 2.

Compute the value of each series.
1. $$\displaystyle{\sum_{k=1}^{50} (3k+2)}=$$
2. $$\displaystyle{\sum_{n=1}^{8} 10^n}=$$
$$3925$$
$$1.11111\times 10^{8}$$
Solution.
1. 3925
2. 111,111,110
If the series is not arithmetic or geometric, we don’t have a formula for the sum, so we must compute the sum directly.

#### Example9.60.

Compute the value of each series.
1. $$\displaystyle \displaystyle{\sum_{m=1}^{5} m^2}$$
2. $$\displaystyle \displaystyle{\sum_{p=1}^{800} 5}$$
Solution.
1. Because the general term, $$m^2\text{,}$$ is neither linear nor exponential, we know that the series is not arithmetic or geometric. However, we can expand the series and evaluate it directly.
\begin{equation*} \begin{aligned}[t] \displaystyle{\sum_{m=1}^{5} m^2}\amp = 1^2+2^2+3^2+4^2+5^2\\ \amp = 1+4+9+16+25=55\\ \end{aligned} \end{equation*}
2. The general term is the number 5. Because the index $$p$$ runs from 1 to 800, we are adding 800 terms, each of which is 5. Thus
\begin{equation*} \begin{aligned}[t] \displaystyle{\sum_{p=1}^{800} 5}\amp = 5+5+5+\cdots+5+5\\ \amp = 800(5) = 4000\\ \end{aligned} \end{equation*}

#### Checkpoint9.61.Practice 3.

Compute the value of each series.
1. $$\displaystyle{\sum_{k=0}^{20} \dfrac{1}{3}}=$$
2. $$\displaystyle{\sum_{k=0}^{3} \dfrac{k}{k+1}}=$$
$$7$$
$$\frac{23}{12}$$
Solution.
1. 7
2. $$\displaystyle \dfrac{23}{12}$$

#### Checkpoint9.62.QuickCheck 2.

Which of these series is neither arithmetic nor geometric?
• $$\displaystyle \displaystyle\sum_{k=5}^{15} \left(\frac{1}{2}k-4\right)$$
• $$\displaystyle \displaystyle\sum_{m=1}^{50} 0.8^m$$
• $$\displaystyle \displaystyle\sum_{n=2}^{12} (n^3+2)$$
• $$\displaystyle \displaystyle\sum_{p=1}^{5} 5(1.5)^{2p}$$
$$\text{Choice 3}$$
Solution.
$$\displaystyle\sum_{n=2}^{12} (n^3+2)$$ is neither arithmetic nor geometric. (The first series is arithmetic; the second and fourth series are both geometric.

### SubsectionInfinite Series

A series with infinitely many terms is called an infinite series. Is it possible to add infinitely many terms and arrive at a finite sum? In some cases, if the terms added are small enough, the answer is yes.
Consider the infinite geometric series
\begin{equation*} \dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\cdots \end{equation*}
The $$n^{\text{th}}$$ partial sum of the series is the sum of its first $$n$$ terms, and is denoted by $$S_n\text{.}$$ Thus,
\begin{align*} S_1\amp = \dfrac{1}{2}\\ S_2 \amp = \dfrac{1}{2}+\dfrac{1}{4} = \dfrac{3}{4}\\ S_3 \amp = \dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8} = \dfrac{7}{8}\\ S_4 \amp = \dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16} = \dfrac{15}{16} \end{align*}
Notice that for each partial sum, the new term added gets smaller and smaller: $$\dfrac{1}{2}\text{,}$$ $$\dfrac{1}{4}\text{,}$$ $$\dfrac{1}{8}\text{,}$$ $$\dfrac{1}{16},$$ and so on. You can also see that as $$n$$ increases and we add more and more terms of the series, the partial sums are getting closer to 1. In fact, as $$n$$ becomes very large, $$S_n$$ gets very close to 1. It seems reasonable that the sum of all the terms of the series is 1.

#### Note9.63.

Will the partial sums ever be greater than 1? No, because each new term $$a_n$$ added is half the difference between $$S_n$$ and 1.
We can make this conjecture more plausible by examining the formula for the $$n^{\text{th}}$$ partial sum of a geometric series,
\begin{align*} S_n\amp = \dfrac{a_{n+1} - a_1}{r-1}\\ \amp = \dfrac{ar^n - a}{r-1} = \dfrac{a - ar^n}{1-r} \amp\amp (r \not= 1) \end{align*}
Look at the second term of the numerator, $$ar^n\text{.}$$ This term is the only part of the formula that depends on $$n\text{.}$$ What happens to $$ar^n$$ as $$n$$ increases? Consider two examples:
• If $$r = \dfrac{1}{2}\text{,}$$ then
\begin{equation*} r^2 = \left(\dfrac{1}{2}\right)^2 = \dfrac{1}{4},\qquad r^3 = \left(\dfrac{1}{2}\right)^3 = \dfrac{1}{8},\qquad r^4 = \left(\dfrac{1}{2}\right)^4 = \dfrac{1}{16}, \end{equation*}
and so on, with $$\left(\dfrac{1}{2}\right)^n$$ becoming smaller and smaller for larger values of $$n\text{.}$$ Each time we multiply by another factor of $$r=\dfrac{1}{2}\text{,}$$ the product gets smaller, because $$\dfrac{1}{2} \lt 1\text{.}$$
• On the other hand, if $$r \gt 1\text{,}$$ multiplying by $$r$$ makes the product larger. If $$r = \dfrac{3}{2}$$ for example, then
\begin{equation*} r^2 = \left(\dfrac{3}{2}\right)^2 = \dfrac{9}{4},\qquad r^3 = \left(\dfrac{3}{2}\right)^3 = \dfrac{27}{8},\qquad r^4 = \left(\dfrac{3}{2}\right)^4 = \dfrac{81}{16}, \end{equation*}
and so on. In this case the powers of $$r$$ are increasing.
In general, we have the following result.
1. If $$0 \lt \abs{r} \lt 1\text{,}$$ then $$r^n$$ gets closer to zero as $$n$$ increases.
2. If $$\abs{r} \gt 1\text{,}$$ then $$r^n$$ does not approach a finite number as $$n$$ increases.
Now let’s return to the formula for $$S_n$$ and write it in the form
\begin{equation*} S_n = \dfrac{a}{1-r} (1-r^n) \end{equation*}
where we have factored $$a$$ from the numerator. There are two cases to consider. If $$\abs{r} \lt 1\text{,}$$ then $$r^n$$ approaches 0, and the factor $$1-r^n$$ gets closer and closer to 1 as $$n$$ grows larger. Consequently, if we compute $$S_n$$ for larger and larger values of $$n\text{,}$$ the sum approaches the value
\begin{equation*} \dfrac{a}{1-r} \end{equation*}
This analysis motivates us to define the sum of an infinite geometric series as follows.

#### Sum of an Infinite Geometric Series.

The sum of an infinite geometric series $$~~\displaystyle{\sum_{k=0}^{\infty} ar^{k-1}}~~$$ is
\begin{equation*} S_{\infty} = \dfrac{a}{1-r}~~~~~~\text{if}~~~~-1 \lt r \lt 1 \end{equation*}
In the second case, if $$\abs{r} \gt 1\text{,}$$ as in the infinite series
\begin{equation*} 3 + 6 + 12 + \cdots \end{equation*}
where $$r = 2\text{,}$$ the terms become larger as $$n$$ increases, and the sum of the series is not a finite number. In this case the series does not have a sum.

#### Checkpoint9.64.QuickCheck 3.

Under what circumstances does the series $$\displaystyle\sum_{k=0}^{\infty} ar^k$$ have a finite sum?
• If $$a \lt 1$$
• If $$a \lt 1$$
• If $$| r | \lt 1$$
• If $$ar \lt r$$
$$\text{Choice 3}$$
Solution.
If $$\abs{r} \lt 1$$

#### Example9.65.

Make a table showing the first five partial sums of each series. Then use the formula to find the sum, if it exists.
1. $$\displaystyle ~~\displaystyle{\sum_{j=0}^{\infty} 30(0.8)^j}~~$$
2. $$\displaystyle ~~\displaystyle{\sum_{m=0}^{\infty} 3(\dfrac{4}{3})^m}~~$$
Solution.
1. First, we evaluate the general term $$30(0.8)^j$$ for $$j=1, 2, \cdots, 5$$ and compute the partial sums. For example,
\begin{equation*} \begin{aligned}[t] S_1\amp = 30(0.8)^1 = 24\\ S_2\amp = 30(0.8)^1+30(0.8)^2 = 43.2\\ \end{aligned} \end{equation*}
and so on. The results, rounded to hundredths, are shown in the table.
 $$n$$ $$1$$ $$2$$ $$3$$ $$4$$ $$5$$ $$S_n$$ $$24$$ $$43.2$$ $$58.56$$ $$70.85$$ $$80.68$$
This is an infinite geometric series with $$a=30$$ and $$r=0.8\text{.}$$ The series has a sum because $$\abs{r} \lt 1\text{.}$$ Thus,
\begin{equation*} S_{\infty}=\dfrac{a}{1-r} = \dfrac{30}{1-0.8} = 150 \end{equation*}
2. We evaluate the general term $$3(\dfrac{4}{3})^m$$ for $$m=1, 2, \cdots, 5$$ and compute the partial sums. For example,
\begin{equation*} \begin{aligned}[t] S_1\amp = 3(\dfrac{4}{3})^1 = 4\\ S_2\amp = 3(\dfrac{4}{3})^1 + 3(\dfrac{4}{3})^2 = \dfrac{28}{3}\\ \end{aligned} \end{equation*}
and so on. The results, rounded to hundredths, are shown in the table.
 $$n$$ $$1$$ $$2$$ $$3$$ $$4$$ $$5$$ $$S_n$$ $$4$$ $$9.33$$ $$16.44$$ $$25.93$$ $$38.57$$
This is a geometric series with $$a=3$$ and $$r = \dfrac{4}{3}\text{.}$$ The infinite series does not have a sum because $$\abs{r} \gt 1\text{.}$$

#### Checkpoint9.66.Practice 4.

Find the sum, if it exists. If the series does not have a sum, enter “no sum”.
1. $$\displaystyle{\sum_{j=0}^{\infty} 13\left(\dfrac{7}{6}\right)^j}=$$
2. $$\displaystyle{\sum_{m=0}^{\infty} 5.9(0.9)^m}=$$
$$\text{no sum}\hbox{, }\text{dne}\hbox{, or }\text{diverges}$$
$$\text{59}$$
Solution.
1. No sum
2. 59

### SubsectionRepeating Decimals

An interesting application of geometric series involves repeating decimals. Recall that the decimal representation of a rational number either terminates, as does $$0.75,$$ or repeats a pattern of digits. For example, you probably recognize $$0.333\overline{3}$$ as the decimal representation of $$\dfrac{1}{3}\text{.}$$ It is easy to find the decimal form of a fraction: we just divide the denominator into the numerator. Is there a way to find the fractional form of a repeating decimal?
Consider the repeating decimal
\begin{equation*} 0.2121\overline{21} \end{equation*}
We can write this number as an infinite geometric series:
\begin{equation*} 0.21 + 0.0021 + 0.000021 + \cdots \end{equation*}
The first term of this series is $$0.21\text{,}$$ or $$\dfrac{21}{100}\text{,}$$ and its common ratio is $$r=0.01\text{,}$$ or $$\dfrac{1}{100}\text{.}$$ Because $$\abs{r} \lt 1\text{,}$$ the series has a sum given by
\begin{equation*} \begin{aligned}[t] S_{\infty}\amp = \dfrac{a}{1-r} = \dfrac{\dfrac{21}{100}}{1-\dfrac{1}{100}}\\ \amp = \dfrac{\dfrac{21}{100}}{\dfrac{99}{100}} = \dfrac{21}{99} = \dfrac{7}{33}\\ \end{aligned} \end{equation*}
Thus, the decimal number $$0.2121\overline{21}$$ is equal to the fraction $$\dfrac{7}{33}\text{.}$$

#### Example9.67.

Find a common fraction equivalent to $$0.3\overline{7}\text{.}$$
Solution.
The decimal can be written as $$0.3 + 0.0\overline{7}\text{.}$$ We will find a fraction equivalent to the repeating decimal $$0.0\overline{7}$$ and add that to $$0.3\text{,}$$ or $$\dfrac{3}{10}\text{.}$$ We write $$0.0\overline{7}$$ as a series:
\begin{equation*} 0.0\overline{7} = \dfrac{7}{100} + \dfrac{7}{1000} + \dfrac{7}{10,000} + \cdots \end{equation*}
This is an infinite geometric series with first term $$\dfrac{7}{100}$$ and common ratio $$\dfrac{1}{10}\text{.}$$ The sum of the series is given by
\begin{equation*} \begin{aligned}[t] S_{\infty}\amp = \dfrac{a}{1-r} = \dfrac{\dfrac{7}{100}}{1-\dfrac{1}{10}}\\ \amp = \dfrac{\dfrac{7}{100}}{\dfrac{9}{10}} = \dfrac{7}{100} \cdot \dfrac{9}{10} = \dfrac{7}{90}\\ \end{aligned} \end{equation*}
Finally, we add $$\dfrac{7}{90}$$ and $$\dfrac{3}{10}$$ to find
\begin{equation*} 0.3\overline{7} = \dfrac{3}{10}+\dfrac{7}{90} = \dfrac{34}{90} \end{equation*}

#### Checkpoint9.68.Practice 5.

Find a common fraction equivalent to $$0.\overline{8}\text{:}$$
$$\frac{8}{9}$$
Solution.
$$\dfrac{8}{9}$$

### SubsectionSection Summary

#### SubsubsectionVocabulary

• Sigma notation
• Index of summation
• Infinite series
• Partial sum

#### SubsubsectionCONCEPTS

1. We can use sigma notation to denote a series.
2. It is possible to add infinitely many terms and arrive at a finite sum if the terms are small enough.
3. The sum of an infinite geometric series $$~~\displaystyle{\sum_{k=0}^{\infty} ar^{k-1}}~~$$ is
\begin{equation*} S_{\infty} = \dfrac{a}{1-r}~~~~~~\text{if}~~~~-1 \lt r \lt 1 \end{equation*}
4. If $$\abs{r} \gt 1$$ in an infinite geometric series, the series does not have a sum.

#### SubsubsectionSTUDY QUESTIONS

1. Explain how to use sigma notation to define a series.
2. What is an infinite series?
3. What is a partial sum of an infinite series?
4. State a formula for evaluating an infinite geometric series. Under what conditions is the formula valid?
5. A repeating decimal can be rewritten as what kind of series?

#### SubsubsectionSKILLS

Practice each skill in the Homework Problems listed.
1. Express sigma notation in expanded form: #1-8
2. Write a series in sigma notation: #12-20
3. Identify a series as arithmetic, geometric, or neither and evaluate: #21-40
4. Evaluate an infinite geometric series: #41-48, #57-60
5. Write a repeating decimal as a common fraction: #49-56

### ExercisesHomework 9.4

#### Exercise Group.

For Problems 1–8, write the sum in expanded form.
##### 1.
$$\displaystyle{\sum_{i=1}^{4} i^2}$$
##### 2.
$$\displaystyle{\sum_{i=1}^{3} (3i-2)}$$
##### 3.
$$\displaystyle{\sum_{j=5}^{7} (j-2)}$$
##### 4.
$$\displaystyle{\sum_{j=2}^{6} (j^2+1)}$$
##### 5.
$$\displaystyle{\sum_{k=1}^{4} k(k+1)}$$
##### 6.
$$\displaystyle{\sum_{k=2}^{6} \dfrac{k}{2}(k+1)}$$
##### 7.
$$\displaystyle{\sum_{m=1}^{4} \dfrac{(-1)^m}{2^m}}$$
##### 8.
$$\displaystyle{\sum_{m=3}^{5} \dfrac{(-1)^{m+1}}{m-2}}$$

#### Exercise Group.

For Problems 9–20, write the series using sigma notation.
##### 9.
$$1+3+5+7$$
##### 10.
$$2+4+6+8$$
##### 11.
$$5+5^3+5^5+5^7$$
##### 12.
$$4^3+4^5+4^7+4^9+4^{11}$$
##### 13.
$$1+4+9+16+25$$
##### 14.
$$1+8+27+64+125$$
##### 15.
$$\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+\dfrac{4}{5}+\dfrac{5}{6}$$
##### 16.
$$\dfrac{2}{1}+\dfrac{3}{2}+\dfrac{4}{3}+\dfrac{5}{4}+\dfrac{6}{5}$$
##### 17.
$$\dfrac{1}{1}+\dfrac{2}{3}+\dfrac{3}{5}+\dfrac{4}{7}+\dfrac{5}{9}+\dfrac{6}{11}$$
##### 18.
$$\dfrac{3}{1}+\dfrac{5}{3}+\dfrac{7}{5}+\dfrac{9}{7}+\dfrac{11}{9}$$
##### 19.
$$\dfrac{1}{1}+\dfrac{2}{2}+\dfrac{4}{3}+\dfrac{8}{4}+\cdots$$
##### 20.
$$\dfrac{1}{2}+\dfrac{3}{4}+\dfrac{9}{6}+\dfrac{27}{8}+\cdots$$

#### Exercise Group.

For Problems 21–40, identify the series as arithmetic, geometric or neither, then evaluate it.
##### 21.
$$\displaystyle{\sum_{i=1}^{6} (i^2+1)}$$
##### 22.
$$\displaystyle{\sum_{i=1}^{5} 3i^2}$$
##### 23.
$$\displaystyle{\sum_{j=1}^{4} \dfrac{1}{j}}$$
##### 24.
$$\displaystyle{\sum_{j=0}^{4} \dfrac{2}{j+1}}$$
##### 25.
$$\displaystyle{\sum_{k=1}^{100} 1}$$
##### 26.
$$\displaystyle{\sum_{k=1}^{300} 3}$$
##### 27.
$$\displaystyle{\sum_{q=1}^{20} 3^q}$$
##### 28.
$$\displaystyle{\sum_{p=1}^{30} 2^p}$$
##### 29.
$$\displaystyle{\sum_{k=1}^{200} k}$$
##### 30.
$$\displaystyle{\sum_{k=1}^{150} k}$$
##### 31.
$$\displaystyle{\sum_{n=1}^{6} n^3}$$
##### 32.
$$\displaystyle{\sum_{n=1}^{7} n^2}$$
##### 33.
$$\displaystyle{\sum_{n=0}^{30} (3n-1)}$$
##### 34.
$$\displaystyle{\sum_{k=0}^{20} (5k+2)}$$
##### 35.
$$\displaystyle{\sum_{k=0}^{25} (5-2k)}$$
##### 36.
$$\displaystyle{\sum_{p=0}^{15} (2-3p)}$$
##### 37.
$$\displaystyle{\sum_{j=0}^{10} 5\cdot 2^j}$$
##### 38.
$$\displaystyle{\sum_{j=0}^{10} 3\cdot 2^j}$$
##### 39.
$$\displaystyle{\sum_{m=0}^{12} 50(1.08)^m}$$
##### 40.
$$\displaystyle{\sum_{m=0}^{18} 300(1.12)^m}$$

#### Exercise Group.

For Problems 41–48,
1. Make a table of values showing the first five partial sums of the series.
2. Evaluate the series algebraically.
##### 41.
$$\displaystyle{\sum_{n=1}^{\infty} \left(\dfrac{1}{2}\right)^n}$$
##### 42.
$$\displaystyle{\sum_{n=1}^{\infty} \left(\dfrac{1}{3}\right)^n}$$
##### 43.
$$\displaystyle{\sum_{k=0}^{\infty} 12(0.15)^{k-1}}$$
##### 44.
$$\displaystyle{\sum_{k=0}^{\infty} 25(0.08)^{k-1}}$$
##### 45.
$$\displaystyle{\sum_{j=0}^{\infty} 4\left(\dfrac{-3}{5}\right)^{j}}$$
##### 46.
$$\displaystyle{\sum_{j=0}^{\infty} 6\left(\dfrac{-2}{5}\right)^{j}}$$
##### 47.
$$\displaystyle{\sum_{n=4}^{\infty} 3\left(\dfrac{1}{2}\right)^{n}}$$
##### 48.
$$\displaystyle{\sum_{n=3}^{\infty} 2\left(\dfrac{1}{3}\right)^{n}}$$

#### Exercise Group.

For Problems 49–56, find a fraction equivalent to the repeating decimal.
##### 49.
$$0.\overline{4}$$
##### 50.
$$0.\overline{6}$$
##### 51.
$$0.\overline{31}$$
##### 52.
$$0.\overline{45}$$
##### 53.
$$0.\overline{410}$$
##### 54.
$$0.\overline{027}$$
##### 55.
$$0.12\overline{8}$$
##### 56.
$$0.8\overline{3}$$

#### 57.

The arc length through which the bob of a pendulum moves is nine-tenths of its preceding arc length. Approximately how far will the bob move before coming to rest if the first arc length is $$12$$ inches?

#### 58.

A force is applied to a particle moving in a straight line in such a fashion that each second it moves only one-half of the distance it moved the preceding second. If the particle moves $$10$$ centimeters the first second, approximately how far will it move before coming to rest?

#### 59.

A ball returns two-thirds of its preceding height on each bounce. If the ball is dropped from a height of $$6$$ feet, approximately what is the total distance the ball travels before coming to rest? (Hint: Compute separately the total distance the ball falls from the total distance it moves upwards.)

#### 60.

If a ball is dropped from a height of $$10$$ feet and returns three-fifths of its preceding height on each bounce, approximately what is the total distance the ball travels before coming to rest? (See the Hint for Problem 59.)