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Study Guides > College Algebra: Co-requisite Course

Series and Their Notations*

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.

Arithmetic Series

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 monthly. The sum of the terms of a sequence is called a series. Consider, for example, the following series.

3+7+11+15+19+⋯3+7+11+15+19+\cdots

The nth n\text{th } partial sum of a series is the sum of a finite number of consecutive terms beginning with the first term. The notation

 Sn \text{ }{S}_{n}\text{ } represents the partial sum. S1=3S2=3+7=10S3=3+7+11=21S4=3+7+11+15=36\begin{array}{l}{S}_{1}=3\\ {S}_{2}=3+7=10\\ {S}_{3}=3+7+11=21\\ {S}_{4}=3+7+11+15=36\end{array}

Summation notation is used to represent series. Summation notation is often known as sigma notation because it uses the Greek capital letter sigma, σ\sigma, to represent the sum. Summation notation includes an explicit formula and specifies the first and last terms in the series. An explicit formula for each term of the series is given to the right of the sigma. A variable called the index of summation is written below the sigma. The index of summation is set equal to the lower limit of summation, which is the number used to generate the first term in the series. The number above the sigma, called the upper limit of summation, is the number used to generate the last term in a series. Explanation of summation notion as described in the text. If we interpret the given notation, we see that it asks us to find the sum of the terms in the series ak=2k{a}_{k}=2k for k=1k=1 through k=5k=5. We can begin by substituting the terms for kk and listing out the terms of this series.

a1=2(1)=2a2=2(2)=4a3=2(3)=6a4=2(4)=8a5=2(5)=10\begin{array}{l}\begin{array}{l}\\ {a}_{1}=2\left(1\right)=2\end{array}\hfill \\ {a}_{2}=2\left(2\right)=4\hfill \\ {a}_{3}=2\left(3\right)=6\hfill \\ {a}_{4}=2\left(4\right)=8\hfill \\ {a}_{5}=2\left(5\right)=10\hfill \end{array}

We can find the sum of the series by adding the terms:

∑k=152k=2+4+6+8+10=30\sum _{k=1}^{5}2k=2+4+6+8+10=30

A General Note: Summation Notation

The sum of the first nn terms of a series can be expressed in summation notation as follows:

∑k=1nak\sum _{k=1}^{n}{a}_{k}

This notation tells us to find the sum of ak{a}_{k} from

k=1k=1 to k=nk=n.

kk is called the index of summation, 1 is the lower limit of summation, and nn is the upper limit of summation.

Q & A

Does the lower limit of summation have to be 1?

No. 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.

How To: Given summation notation for a series, evaluate the value.

  1. Identify the lower limit of summation.
  2. Identify the upper limit of summation.
  3. Substitute each value of kk from the lower limit to the upper limit into the formula.
  4. Add to find the sum.

Example: Using Summation Notation

Evaluate ∑k=37k2\sum _{k=3}^{7}{k}^{2}.

Answer: 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 k2{k}^{2} from k=3k=3 to k=7k=7. We find the terms of the series by substituting k=3,4,5,6k=3\text{,}4\text{,}5\text{,}6, and 77 into the function k2{k}^{2}. We add the terms to find the sum.

∑k=37k2=32+42+52+62+72=9+16+25+36+49=135\begin{array}{ll}\sum _{k=3}^{7}{k}^{2}\hfill & ={3}^{2}+{4}^{2}+{5}^{2}+{6}^{2}+{7}^{2}\hfill \\ \hfill & =9+16+25+36+49\hfill \\ \hfill & =135\hfill \end{array}

Try It

Evaluate ∑k=25(3k−1)\sum _{k=2}^{5}\left(3k - 1\right).

Answer: 38

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, dd. The sum of the terms of an arithmetic sequence is called an arithmetic series. We can write the sum of the first nn terms of an arithmetic series as:

Sn=a1+(a1+d)+(a1+2d)+...+(an−d)+an{S}_{n}={a}_{1}+\left({a}_{1}+d\right)+\left({a}_{1}+2d\right)+...+\left({a}_{n}-d\right)+{a}_{n}.

We can also reverse the order of the terms and write the sum as

Sn=an+(an−d)+(an−2d)+...+(a1+d)+a1{S}_{n}={a}_{n}+\left({a}_{n}-d\right)+\left({a}_{n}-2d\right)+...+\left({a}_{1}+d\right)+{a}_{1}.

If we add these two expressions for the sum of the first nn terms of an arithmetic series, we can derive a formula for the sum of the first nn terms of any arithmetic series.

Sn=a1+(a1+d)+(a1+2d)+...+(an−d)+an+Sn=an+(an−d)+(an−2d)+...+(a1+d)+a12Sn=(a1+an)+(a1+an)+...+(a1+an)\frac{\begin{array}{l}{S}_{n}={a}_{1}+\left({a}_{1}+d\right)+\left({a}_{1}+2d\right)+...+\left({a}_{n}-d\right)+{a}_{n}\hfill \\ +{S}_{n}={a}_{n}+\left({a}_{n}-d\right)+\left({a}_{n}-2d\right)+...+\left({a}_{1}+d\right)+{a}_{1}\hfill \end{array}}{2{S}_{n}=\left({a}_{1}+{a}_{n}\right)+\left({a}_{1}+{a}_{n}\right)+...+\left({a}_{1}+{a}_{n}\right)}

Because there are nn terms in the series, we can simplify this sum to

2Sn=n(a1+an)2{S}_{n}=n\left({a}_{1}+{a}_{n}\right).

We divide by 2 to find the formula for the sum of the first nn terms of an arithmetic series.

Sn=n(a1+an)2{S}_{n}=\frac{n\left({a}_{1}+{a}_{n}\right)}{2}

This is generally referred to as the Partial Sum of the series.

A General Note: Formula for the Partial Sum of an Arithmetic Series

An arithmetic series is the sum of the terms of an arithmetic sequence. The formula for the partial sum of an arithmetic sequence is

Sn=n(a1+an)2{S}_{n}=\frac{n\left({a}_{1}+{a}_{n}\right)}{2}

How To: Given terms of an arithmetic series, find the partial sum

  1. Identify a1{a}_{1} and an{a}_{n}.
  2. Determine nn.
  3. Substitute values for a1, an{a}_{1}\text{, }{a}_{n}, and nn into the formula Sn=n(a1+an)2{S}_{n}=\frac{n\left({a}_{1}+{a}_{n}\right)}{2}.
  4. Simplify to find Sn{S}_{n}.

Example: Finding the partial sum of an Arithmetic Series

Find the partial sum of each arithmetic series.
  1. 5 + 8 + 11 + 14 + 17 + 20 + 23 + 26 + 29 + 32\text{5 + 8 + 11 + 14 + 17 + 20 + 23 + 26 + 29 + 32}
  2. 20 + 15 + 10 +…+ -50\text{20 + 15 + 10 +\ldots + -50}
  3. ∑k=1123k−8\sum _{k=1}^{12}3k - 8

Answer:

  1. We are given a1=5{a}_{1}=5 and an=32{a}_{n}=32.Count the number of terms in the sequence to find n=10n=10.Substitute values for a1,an,{a}_{1},{a}_{n}\text{\hspace{0.17em},} and nn into the formula and simplify. Sn=n(a1+an)2S10=10(5+32)2=185\begin{array}{l}\begin{array}{l}\hfill \\ {S}_{n}=\frac{n\left({a}_{1}+{a}_{n}\right)}{2}\hfill \end{array}\hfill \\ {S}_{10}=\frac{10\left(5+32\right)}{2}=185\hfill \end{array}
  2. We are given a1=20{a}_{1}=20 and an=−50{a}_{n}=-50.Use the formula for the general term of an arithmetic sequence to find nn. an=a1+(n−1)d−50=20+(n−1)(−5)−70=(n−1)(−5)14=n−115=n\begin{array}{l}{a}_{n}={a}_{1}+\left(n - 1\right)d\hfill \\ -50=20+\left(n - 1\right)\left(-5\right)\hfill \\ -70=\left(n - 1\right)\left(-5\right)\hfill \\ 14=n - 1\hfill \\ 15=n\hfill \end{array} Substitute values for a1,an,n{a}_{1},{a}_{n}\text{,}n into the formula and simplify.Sn=n(a1+an)2S15=15(20−50)2=−225\begin{array}{l}\begin{array}{l}\\ {S}_{n}=\frac{n\left({a}_{1}+{a}_{n}\right)}{2}\end{array}\hfill \\ {S}_{15}=\frac{15\left(20 - 50\right)}{2}=-225\hfill \end{array}
  3. To find a1{a}_{1}, substitute k=1k=1 into the given explicit formula. ak=3k−8 a1=3(1)−8=−5\begin{array}{l}{a}_{k}=3k - 8\hfill \\ \text{ }{a}_{1}=3\left(1\right)-8=-5\hfill \end{array} We are given that n=12n=12. To find a12{a}_{12}, substitute k=12k=12 into the given explicit formula.  ak=3k−8a12=3(12)−8=28\begin{array}{l}\text{ }{a}_{k}=3k - 8\hfill \\ {a}_{12}=3\left(12\right)-8=28\hfill \end{array} Substitute values for a1,an{a}_{1},{a}_{n}, and nn into the formula and simplify.  Sn=n(a1+an)2S12=12(−5+28)2=138\begin{array}{l}\text{ }{S}_{n}=\frac{n\left({a}_{1}+{a}_{n}\right)}{2}\hfill \\ {S}_{12}=\frac{12\left(-5+28\right)}{2}=138\hfill \end{array}

 

Try It

Use the formula to find the partial 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\text{1}\text{.4 + 1}\text{.6 + 1}\text{.8 + 2}\text{.0 + 2}\text{.2 + 2}\text{.4 + 2}\text{.6 + 2}\text{.8 + 3}\text{.0 + 3}\text{.2 + 3}\text{.4}

Answer: 26.4\text{26}\text{.4}

13 + 21 + 29 + …+ 69\text{13 + 21 + 29 + }\dots \text{+ 69}

Answer: 328\text{328}

∑k=1105−6k\sum _{k=1}^{10}5 - 6k

Answer: -280\text{-280}

Example: Solving Application Problems with Arithmetic Series

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?

Answer: This problem can be modeled by an arithmetic series with a1=12{a}_{1}=\frac{1}{2} and d=14d=\frac{1}{4}. We are looking for the total number of miles walked after 8 weeks, so we know that n=8n=8, and we are looking for S8{S}_{8}. To find a8{a}_{8}, we can use the explicit formula for an arithmetic sequence.

an=a1+d(n−1)a8=12+14(8−1)=94\begin{array}{l}\begin{array}{l}\\ {a}_{n}={a}_{1}+d\left(n - 1\right)\end{array}\hfill \\ {a}_{8}=\frac{1}{2}+\frac{1}{4}\left(8 - 1\right)=\frac{9}{4}\hfill \end{array}

We can now use the formula for arithmetic series.

Sn=n(a1+an)2 S8=8(12+94)2=11\begin{array}{l} {S}_{n}=\frac{n\left({a}_{1}+{a}_{n}\right)}{2}\hfill \\ \text{ }{S}_{8}=\frac{8\left(\frac{1}{2}+\frac{9}{4}\right)}{2}=11\hfill \end{array}

She will have walked a total of 11 miles.

Try It

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?

Answer: $2,025

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. Recall that a geometric sequence is a sequence in which the ratio of any two consecutive terms is the common ratio, rr. We can write the sum of the first nn terms of a geometric series as

Sn=a1+ra1+r2a1+...+rn−1a1{S}_{n}={a}_{1}+r{a}_{1}+{r}^{2}{a}_{1}+...+{r}^{n - 1}{a}_{1}.

Just as with arithmetic series, we can do some algebraic manipulation to derive a formula for the sum of the first nn terms of a geometric series. We will begin by multiplying both sides of the equation by rr.

rSn=ra1+r2a1+r3a1+...+rna1r{S}_{n}=r{a}_{1}+{r}^{2}{a}_{1}+{r}^{3}{a}_{1}+...+{r}^{n}{a}_{1}

Next, we subtract this equation from the original equation.

 Sn=a1+ra1+r2a1+...+rn−1a1−rSn=−(ra1+r2a1+r3a1+...+rna1)(1−r)Sn=a1−rna1\begin{array}{l}\\ \frac{\begin{array}{l}\text{ }{S}_{n}={a}_{1}+r{a}_{1}+{r}^{2}{a}_{1}+...+{r}^{n - 1}{a}_{1}\hfill \\ -r{S}_{n}=-\left(r{a}_{1}+{r}^{2}{a}_{1}+{r}^{3}{a}_{1}+...+{r}^{n}{a}_{1}\right)\hfill \end{array}}{\left(1-r\right){S}_{n}={a}_{1}-{r}^{n}{a}_{1}}\end{array}

Notice that when we subtract, all but the first term of the top equation and the last term of the bottom equation cancel out. To obtain a formula for Sn{S}_{n}, divide both sides by

(1−r)\left(1-r\right). Sn=a1(1−rn)1−r r≠1{S}_{n}=\frac{{a}_{1}\left(1-{r}^{n}\right)}{1-r}\text{ r}\ne \text{1}

A General Note: Formula for the Sum of the First n Terms of a Geometric Series

A geometric series is the sum of the terms in a geometric sequence. The formula for the sum of the first nn terms of a geometric sequence is represented as

Sn=a1(1−rn)1−r r≠1{S}_{n}=\frac{{a}_{1}\left(1-{r}^{n}\right)}{1-r}\text{ r}\ne \text{1}

How To: Given a geometric series, find the sum of the first n terms.

  1. Identify a1,r,andn{a}_{1},r,\text{and}n.
  2. Substitute values for a1,r{a}_{1},r, and nn into the formula Sn=a1(1−rn)1−r{S}_{n}=\frac{{a}_{1}\left(1-{r}^{n}\right)}{1-r}.
  3. Simplify to find Sn{S}_{n}.

Example: Finding the First n Terms of a Geometric Series

Use the formula to find the indicated partial sum of each geometric series.
  1. S11{S}_{11} for the series  8 + -4 + 2 + …\text{ 8 + -4 + 2 + }\dots
  2. ∑ k=163⋅2k\underset{6}{\overset{k=1}{{\sum }^{\text{ }}}}3\cdot {2}^{k}

Answer:

  1. a1=8{a}_{1}=8, and we are given that n=11n=11.We can find rr by dividing the second term of the series by the first. r=−48=−12r=\frac{-4}{8}=-\frac{1}{2} Substitute values for a1,r,andn{a}_{1}, r, \text{and} n into the formula and simplify. Sn=a1(1−rn)1−rS11=8(1−(−12)11)1−(−12)≈5.336\begin{array}{l}{S}_{n}=\frac{{a}_{1}\left(1-{r}^{n}\right)}{1-r}\hfill \\ {S}_{11}=\frac{8\left(1-{\left(-\frac{1}{2}\right)}^{11}\right)}{1-\left(-\frac{1}{2}\right)}\approx 5.336\hfill \end{array}
  2. Find a1{a}_{1} by substituting k=1k=1 into the given explicit formula. a1=3⋅21=6{a}_{1}=3\cdot {2}^{1}=6 We can see from the given explicit formula that r=2r=2. The upper limit of summation is 6, so n=6n=6.Substitute values for a1,r{a}_{1},r, and nn into the formula, and simplify. Sn=a1(1−rn)1−rS6=6(1−26)1−2=378\begin{array}{l}{S}_{n}=\frac{{a}_{1}\left(1-{r}^{n}\right)}{1-r}\hfill \\ {S}_{6}=\frac{6\left(1-{2}^{6}\right)}{1 - 2}=378\hfill \end{array}

Try It

Use the formula to find the indicated partial sum of each geometric series. S20[/latex]fortheseries[latex] 1,000 + 500 + 250 + …{S}_{20}[/latex] for the series [latex]\text{ 1,000 + 500 + 250 + }\dots

Answer: ≈2,000.00\approx 2,000.00

∑k=183k\sum _{k=1}^{8}{3}^{k}

Answer: 9,840

Example: Solving an Application Problem with a Geometric Series

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.

Answer: The problem can be represented by a geometric series with a1=26,750{a}_{1}=26,750; n=5n=5; and r=1.016r=1.016. Substitute values for a1{a}_{1}, rr, and nn into the formula and simplify to find the total amount earned at the end of 5 years.

Sn=a1(1−rn)1−rS5=26,750(1−1.0165)1−1.016≈138,099.03\begin{array}{l}{S}_{n}=\frac{{a}_{1}\left(1-{r}^{n}\right)}{1-r}\hfill \\ {S}_{5}=\frac{26\text{,}750\left(1-{1.016}^{5}\right)}{1 - 1.016}\approx 138\text{,}099.03\hfill \end{array}

He will have earned a total of $138,099.03 by the end of 5 years.

Try It

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?

Answer: $275,513.31

Annuities

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 interest (APR) rate 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 a1=50{a}_{1}=50 and r=100.5r=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 nn deposits using the formula for the sum of the first nn terms of a geometric series. In 6 years, there are 72 months, so n=72n=72. We can substitute a1=50,r=1.005,andn=72{a}_{1}=50, r=1.005, \text{and} n=72 into the formula, and simplify to find the value of the annuity after 6 years.

S72=50(1−1.00572)1−1.005≈4,320.44{S}_{72}=\frac{50\left(1-{1.005}^{72}\right)}{1 - 1.005}\approx 4\text{,}320.44

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\left(50\right) = $3,600. This means that because of the annuity, the couple earned $720.44 interest in their college fund.

How To: Given an initial deposit and an interest rate, find the value of an annuity.

  1. Determine a1{a}_{1}, the value of the initial deposit.
  2. Determine nn, the number of deposits.
  3. Determine rr.
    1. Divide the annual interest rate by the number of times per year that interest is compounded.
    2. Add 1 to this amount to find rr.
  4. Substitute values for a1,r,andn{a}_{1}\text{,}r,\text{and}n into the formula for the sum of the first nn terms of a geometric series, Sn=a1(1−rn)1−r{S}_{n}=\frac{{a}_{1}\left(1-{r}^{n}\right)}{1-r}.
  5. Simplify to find Sn{S}_{n}, the value of the annuity after nn deposits.

Example: Solving an Annuity Problem

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?

Answer: The value of the initial deposit is $100, so a1=100{a}_{1}=100. A total of 120 monthly deposits are made in the 10 years, so n=120n=120. To find rr, divide the annual interest rate by 12 to find the monthly interest rate and add 1 to represent the new monthly deposit.

r=1+0.0912=1.0075r=1+\frac{0.09}{12}=1.0075

Substitute a1=100,r=1.0075,andn=120{a}_{1}=100\text{,}r=1.0075\text{,}\text{and}n=120 into the formula for the sum of the first nn terms of a geometric series, and simplify to find the value of the annuity.

S120=100(1−1.0075120)1−1.0075≈19,351.43{S}_{120}=\frac{100\left(1-{1.0075}^{120}\right)}{1 - 1.0075}\approx 19\text{,}351.43

So the account has $19,351.43 after the last deposit is made.

Try It

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?

Answer: $92,408.18

 

Key Equations

sum of the first nn terms of an arithmetic series Sn=n(a1+an)2{S}_{n}=\frac{n\left({a}_{1}+{a}_{n}\right)}{2}
sum of the first nn terms of a geometric series Sn=a1(1−rn)1−r⋅r≠1{S}_{n}=\frac{{a}_{1}\left(1-{r}^{n}\right)}{1-r}\cdot r\ne 1
sum of an infinite geometric series with −1<r< 1-1<r<\text{ }1 Sn=a11−r⋅r≠1{S}_{n}=\frac{{a}_{1}}{1-r}\cdot r\ne 1

Key Concepts

  • The sum of the terms in a sequence is called a series.
  • A common notation for series is called summation notation, which uses the Greek letter sigma to represent the sum.
  • The sum of the terms in an arithmetic sequence is called an arithmetic series.
  • The sum of the first nn terms of an arithmetic series can be found using a formula.
  • The sum of the terms in a geometric sequence is called a geometric series.
  • The sum of the first nn terms of a geometric series can be found using a formula.
  • The sum of an infinite series exists if the series is geometric with −1<r<1-1<r<1.
  • If the sum of an infinite series exists, it can be found using a formula.
  • An annuity is an account into which the investor makes a series of regularly scheduled payments. The value of an annuity can be found using geometric series.

Glossary

annuity an investment in which the purchaser makes a sequence of periodic, equal payments arithmetic series the sum of the terms in an arithmetic sequence diverge a series is said to diverge if the sum is not a real number geometric series the sum of the terms in a geometric sequence index of summation in summation notation, the variable used in the explicit formula for the terms of a series and written below the sigma with the lower limit of summation infinite series the sum of the terms in an infinite sequence lower limit of summation the number used in the explicit formula to find the first term in a series nth partial sum the sum of the first nn terms of a sequence series the sum of the terms in a sequence summation notation a notation for series using the Greek letter sigma; it includes an explicit formula and specifies the first and last terms in the series upper limit of summation the number used in the explicit formula to find the last term in a series

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