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Series and Their Notations

By the end of this section, you will be able to:

  • Use summation notation
  • Use the formula for the sum of the first nn terms of an arithmetic series
  • Use the formula for the sum of the first nn terms of a geometric series
  • Use the formula for the sum of an infinite geometric series
  • Solve annuity problems

A parent decides to start a college fund for their daughter. They plan to invest $50 in the fund each month. The fund pays 6%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 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+\ldots

The nnth partial sum of a series is the sum of a finite number of consecutive terms beginning with the first term. The notation SnS_n represents the partial sum.

S1=3S2=3+7=10S3=3+7+11=21S4=3+7+11+15=36 \begin{array}{lrcl} & S_1 &=& 3 \\[4pt] & S_2 &=& 3+7=10 \\[4pt] & S_3 &=& 3+7+11=21 \\[4pt] & 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.

If we interpret the given notation, we see that it asks us to find the sum of the terms in the series ak=2ka_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}{lrcl} & a_1 &=& 2(1)=2 \\[4pt] & a_2 &=& 2(2)=4 \\[4pt] & a_3 &=& 2(3)=6 \\[4pt] & a_4 &=& 2(4)=8 \\[4pt] & a_5 &=& 2(5)=10 \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

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 aka_k from k=1k=1 to k=nk=n. kk is called the index of summation, 11 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 11?

No. The lower limit of summation can be any number, but 11 is frequently used. We will look at examples with lower limits of summation other than 11.

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. Evaluate k=37k2\sum_{k=3}^{7}k^2.

Solution. According to the notation, the lower limit of summation is 33 and the upper limit is 77. So we need to find the sum of k2k^2 from k=3k=3 to k=7k=7. We find the terms of the series by substituting k=3,4,5,6,k=3,4,5,6, and 77 into the function k2k^2. We add the terms to find the sum.

k=37k2=32+42+52+62+72=9+16+25+36+49=135 \begin{array}{lrcl} & \sum_{k=3}^{7}k^2 &=& 3^2+4^2+5^2+6^2+7^2 \\[4pt] & &=& 9+16+25+36+49 \\[4pt] & &=& 135 \end{array}

Evaluatek=25(3k1)\sum_{k=2}^{5}(3k-1).

Using the Formula for 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)++(and)+anS_n=a_1+(a_1+d)+(a_1+2d)+\ldots+(a_n-d)+a_n

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

Sn=an+(and)+(an2d)++(a1+d)+a1S_n=a_n+(a_n-d)+(a_n-2d)+\ldots+(a_1+d)+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.

$$ \begin{array}{lrcl} & S_n &=& a_1+(a_1+d)+(a_1+2d)+\ldots+(a_n-d)+a_n \[4pt]

  • & S_n &=& a_n+(a_n-d)+(a_n-2d)+\ldots+(a_1+d)+a_1 \[4pt] \hline & 2S_n &=& (a_1+a_n)+(a_1+a_n)+\ldots+(a_1+a_n) \end{array} $$

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

2Sn=n(a1+an)2S_n=n(a_1+a_n)

We divide by 22 to find the formula for the sum of the first nn terms of an arithmetic series.

Sn=n(a1+an)2S_n=\tfrac{n(a_1+a_n)}{2}

Formula for the Sum of the First nn Terms of an Arithmetic Series. An arithmetic series is the sum of the terms of an arithmetic sequence. The formula for the sum of the first nn terms of an arithmetic sequence is

Sn=n(a1+an)2S_n=\tfrac{n(a_1+a_n)}{2}

How To: given terms of an arithmetic series, find the sum of the first nn terms.

  1. Identify a1a_1 and ana_n.
  2. Determine nn.
  3. Substitute values for a1a_1, ana_n, and nn into the formula Sn=n(a1+an)2S_n=\tfrac{n(a_1+a_n)}{2}.
  4. Simplify to find SnS_n.

Example. Find the sum of each arithmetic series.

a. 5+8+11+14+17+20+23+26+29+325+8+11+14+17+20+23+26+29+32 b. 20+15+10++(50)20+15+10+\ldots+(-50) c. k=112(3k8)\sum_{k=1}^{12}(3k-8)

Solution.

a. We are given a1=5a_1=5 and an=32a_n=32. Count the number of terms in the sequence to find n=10n=10. Substitute values for a1a_1, ana_n, and nn into the formula and simplify.

Sn=n(a1+an)2S10=10(5+32)2=185 \begin{array}{lrcl} & S_n &=& \tfrac{n(a_1+a_n)}{2} \\[4pt] & S_{10} &=& \tfrac{10(5+32)}{2}=185 \end{array}

b. We are given a1=20a_1=20 and an=50a_n=-50. Use the formula for the general term of an arithmetic sequence to find nn.

an=a1+(n1)d50=20+(n1)(5)70=(n1)(5)14=n115=n \begin{array}{lrcl} & a_n &=& a_1+(n-1)d \\[4pt] & -50 &=& 20+(n-1)(-5) \\[4pt] & -70 &=& (n-1)(-5) \\[4pt] & 14 &=& n-1 \\[4pt] & 15 &=& n \end{array}

Substitute values for a1a_1, ana_n, nn into the formula and simplify.

Sn=n(a1+an)2S15=15(2050)2=225 \begin{array}{lrcl} & S_n &=& \tfrac{n(a_1+a_n)}{2} \\[4pt] & S_{15} &=& \tfrac{15(20-50)}{2}=-225 \end{array}

c. To find a1a_1, substitute k=1k=1 into the given explicit formula.

ak=3k8a1=3(1)8=5 \begin{array}{lrcl} & a_k &=& 3k-8 \\[4pt] & a_1 &=& 3(1)-8=-5 \end{array}

We are given that n=12n=12. To find a12a_{12}, substitute k=12k=12 into the given explicit formula.

ak=3k8a12=3(12)8=28 \begin{array}{lrcl} & a_k &=& 3k-8 \\[4pt] & a_{12} &=& 3(12)-8=28 \end{array}

Substitute values for a1a_1, ana_n, and nn into the formula and simplify.

Sn=n(a1+an)2S12=12(5+28)2=138 \begin{array}{lrcl} & S_n &=& \tfrac{n(a_1+a_n)}{2} \\[4pt] & S_{12} &=& \tfrac{12(-5+28)}{2}=138 \end{array}

Find the sum1.4+1.6+1.8+2.0+2.2+2.4+2.6+2.8+3.0+3.2+3.41.4+1.6+1.8+2.0+2.2+2.4+2.6+2.8+3.0+3.2+3.4.

Find the sum13+21+29++6913+21+29+\ldots+69.

Evaluatek=110(56k)\sum_{k=1}^{10}(5-6k).

Example. 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 88 weeks, what will be the total number of miles she has walked?

Solution. This problem can be modeled by an arithmetic series with a1=12a_1=\tfrac12 and d=14d=\tfrac14. 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 S8S_8. To find a8a_8, we can use the explicit formula for an arithmetic sequence.

an=a1+d(n1)a8=12+14(81)=94 \begin{array}{lrcl} & a_n &=& a_1+d(n-1) \\[4pt] & a_8 &=& \tfrac12+\tfrac14(8-1)=\tfrac94 \end{array}

We can now use the formula for arithmetic series.

Sn=n(a1+an)2S8=8(12+94)2=11 \begin{array}{lrcl} & S_n &=& \tfrac{n(a_1+a_n)}{2} \\[4pt] & S_8 &=& \tfrac{8\left(\tfrac12+\tfrac94\right)}{2}=11 \end{array}

She will have walked a total of 1111 miles.

A man earns $100 in the first week of June. Each week, he earns $12.50 more than the previous week. After1212weeks, how much has he earned in total? Enter the number only, without a dollar sign.

Using the Formula for 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++rn1a1S_n=a_1+ra_1+r^2a_1+\ldots+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++rna1rS_n=ra_1+r^2a_1+r^3a_1+\ldots+r^na_1

Next, we subtract this equation from the original equation.

$$ \begin{array}{lrcl} & S_n &=& a_1+ra_1+r^2a_1+\ldots+r^{n-1}a_1 \[4pt]

  • & rS_n &=& ra_1+r^2a_1+r^3a_1+\ldots+r^na_1 \[4pt] \hline & (1-r)S_n &=& a_1-r^na_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 SnS_n, divide both sides by (1r)(1-r).

Sn=a1(1rn)1r,r1S_n=\tfrac{a_1(1-r^n)}{1-r},\quad r\ne1

Formula for the Sum of the First nn 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(1rn)1r,r1S_n=\tfrac{a_1(1-r^n)}{1-r},\quad r\ne1

How To: given a geometric series, find the sum of the first nn terms.

  1. Identify a1a_1, rr, and nn.
  2. Substitute values for a1a_1, rr, and nn into the formula Sn=a1(1rn)1rS_n=\tfrac{a_1(1-r^n)}{1-r}.
  3. Simplify to find SnS_n.

Example. Use the formula to find the indicated partial sum of each geometric series.

a. S11S_{11} for the series 8+(4)+2+8+(-4)+2+\ldots b. k=1632k\sum_{k=1}^{6}3\cdot2^k

Solution.

a. a1=8a_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=\tfrac{-4}{8}=-\tfrac12

Substitute values for a1a_1, rr, and nn into the formula and simplify.

Sn=a1(1rn)1rS11=8(1(12)11)1(12)5.336 \begin{array}{lrcl} & S_n &=& \tfrac{a_1(1-r^n)}{1-r} \\[4pt] & S_{11} &=& \tfrac{8\left(1-\left(-\tfrac12\right)^{11}\right)}{1-\left(-\tfrac12\right)}\approx5.336 \end{array}

b. Find a1a_1 by substituting k=1k=1 into the given explicit formula.

a1=321=6a_1=3\cdot2^1=6

We can see from the given explicit formula that r=2r=2. The upper limit of summation is 66, so n=6n=6. Substitute values for a1a_1, rr, and nn into the formula, and simplify.

Sn=a1(1rn)1rS6=6(126)12=378 \begin{array}{lrcl} & S_n &=& \tfrac{a_1(1-r^n)}{1-r} \\[4pt] & S_6 &=& \tfrac{6(1-2^6)}{1-2}=378 \end{array}

FindS20S_{20}for the series1000+500+250+1000+500+250+\ldots, rounded to the nearest cent. Enter the number only, without a dollar sign.

Evaluatek=183k\sum_{k=1}^{8}3^k.

Example. At a new job, an employee’s starting salary is $26{,}750. He receives a 1.6%1.6\% annual raise. Find his total earnings at the end of 55 years.

Solution. The problem can be represented by a geometric series with a1=26,750a_1=26{,}750; n=5n=5; and r=1.016r=1.016. Substitute values for a1a_1, rr, and nn into the formula and simplify to find the total amount earned at the end of 5 years.

Sn=a1(1rn)1rS5=26,750(11.0165)11.016138,099.03 \begin{array}{lrcl} & S_n &=& \tfrac{a_1(1-r^n)}{1-r} \\[4pt] & S_5 &=& \tfrac{26{,}750(1-1.016^5)}{1-1.016}\approx138{,}099.03 \end{array}

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 a2%2\%annual raise. How much will she have earned by the end of88years, rounded to the nearest cent? Enter the number only, without a dollar sign.

Using the Formula for the Sum of an Infinite Geometric Series

Thus far, we have looked only at 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 nn terms. An infinite series is the sum of the terms of an infinite sequence. An example of an infinite series is 2+4+6+8+2+4+6+8+\ldots

This series can also be written in summation notation as k=12k\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. When the sum is not a real number, we say the series diverges.

Determining Whether the Sum of an Infinite Geometric Series is Defined

If the terms of an infinite geometric sequence approach 00, the sum of an infinite geometric series can be defined. The terms in this series approach 00:

1+0.2+0.04+0.008+0.0016+1+0.2+0.04+0.008+0.0016+\ldots

The common ratio r=0.2r=0.2. As nn gets very large, the values of rnr^n get very small and approach 00. Each successive term affects the sum less than the preceding term. As each succeeding term gets closer to 00, the sum of the terms approaches a finite value. The terms of any infinite geometric series with 1<r<1-1<r<1 approach 00; the sum of a geometric series is defined when 1<r<1-1<r<1.

Determining Whether the Sum of an Infinite Geometric Series is Defined. The sum of an infinite series is defined if the series is geometric and 1<r<1-1<r<1.

How To: given the first several terms of an infinite series, determine if the sum of the series exists.

  1. Find the ratio of the second term to the first term.
  2. Find the ratio of the third term to the second term.
  3. Continue this process to ensure the ratio of a term to the preceding term is constant throughout. If so, the series is geometric.
  4. If a common ratio, rr, was found in step 3, check to see if 1<r<1-1<r<1. If so, the sum is defined. If not, the sum is not defined.

Example. Determine whether the sum of each infinite series is defined.

a. 12+8+4+12+8+4+\ldots b. 34+12+13+\tfrac34+\tfrac12+\tfrac13+\ldots c. k=127(13)k\sum_{k=1}^{\infty}27\cdot\left(\tfrac13\right)^k d. k=15k\sum_{k=1}^{\infty}5k

Solution.

a. The ratio of the second term to the first is 23\tfrac23, which is not the same as the ratio of the third term to the second, 12\tfrac12. The series is not geometric.

b. 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 23\tfrac23. The sum of the infinite series is defined.

c. The given formula is exponential with a base of 13\tfrac13; the series is geometric with a common ratio of 13\tfrac13. The sum of the infinite series is defined.

d. The given formula is not exponential; the series is not geometric because the terms are increasing, and so cannot yield a finite sum.

Determine whether the sum of the infinite series13+12+34+98+\tfrac13+\tfrac12+\tfrac34+\tfrac98+\ldotsis defined.

Determine whether the sum of the infinite series24+(12)+6+(3)+24+(-12)+6+(-3)+\ldotsis defined.

Determine whether the sum of the infinite seriesk=115(0.3)k\sum_{k=1}^{\infty}15\cdot(-0.3)^kis defined.

Finding Sums of Infinite Series

When the sum of an infinite geometric series exists, we can calculate the sum. The formula for the sum of an infinite series is related to the formula for the sum of the first nn terms of a geometric series.

Sn=a1(1rn)1rS_n=\tfrac{a_1(1-r^n)}{1-r}

We will examine an infinite series with r=12r=\tfrac12. What happens to rnr^n as nn increases?

(12)2=14(12)3=18(12)4=116 \begin{array}{lrcl} & \left(\tfrac12\right)^2 &=& \tfrac14 \\[4pt] & \left(\tfrac12\right)^3 &=& \tfrac18 \\[4pt] & \left(\tfrac12\right)^4 &=& \tfrac{1}{16} \end{array}

The value of rnr^n decreases rapidly. What happens for greater values of nn?

(12)10=11,024(12)20=11,048,576(12)30=11,073,741,824 \begin{array}{lrcl} & \left(\tfrac12\right)^{10} &=& \tfrac{1}{1{,}024} \\[4pt] & \left(\tfrac12\right)^{20} &=& \tfrac{1}{1{,}048{,}576} \\[4pt] & \left(\tfrac12\right)^{30} &=& \tfrac{1}{1{,}073{,}741{,}824} \end{array}

As nn gets very large, rnr^n gets very small. We say that, as nn increases without bound, rnr^n approaches 00. As rnr^n approaches 00, 1rn1-r^n approaches 11. When this happens, the numerator approaches a1a_1. This gives us a formula for the sum of an infinite geometric series.

Formula for the Sum of an Infinite Geometric Series. The formula for the sum of an infinite geometric series with 1<r<1-1<r<1 is

S=a11rS=\tfrac{a_1}{1-r}

How To: given an infinite geometric series, find its sum.

  1. Identify a1a_1 and rr.
  2. Confirm that 1<r<1-1<r<1.
  3. Substitute values for a1a_1 and rr into the formula, S=a11rS=\tfrac{a_1}{1-r}.
  4. Simplify to find SS.

Example. Find the sum, if it exists, for the following:

a. 10+9+8+7+10+9+8+7+\ldots b. 248.6+99.44+39.776+248.6+99.44+39.776+\ldots c. k=14,374(13)k1\sum_{k=1}^{\infty}4{,}374\cdot\left(-\tfrac13\right)^{k-1} d. k=119(43)k\sum_{k=1}^{\infty}\tfrac19\cdot\left(\tfrac43\right)^k

Solution.

a. There is not a constant ratio; the series is not geometric.

b. There is a constant ratio; the series is geometric. a1=248.6a_1=248.6 and r=99.44248.6=0.4r=\tfrac{99.44}{248.6}=0.4, so the sum exists. Substitute a1=248.6a_1=248.6 and r=0.4r=0.4 into the formula and simplify to find the sum:

S=a11rS=248.610.4=414.3 \begin{array}{lrcl} & S &=& \tfrac{a_1}{1-r} \\[4pt] & S &=& \tfrac{248.6}{1-0.4}=414.\overline3 \end{array}

c. The formula is exponential, so the series is geometric with r=13r=-\tfrac13. Find a1a_1 by substituting k=1k=1 into the given explicit formula:

a1=4,374(13)11=4,374a_1=4{,}374\cdot\left(-\tfrac13\right)^{1-1}=4{,}374

Substitute a1=4,374a_1=4{,}374 and r=13r=-\tfrac13 into the formula, and simplify to find the sum:

S=a11rS=4,3741(13)=3,280.5 \begin{array}{lrcl} & S &=& \tfrac{a_1}{1-r} \\[4pt] & S &=& \tfrac{4{,}374}{1-\left(-\tfrac13\right)}=3{,}280.5 \end{array}

d. The formula is exponential, so the series is geometric, but r>1r>1. The sum does not exist.

Example. Find an equivalent fraction for the repeating decimal 0.30.\overline3.

Solution. We notice the repeating decimal 0.3=0.3330.\overline3=0.333\ldots so we can rewrite the repeating decimal as a sum of terms.

0.3=0.3+0.03+0.003+0.\overline3=0.3+0.03+0.003+\ldots

Looking for a pattern, we rewrite the sum, noticing that we see the first term multiplied by 0.10.1 in the second term, and the second term multiplied by 0.10.1 in the third term. Notice the pattern; we multiply each consecutive term by a common ratio of 0.10.1 starting with the first term of 0.30.3. So, substituting into our formula for an infinite geometric sum, we have

Sn=a11r=0.310.1=0.30.9=13S_n=\tfrac{a_1}{1-r}=\tfrac{0.3}{1-0.1}=\tfrac{0.3}{0.9}=\tfrac13

Find the sum, if it exists:2+23+29+2+\tfrac23+\tfrac29+\ldots

Determine whether the sum of the infinite seriesk=1(0.76k+1)\sum_{k=1}^{\infty}(0.76k+1)is defined.

Find the sum, if it exists, as a fraction:k=1(38)k\sum_{k=1}^{\infty}\left(-\tfrac38\right)^k.

Solving Annuity Problems

At the beginning of the section, we looked at a problem in which a parent 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 parent invests $50 each month. This is the value of the initial deposit. The account paid 6%6\% annual interest, compounded monthly. To find the interest rate per payment period, we need to divide the 6%6\% annual percentage interest (APR) rate by 1212. So the monthly interest rate is 0.5%0.5\%. We can multiply the amount in the account each month by 100.5%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=50a_1=50 and r=100.5%=1.005r=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 7272 months, so n=72n=72. We can substitute a1=50a_1=50, r=1.005r=1.005, and n=72n=72 into the formula, and simplify to find the value of the annuity after 6 years.

S72=50(11.00572)11.0054,320.44S_{72}=\tfrac{50(1-1.005^{72})}{1-1.005}\approx4{,}320.44

After the last deposit, the parent will have a total of $4{,}320.44 in the account. Notice, the parent made 7272 payments of $50 each, for a total of 72(50)=3,60072(50)=3{,}600 dollars, or $3{,}600. This means that because of the annuity, the parent 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 a1a_1, the value of the initial deposit.
  2. Determine nn, the number of deposits.
  3. Determine rr. Divide the annual interest rate by the number of times per year that interest is compounded. Add 11 to this amount to find rr.
  4. Substitute values for a1a_1, rr, and nn into the formula for the sum of the first nn terms of a geometric series, Sn=a1(1rn)1rS_n=\tfrac{a_1(1-r^n)}{1-r}.
  5. Simplify to find SnS_n, the value of the annuity after nn deposits.

Example. A deposit of $100 is placed into a college fund at the beginning of every month for 1010 years. The fund earns 9%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. The value of the initial deposit is $100, so a1=100a_1=100. A total of 120120 monthly deposits are made in the 10 years, so n=120n=120. To find rr, divide the annual interest rate by 1212 to find the monthly interest rate and add 11 to represent the new monthly deposit.

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

Substitute a1=100a_1=100, r=1.0075r=1.0075, and n=120n=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(11.0075120)11.007519,351.43S_{120}=\tfrac{100(1-1.0075^{120})}{1-1.0075}\approx19{,}351.43

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 earns6%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 for1010years, rounded to the nearest cent? Enter the number only, without a dollar sign.

Media. Access these online resources for additional instruction and practice with series.

Key equations

sum of the first nn terms of an arithmetic seriesSn=n(a1+an)2S_n=\tfrac{n(a_1+a_n)}{2}
sum of the first nn terms of a geometric seriesSn=a1(1rn)1r, r1S_n=\tfrac{a_1(1-r^n)}{1-r},\ r\ne1
sum of an infinite geometric series with 1<r<1-1<r<1Sn=a11r, r1S_n=\tfrac{a_1}{1-r},\ r\ne1

Key concepts

  • The sum of the terms in a sequence is called a series.
  • A common notation for series is summation notation, which uses the Greek letter sigma, Σ\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.

Practice

Use summation notation

Express the following description of a sum using summation notation: the sum that results from adding the number44five times.

Express the arithmetic sum10+18+26++16210+18+26+\ldots+162using summation notation.

Use the formula for the sum of the first nn terms of an arithmetic series

Use the formula for the sum of the firstnnterms of an arithmetic series to find the sum3.2+3.4+3.6++5.63.2+3.4+3.6+\ldots+5.6.

A boulder rolled down a mountain, traveling66feet in the first second. Each successive second, its distance increased by88feet. How far did the boulder travel after1010seconds?

How many terms must be added before the arithmetic series1357-1-3-5-7-\ldotshas a sum less than75-75?

Use the formula for the sum of the first nn terms of a geometric series

Use the formula for the sum of the firstnnterms of a geometric series to find the sum9+3+1+13+199+3+1+\tfrac13+\tfrac19, as a fraction.

Use the formula for the sum of the firstnnterms of a geometric series to findS7S_7for the series0.42+1050+0.4-2+10-50+\ldots

Use the formula for the sum of an infinite geometric series

Find the sum, if it exists, of the infinite geometric series2+1.6+1.28+1.024+2+1.6+1.28+1.024+\ldots

Find the sum, if it exists, of the infinite geometric series114116164-1-\tfrac14-\tfrac{1}{16}-\tfrac{1}{64}-\ldots, as a fraction.

A pendulum travels a distance of33feet on its first swing. On each successive swing, it travels34\tfrac34the distance of the previous swing. What is the total distance traveled by the pendulum when it stops swinging?

Solve annuity problems

Find the value of an annuity with a monthly deposit of $150,2424total deposits, and an interest rate of3%3\%compounded monthly, rounded to the nearest cent. Enter the number only, without a dollar sign.

Karl has two years to save $10{,}000 to buy a used car when he graduates. To the nearest dollar, what would his monthly deposits need to be if he invests in an account offering a4.2%4.2\%annual interest rate that compounds monthly? Enter the number only, without a dollar sign.


This section is adapted from Precalculus 2e, Section 11.4: Series and Their Notations by Jay Abramson and OpenStax, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: omitted a coreq-skills block the pinned CNXML prepends before the section proper (its own two “Objective N” reviews of summation notation and arithmetic-series sums, keyed to Intermediate Algebra 12.1.5 and 12.2.3, each with its own worked example and “Practice Makes Perfect” exercise set, including one exercise whose printed sequence “7, 10, 13, 13, 19,…” repeats a term and cannot be arithmetic as printed) — confirmed against the rendered PDF, page 1124 (true PDF index 1134), where “11.4 Series and Their Notations” and its Learning Objectives begin at the top of the page with no corequisite-skills material before them; the same prepended-block pattern is already logged in this book’s errata for earlier chapters, and this section joins that list without needing its own entry, since the flawed exercise lives entirely inside the omitted block and was never transcribed onto this page. Omitted three purely-typographic illustrations: two of summation-notation vocabulary (labelled call-outs pointing arrows from “upper limit of summation,” “lower limit of summation,” “index of summation,” and “explicit formula for kkth term” to the parts of k=152k\sum_{k=1}^{5}2k) and one that brackets the first two terms of the repeating-decimal expansion 0.3=0.3+(0.1)(0.3)+0.\overline{3}=0.3+(0.1)(0.3)+\ldots as “First Term” and “Second Term”, since the surrounding prose, transcribed in full, already states each label in words. Kept the “Media” callout’s introductory sentence but omitted its three external resource links, matching house precedent elsewhere in this book. Every retained Try It is a real fillin or multiplechoice component; the four “determine whether the sum of the infinite series is defined” Try Its are posed as multiple choice among the same three categorical outcomes (not geometric; geometric and defined; geometric but not defined) used in the worked example directly above them, since a categorical judgment is not a fill-in answer. Every “evaluate the sum”/“find the sum” fill-in that prints a fully expanded or summation-notation span declares decimal or fraction (with “as a fraction” stated in the question) so the engine’s own evaluation of the printed span cannot pass as a retype; every “write in summation notation” fill-in declares summation, with a hint to parenthesize a sum-shaped summand where the source’s own key needs it (10+18+26+…+162). Annuity answers are decimal, keyed to the source’s own rounding (nearest cent or nearest dollar as each problem states), with “enter the number only, without a dollar sign” in the question since a \$ in a response is invalid to the grading engine; every annuity value was independently computed at full precision and against the source’s own stated rounding, and the two rounds agreed in every case used here. The Key Equations table’s third row is reproduced as the pinned CNXML prints it — Sn=a11r, r1S_n=\tfrac{a_1}{1-r},\ r\ne1 under the heading “sum of an infinite geometric series with 1<r<1-1<r<1” — even though the module’s own body text states the formula as S=a11rS=\tfrac{a_1}{1-r} (no subscript, since the sum is not a partial sum) with the 1<r<1-1<r<1 condition, so the row carries the finite-series boilerplate by copy; the printed book omits this section’s Key Equations entirely, so the CNXML row is the only source spelling and it is kept as printed and recorded as a source note rather than silently repaired. Twelve selected end-of-section exercises were adapted into interactive Practice components, one or more groups per objective, every one independently re-derived (including by running the arithmetic in Node) rather than read off the source key; three of them (the boulder, pendulum, and Karl’s-car-fund exercises) sit on PDF page 1147, one page past this run’s assigned 1134–1146 range, flagged for the parent’s awareness since they are still this section’s own end-of-section exercises (module m49447, “Extensions” and “Real-World Applications” subsections) rather than the next section’s.