Subsection Summation 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.
Example 9.54.
Use sigma notation to represent the sum of the first 20 terms of the sequence
\begin{equation*}
1, 2, 7, \cdots, k^22, \cdots
\end{equation*}
Solution.
The general term of the sequence is \(k^22\text{,}\) 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^22)
\end{equation*}
Checkpoint 9.55. Practice 1.
Checkpoint 9.56. QuickCheck 1.
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*}
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.
Example 9.58.
Compute the value of each series.
\(\displaystyle \displaystyle{\sum_{i=1}^{13} (905i)}\)
\(\displaystyle \displaystyle{\sum_{k=0}^{9} 2^k}\)
Solution.
Because the general term \(905i\) 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 \(a_{13} = 905(13) = 25\) Thus,
\begin{equation*}
\sum_{i=1}^{13} (905i) = \dfrac{13}{2}(85+25) = 715
\end{equation*}
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(12^{10})}{12} = 1023
\end{equation*}
Checkpoint 9.59. Practice 2.
If the series is not arithmetic or geometric, we don’t have a formula for the sum, so we must compute the sum directly.
Example 9.60.
Compute the value of each series.
\(\displaystyle \displaystyle{\sum_{m=1}^{5} m^2}\)
\(\displaystyle \displaystyle{\sum_{p=1}^{800} 5}\)
Solution.
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*}
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*}
Checkpoint 9.61. Practice 3.
Checkpoint 9.62. QuickCheck 2.
Subsection Infinite 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.
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}{r1}\\
\amp = \dfrac{ar^n  a}{r1} = \dfrac{a  ar^n}{1r} \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.
If \(0 \lt \abs{r} \lt 1\text{,}\) then \(r^n\) gets closer to zero as \(n\) increases.
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}{1r} (1r^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 \(1r^n\) 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}{1r}
\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 \(~~\displaystyle{\sum_{k=0}^{\infty} ar^{k1}}~~\) is
\begin{equation*}
S_{\infty} = \dfrac{a}{1r}~~~~~~\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.
Checkpoint 9.64. QuickCheck 3.
Example 9.65.
Make a table showing the first five partial sums of each series. Then use the formula to find the sum, if it exists.
\(\displaystyle ~~\displaystyle{\sum_{j=0}^{\infty} 30(0.8)^j}~~\)
\(\displaystyle ~~\displaystyle{\sum_{m=0}^{\infty} 3(\dfrac{4}{3})^m}~~\)
Solution.

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}{1r} = \dfrac{30}{10.8} = 150
\end{equation*}

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 \(\abs{r} \gt 1\text{.}\)
Checkpoint 9.66. Practice 4.
Subsection Repeating 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}{1r} = \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 \(\dfrac{7}{33}\text{.}\)
Example 9.67.
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}{1r} = \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*}
Checkpoint 9.68. Practice 5.
Exercises Homework 9.4
Exercise Group.
For Problems 1–8, write the sum in expanded form.
1.
\(\displaystyle{\sum_{i=1}^{4} i^2}\)
2.
\(\displaystyle{\sum_{i=1}^{3} (3i2)}\)
3.
\(\displaystyle{\sum_{j=5}^{7} (j2)}\)
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.
\(\displaystyle{\sum_{m=3}^{5} \dfrac{(1)^{m+1}}{m2}}\)
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.
\(\displaystyle{\sum_{n=0}^{30} (3n1)}\)
34.
\(\displaystyle{\sum_{k=0}^{20} (5k+2)}\)
35.
\(\displaystyle{\sum_{k=0}^{25} (52k)}\)
36.
\(\displaystyle{\sum_{p=0}^{15} (23p)}\)
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,
Make a table of values showing the first five partial sums of the series.
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.
\(\displaystyle{\sum_{k=0}^{\infty} 12(0.15)^{k1}}\)
44.
\(\displaystyle{\sum_{k=0}^{\infty} 25(0.08)^{k1}}\)
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 ninetenths 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 onehalf 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 twothirds 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 threefifths 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.)