Opening the book…
- Note to the Instructor (click to expand)
- Series are necessary for both Differential and Integral Calculus. Be as thorough as possible with this material and drive home the importance of the material.
The following is a list of learning objectives for this section.
- Learning Objectives (click to expand)
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- Use summation notation.
- Use the formula for the sum of the first \(n\) terms of an arithmetic series.
- Use the formula for the sum of the first \(n\) terms of a geometric series.
- Use the formula for the sum of an infinite geometric series.
- Solve annuity problems.
A couple decides to start a college fund for their daughter. They plan to invest $50 in the fund each month. The fund pays 6% annual interest, compounded monthly. How much money will they have saved when their daughter is ready to start college in 6 years? In this section, we will learn how to answer this question. To do so, we need to consider the amount of money invested and the amount of interest earned.
Using Summation Notation
To find the total amount of money in the college fund and the sum of the amounts deposited, we need to add the amounts deposited each month and the amounts earned in interest monthly. The sum of the terms of a sequence is called a series.
Consider, for example, the following sequence.\[\{a_n \} = \{ 3, 7, 11, 15, 19, \ldots \}\nonumber\]The series for this sequence is the infinite summation\[3+7+11+15+19+\ldots\nonumber\]Obviously, adding terms of a sequence forever is impossible, so we begin our excursion into summations of sequences by asking ourselves what would happen if we only added up, say, the first \(n\) terms of the sequence. This is called the \(n^{\text{th}}\) partial sum.
For example, with the series above, the first few partial sums are as follows:\[\begin{array}{rcl} S_1 & = & 3 \\[6pt] S_2 & = & 3 + 7 = 10 \\[6pt] S_3 & = & 3 + 7 + 11 = 21 \\[6pt] S_4 & = & 3 + 7 + 11 + 15 = 36 \\[6pt] \end{array}\nonumber\]Since partial sums can get pretty bulky to write, we introduce a new notation. Summation notation is used to represent series.
The index variable is also called the index of summation.
If we interpret the given notation, we see that it asks us to find the sum of the terms in the series \(a_k =2k\) for \(k=1\) through \(k=5\). We can begin by substituting the terms for \(k\) and listing out the terms of this series.\[\begin{array}{rcl} a_1 & = & 2(1) = 2 \\[6pt] a_2 & = & 2(2) = 4 \\[6pt] a_3 & = & 2(3) = 6 \\[6pt] a_4 & = & 2(4) = 8 \\[6pt] a_5 & = & 2(5) = 10 \\[6pt] \end{array}\nonumber\]We can find the sum of the series by adding the terms:\[\sum_{k = 1}^{5} 2k = 2+4+6+8+10=30.\nonumber\]
Does the lower limit of summation have to be 1?
No (and this is why our definition didn't start with the lower limit set as 1). The lower limit of summation can be any number, but 1 is frequently used. We will look at examples with lower limits of summation other than 1.
Evaluate \(\displaystyle \sum_{k=3}^{7} k^2\).
- Solution
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According to the notation, the lower limit of summation is 3 and the upper limit is 7. So we need to find the sum of \(k^2\) from \(k=3\) to \(k=7\). We find the terms of the series by substituting \(k=3, 4, 5, 6,\) and \(7\) into the function \(k^2\). We add the terms to find the sum.\[\begin{array}{rcl} \displaystyle \sum_{k = 3}^{7} k^2 & = & 3^2 + 4^2 + 5^2 + 6^2 + 7^2 \\[6pt] & = & 9 + 16 + 25 + 36 + 49 \\[6pt] & = & 135 \\[6pt] \end{array}\nonumber\]
Evaluate \(\displaystyle \sum_{k=2}^{5} (3k-1)\).
Arithmetic Series
Just as we studied special types of sequences, we will look at special types of series. Recall that an arithmetic sequence is a sequence in which the difference between any two consecutive terms is the common difference, \(d\). That is, if \(d\) is the common difference and \(a_1\) is the first term for the arithmetic sequence, then \(a_2 - a_1 = d\). Moreover, in general, \(a_n - a_{n -1} = d\). Since this difference is constant between any two terms in the arithmetic sequence, we could write the sequence terms as follows:\[\begin{array}{rcl} a_2 & = & a_1 + d \\[6pt] a_3 & = & a_1 + 2d \\[6pt] a_4 & = & a_1 + 3d \\[6pt] \vdots & = & \vdots \\[6pt] a_n & = & a_1 + (n - 1)d \\[6pt] \end{array}\nonumber\]
Arithmetic series are special in that we have a closed formula for their \(n^{\text{th}}\) partial sum.
That is, the sum of the first \(n\) terms of an arithmetic sequence is the product of \(n\) and the average of the first and \(n^{\text{th}}\) terms of the sequence.
Find the sum of each arithmetic series.
- \(5 + 8 + 11 + 14 + 17 + 20 + 23 + 26 + 29 + 32\)
- \(20 + 15 + 10 + \ldots + (-50)\)
- \(\displaystyle \sum_{k=1}^{12} 3k-8\)
- Solutions
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We are given \(a_1 = 5\) and \(a_n = 32\). A quick count shows the number of terms to be 10. Thus, \(n = 10\).
Substitute values for \(a_1\), \(a_n\), and \(n\) into the formula and simplify.\[S_{10} = 10 \cdot \dfrac{5+32}{2} = 5 \cdot (37) = 185.\nonumber\]
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We are given \(a_1 =20\) and \(a_n =-50\). The sequence given suggests that \(d = -5\). Use the formula for the general term of an arithmetic sequence to find \(n\).\[\begin{array}{rrclcl} & a_n & = & a_1 + (n - 1)d & & \\[6pt] \implies & -50 & = & 20 + (n - 1)(-5) & \quad & \left( \text{substituting} \right) \\[6pt] \implies & -70 & = & (n - 1)(-5) & \quad & \left( \text{subtracting }20\text{ from both sides} \right) \\[6pt] \implies & 14 & = & n - 1 & \quad & \left( \text{dividing both sides by }-5 \right) \\[6pt] \implies & 15 & = & n & \quad & \left( \text{adding }1\text{ to both sides} \right) \\[6pt] \end{array}\nonumber\]Substitute values for \(a_1\), \(a_n\), and \(n\) into the formula and simplify.\[S_{15} = 15 \cdot \dfrac{20-50}{2} = 15 \cdot (-15) = -225.\nonumber\]
- To find \(a_1\), substitute \(k=1\) into the given explicit formula.\[\begin{array}{rrcl} & a_k & = & 3k - 8 \\[6pt] \implies & a_1 & = & 3(1) - 8 = -5 \\[6pt] \end{array}\nonumber\]We are given that \(n=12\). To find \(a_{12}\), substitute \(k=12\) into the given explicit formula.\[\begin{array}{rrcl} & a_k & = & 3k - 8 \\[6pt] \implies & a_{12} & = & 3(12) - 8 = 28 \\[6pt] \end{array}\nonumber\]Substitute values for \(a_1\), \(a_n\), and \(n\) into the formula and simplify.\[\begin{array}{rrcl} & S_n & = & n \cdot \dfrac{a_1 + a_n}{2} \\[6pt] \implies & S_{12} & = & 12 \cdot \dfrac{-5 + 28}{2} = 6 \cdot 23 = 138 \\[6pt] \end{array}\nonumber\]
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Find the sum of each arithmetic series.
- \(1.4 + 1.6 + 1.8 + 2.0 + 2.2 + 2.4 + 2.6 + 2.8 + 3.0 + 3.2 + 3.4\)
- \(13 + 21 + 29 + \ldots + 69\)
- \(\displaystyle \sum_{k=1}^{10} 5 - 6k\)
On the Sunday after a minor surgery, a woman is able to walk a half-mile. Each Sunday, she walks an additional quarter-mile. After 8 weeks, what will be the total number of miles she has walked?
- Solution
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This problem can be modeled by an arithmetic series with \(a_1 = \frac{1}{2}\) and \(d = \frac{1}{4}\). We are looking for the total number of miles walked after 8 weeks, so we know that \(n = 8\), and we are looking for \(S_8\). To find \(a_8\), we can use the explicit formula for an arithmetic sequence.\[\begin{array}{rrcl} & a_n & = & a_1 + d(n - 1) \\[6pt] \implies & a_8 & = & \dfrac{1}{2} + \dfrac{1}{4}(8 - 1) = \dfrac{9}{4} \\[6pt] \end{array}\nonumber\]We can now use the formula for arithmetic series.\[\begin{array}{rrcl} & S_n & = & n \cdot \dfrac{a_1 + a_n}{2} \\[6pt] \implies & S_8 & = & 8 \cdot \dfrac{\frac{1}{2} + \frac{9}{4}}{2} = 4 \cdot \left( \dfrac{11}{4} \right) = 11 \\[6pt] \end{array}\nonumber\]She will have walked a total of 11 miles.
A man earns $100 in the first week of June. Each week, he earns $12.50 more than the previous week. After 12 weeks, how much has he earned?
Geometric Series
Just as the sum of the terms of an arithmetic sequence is called an arithmetic series, the sum of the terms in a geometric sequence is called a geometric series.
Likewise, we can derive a closed formula for the \(n^{\text{th}}\) partial sum of a geometric series.
Notice that this formula does not exist when \(r = 1\). This is not an issue because geometric sequences (which geometric series are based upon) are required to have common ratios not equal to 1.
Use the formula to find the indicated partial sum of each geometric series.
- \(S_{11}\) for the series \(8 + (-4) + 2 + \ldots\)
- \(\displaystyle \sum_{k = 1}^6 3 \cdot 2^k\)
- Solutions
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\(a_1 =8\), and we are given that \(n=11\). We can find \(r\) by dividing the second term of the series by the first.\[r= \dfrac{-4}{8} = -\dfrac{1}{2}.\nonumber\]Substitute values for \(a_1\), \(r\), and \(n\) into the formula and simplify.\[\begin{array}{rrcl} & S_n & = & \dfrac{a_1 (1 - r^n)}{1 - r} \\[6pt] \implies & S_{11} & = & \dfrac{8\left( 1 - \left( -\frac{1}{2} \right)^{11} \right)}{1 - \left( -\frac{1}{2} \right)} \approx 5.336 \\[6pt] \end{array}\nonumber\]
- Find \(a_1\) by substituting \(k=1\) into the given explicit formula.\[a_1 =3 \cdot 2^1 = 6.\nonumber\]We can see from the given explicit formula that \(r=2\). The upper limit of summation is 6, so \(n=6\).
Substitute values for \(a_1\), \(r\), and \(n\) into the formula, and simplify.\[\begin{array}{rrcl} & S_n & = & \dfrac{a_1 (1 - r^n)}{1 - r} \\[6pt] \implies & S_{6} & = & \dfrac{6\left( 1 - \left( 2 \right)^{6} \right)}{1 - 2} = 378 \\[6pt] \end{array}\nonumber\]
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Use the formula to find the indicated partial sum of each geometric series.
- \(S_{20}\) for the series \(1,000 + 500 + 250 + \ldots\)
- \(\displaystyle \sum_{k = 1}^8 3^k\)
At a new job, an employee's starting salary is $26,750. He receives a 1.6% annual raise. Find his total earnings at the end of 5 years.
- Solution
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The problem can be represented by a geometric series with \(a_1 =26,750\), \(n=5\), and \(r = 1.016\). Substitute values for \(a_1\), \(r\), and \(n\) into the formula and simplify to find the total amount earned at the end of 5 years.\[\begin{array}{rrcl} & S_n & = & \dfrac{a_1 (1 - r^n)}{1 - r} \\[6pt] \implies & S_{5} & = & \dfrac{26,750 \left( 1 - 1.016^{5} \right)}{1 - 1.016} \approx 138,099.03 \\[6pt] \end{array}\nonumber\]He will have earned a total of $138,099.03 by the end of 5 years.
At a new job, an employee's starting salary is $32,100. She receives a 2% annual raise. How much will she have earned by the end of 8 years?
Infinite Series
Thus far, we have only considered partial sums of series. These are also known as finite series. Sometimes, however, we are interested in the sum of the terms of an infinite sequence rather than the sum of only the first \(n\) terms.
An example of an infinite series is \(2+4+6+8+\ldots\).
This series can also be written in summation notation as \(\displaystyle \sum_{k=1}^{\infty} 2k\), where the upper limit of summation is infinity. Because the terms are not tending to zero, the sum of the series increases without bound as we add more terms. Therefore, the sum of this infinite series is not defined; however, some infinite series tend to a finite value. A major focus of Calculus is determining which infinite series tend to finite values. As such, we need to introduce new language.
The study of Calculus will present you with many concepts that seem counterintuitive. For example, it seems reasonable to state that an infinite series where the terms are tending to zero must become finite.
For example, consider the series\[1 + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \cdots\nonumber\]The terms are tending to zero as we keep writing out the series. In fact, by the time we get to the thousandth term, we are only adding \(\frac{1}{1000}\). Therefore, it seems logical to state that this summation would approach a finite value. After all, we are essentially adding zero as time goes on.
Unfortunately, this is not true!
In general, if the terms of a series tend to zero, we cannot guarantee that the sum of those terms will become finite. The series\[\sum_{n = 1}^{\infty} \dfrac{1}{n} \quad \text{and} \quad \sum_{n = 1}^{\infty} \dfrac{(-1)^n}{n}\nonumber\]have terms that tend to zero as \(n\) tends to get large; however, the first of these series is divergent, while the second is convergent (in fact, eerily, it converges to \(\ln(2)\)).
In general, finding the sum of a convergent (infinite) series is a task left for Calculus.
Infinite Geometric Series
Among infinite series, infinite geometric series claim a special place. This is because, when an infinite geometric series converges, we can calculate its sum—this is often not true for other types of convergent infinite series. The formula for the sum of an infinite geometric series is related to the formula for the sum of the first \(n\) terms of the geometric series:\[S_n = \dfrac{a_1(1 - r^n)}{1 - r}.\nonumber\]We will examine an infinite series with \(r = \frac{1}{2}\). What happens to \(r^n\) as \(n\) increases?\[\begin{array}{rcl} \left( \dfrac{1}{2} \right)^2 & = & \dfrac{1}{4} \\[6pt] \left( \dfrac{1}{2} \right)^3 & = & \dfrac{1}{8} \\[6pt] \left( \dfrac{1}{2} \right)^4 & = & \dfrac{1}{16} \\[6pt] \end{array}\nonumber\]The value of \(r^n\) decreases rapidly. What happens for greater values of \(n\)?\[\begin{array}{rcl} \left( \dfrac{1}{2} \right)^{10} & = & \dfrac{1}{1,024} \\[6pt] \left( \dfrac{1}{2} \right)^{20} & = & \dfrac{1}{1,048,576} \\[6pt] \left( \dfrac{1}{2} \right)^{30} & = & \dfrac{1}{1,073,741,824} \\[6pt] \end{array}\nonumber\]As \(n\) gets very large, \(r^n\) gets very small. We say that, as \(n\) increases without bound, \(r^n\) approaches 0. Borrowing notation from our exploration of rational functions, as \(n \to \infty\), \(r^n \to 0\). As \(r^n\) approaches 0, \(1- r^n\) approaches 1. When this happens, the numerator approaches \(a_1\). This gives us a formula for the sum of a convergent infinite geometric series.
Determine whether the sum of each infinite series is defined.
- \(\frac{3}{4} + \frac{1}{2} + \frac{1}{3} + \ldots\)
- \(\displaystyle \sum_{k=1}^{\infty} 27 \cdot \left( \frac{1}{3} \right)^k\)
- Solutions
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- The ratio of the second term to the first is the same as the ratio of the third term to the second. The series is geometric with a common ratio of \(\frac{2}{3}\). Since \(-1 < \frac{2}{3} < 1\), this series must converge.
- The given formula is exponential with a base of \(\frac{1}{3}\). Therefore, the series is geometric with a common ratio of \(\frac{1}{3}\), and so the infinite series is convergent.
Determine whether the sum of the infinite series is defined.
- \(24 + (-12) + 6 + (-3) + \ldots\)
- \(\frac{1}{3} + \frac{1}{2} + \frac{3}{4} + \frac{9}{8} + \ldots\)
- \(\displaystyle \sum_{k=1}^{\infty} 15 \cdot \left( -0.3 \right)^k\)
Find the sum, if it exists, for the following:
- \(10+9+8+7+ \ldots\)
- \(248.6+99.44+39.776+ \ldots\)
- \(\displaystyle \sum_{k=1}^{\infty} 4,374 \cdot \left( -\frac{1}{3} \right)^{k-1}\)
- \(\displaystyle \sum_{k=1}^{\infty} \frac{1}{9} \cdot \left( \frac{4}{3} \right)^k\)
- Solutions
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- Since there is not a constant ratio between consecutive terms of the series, the series is not geometric. As such, even if it did converge (which it doesn't), we don't have a theorem to help us find its value.
- There is a constant ratio; the series is geometric. \(a_1 = 248.6\) and \(r = \frac{99.44}{248.6} = 0.4\), which is between \(-1\) and \(1\). Therefore, the geometric series must converge. Substitute \(a_1 =248.6\) and \(r=0.4\) into the formula and simplify to find the sum:\[\begin{array}{rrcl} & S & = & \dfrac{a_1}{1 - r} \\[6pt] \implies & S & = & \dfrac{248.6}{1 - 0.4} = 414.\overline{3} \\[6pt] \end{array}\nonumber\]
- The formula is exponential, so the series is geometric with \(r = -\frac{1}{3}\). Find \(a_1\) by substituting \(k=1\) into the given explicit formula:\[a_1 =4,374 \cdot \left( -\dfrac{1}{3} \right)^{1-1} = 4,374\nonumber\]Substitute \(a_1 = 4,374\) and \(r = -\frac{1}{3}\) into the formula, and simplify to find the sum:\[\begin{array}{rrcl} & S & = & \dfrac{a_1}{1 - r} \\[6pt] \implies & S & = & \dfrac{4,374}{1 - \left( -\frac{1}{3} \right)} = 3,280.5 \\[6pt] \end{array}\nonumber\]
- The formula is exponential, so the series is geometric, but \(r \geq 1\). The series must diverge.
Find the sum, if it exists.
- \(2 + \frac{2}{3} + \frac{2}{9} + \ldots\)
- \(\displaystyle \sum_{k=1}^{\infty} 0.76 k + 1\)
- \(\displaystyle \sum_{k=1}^{\infty} \left( -\frac{3}{8} \right)^k\)
A common "classroom" use of convergent infinite geometric series is to find fractional equivalents of repeating decimal numbers.
Find an equivalent fraction for the repeating decimal \(0.\overline{3}\).
- Solution
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We notice the repeating decimal \(0.\overline{3} = 0.333 \ldots\), so we can rewrite the repeating decimal as a sum of terms.\[0.\overline{3} = 0.3 + 0.03 + 0.003 + \ldots\nonumber\]Looking for a pattern, we rewrite the sum, noticing that \(0.3 = 0.3 \cdot 1 = 0.3 \cdot (0.1)^0\), \(0.03 = 0.3 \cdot 0.1 = 0.3 \cdot (0.1)^1\), \(0.003 = 0.3 \cdot 0.01 = 0.3 \cdot (0.1)^2\), and so on.\[0.\overline{3} = 0.3 + 0.3(0.1)^1 + 0.3(0.1)^2 + \cdots\nonumber\]Notice the pattern? We multiply each consecutive term by a common ratio of \(0.1\) starting with the first term of 0.3. So, substituting into our formula for an infinite geometric sum, we have\[S_n = \dfrac{a_1}{1-r} = \dfrac{0.3}{1-0.1} = \dfrac{0.3}{0.9} = \dfrac{1}{3}.\nonumber\]
Solving Annuity Problems
At the beginning of the section, we looked at a problem in which a couple invested a set amount of money each month into a college fund for six years. An annuity is an investment in which the purchaser makes a sequence of periodic, equal payments. To find the amount of an annuity, we need to find the sum of all the payments and the interest earned. In the example, the couple invests $50 each month. This is the value of the initial deposit. The account paid 6% annual interest, compounded monthly. To find the interest rate per payment period, we need to divide the 6% annual percentage rate (APR) by 12. So the monthly interest rate is 0.5%. We can multiply the amount in the account each month by 100.5% to find the value of the account after interest has been added.
We can find the value of the annuity right after the last deposit by using a geometric series with \(a_1 =50\) and \(r=100.5\%=1.005\). After the first deposit, the value of the annuity will be $50. Let us see if we can determine the amount in the college fund and the interest earned.
We can find the value of the annuity after \(n\) deposits using the formula for the sum of the first \(n\) terms of a geometric series. In 6 years, there are 72 months, so \(n=72\). We can substitute \(a_1 =50\), \(r=1.005\), and \(n=72\) into the formula, and simplify to find the value of the annuity after 6 years.\[S_{72} = \dfrac{50(1- 1.005^{72})}{1-1.005} \approx 4,320.44.\nonumber\]After the last deposit, the couple will have a total of $4,320.44 in the account. Notice, the couple made 72 payments of $50 each for a total of \(72(50) = \$3,600\). This means that because of the annuity, the couple earned $720.44 interest in their college fund.
A deposit of $100 is placed into a college fund at the beginning of every month for 10 years. The fund earns 9% annual interest, compounded monthly, and paid at the end of the month. How much is in the account right after the last deposit?
- Solution
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The value of the initial deposit is $100, so \(a_1 =100\). A total of 120 monthly deposits are made in the 10 years, so \(n=120\). To find \(r\), divide the annual interest rate by 12 to find the monthly interest rate and add 1 to represent the new monthly deposit.\[r = 1 + \dfrac{0.09}{12} = 1.0075.\nonumber\]Substitute \(a_1 =100\), \(r=1.0075\), and \(n=120\) into the formula for the sum of the first \(n\) terms of a geometric series, and simplify to find the value of the annuity.\[S_{120} = \dfrac{100(1- 1.0075^{120} )}{1-1.0075} \approx 19,351.43\nonumber\]So the account has $19,351.43 after the last deposit is made.
At the beginning of each month, $200 is deposited into a retirement fund. The fund earns 6% annual interest, compounded monthly, and paid into the account at the end of the month. How much is in the account if deposits are made for 10 years?
This page titled 14.4: Series and Their Notations was last modified on Tue, 08 Jul 2025 17:47:52 GMT and is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Roy Simpson, Cosumnes River College (OpenStax) via source content that was edited to the style and standards of the LibreTexts platform.
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