Sequences and Series

Series

Definition and Notation

A series is the sum of terms of a sequence. A sum of terms can be written using summation notation.

There are times when it makes sense to add the numbers in a sequence. The sum of terms of a sequence is called a series. A finite series is a sum of the terms in a finite sequence.

A sum can be written as an addition expression:
a1+a2+a3+a4+a5+a6+a7+a8+a9a_1+ a_2+ a_3+ a_4+ a_5+ a_6+ a_7+ a_8+ a_9
It can also be written using summation notation. Summation notation is an abbreviated way of writing a sum that uses the Greek letter sigma. This notation is read as "the sum of aa sub nn as nn goes from 1 to 9”:
n=19an\sum_{n=1}^{9}a_n
When possible, the explicit rule for the sequence is often used in place of ana_n.
Step-By-Step Example
Writing a Series in Summation Notation
Write the series in summation notation:
4+8+12+16+204+8+12+16+20
Step 1
Write the explicit rule for the sequence.
4+8+12+16+204+8+12+16+20
Identify the common difference. Subtract each term from the next term.
84=4128=41612=42016=4\begin{gathered}8-4=4\\12-8=4\\16-12=4\\20-16=4\end{gathered}
Each term is 4 more than the previous term, so the common difference dd is 4.
Step 2
Write the explicit rule.
an=a1+d(n1)a_n=a_{1}+d(n-1)
Substitute a1=4a_1=4 and d=4d=4.
an=4+4(n1)an=4n\begin{aligned}a_n&=4+4(n-1)\\a_{n}&=4n\end{aligned}
Solution
Write the notation for the first five terms. The variable nn describes the term number, which goes from 1 to 5.
n=154n\sum_{n=1}^{5}4n
This is read as "the sum of 4n4n as nn goes from 1 to 5."

Finite Arithmetic Series

A formula can be used to find the sum of a finite arithmetic series.

An arithmetic series is the sum of terms in an arithmetic sequence. The sum can be found by adding the terms in a sequence, but this may become tedious for a large number of terms.

The formula for the sum of the first nn terms in a finite arithmetic series a1+a2++ana_1+a_2+…+a_n is:
k=1nak=n2(a1+an)\sum_{k=1}^{n}a_k= \frac{n}{2} \left(a_1+a_n \right)
The variable kk is called an index, or counter, for the position of each term that is added.
Step-By-Step Example
Determining the Sum of an Arithmetic Series
Identify the sum of the first 12 terms of the series:
28+25+22+19+28+ 25+ 22+19+…
Step 1
Write an explicit rule for the sequence.
an=a1+d(n1)a_n=a_1+d(n-1)
The common difference is –3, and the first term is 28.
an=283(n1)=313n\begin{aligned}a_{n}&=28-3(n-1)\\&=31-3n\end{aligned}
Step 2
Use the explicit rule to find the 12th term.
an=313na12=313(12)a12=3136a12=5\begin{aligned}a_{n}&=31-3n\\a_{12}&=31-3(12)\\a_{12}&=31-36\\a_{12}&=-5\end{aligned}
Step 3

Substitute into the formula for a finite arithmetic series and simplify.

Substitute n=12n=12, ak=313na_k=31-3n, a1=28a_1=28, and an=5a_n=-5.
k=1nak=n2(a1+an)k=112(313n)=122(28+(5))k=112(313n)=122(23)k=112(313n)=2762k=112(313n)=138\begin{aligned}\sum_{k=1}^{n}a_k&= \frac{n}{2} \left(a_1+a_n \right)\\ \sum_{k=1}^{12}(31-3n)&= \frac{12}{2} (28+(-5))\\ \sum_{k=1}^{12}(31-3n)&= \frac{12}{2} (23)\\ \sum_{k=1}^{12}(31-3n)&= \frac{276}{2}\\ \sum_{k=1}^{12}(31-3n)&=138\end{aligned}
Solution
The sum of the first 12 terms is 138.
Step-By-Step Example
Applying the Arithmetic Series
Martina started a running program. In her first week she ran a total of 10 miles. Each week after that, she will run an additional 2 miles. How many total miles will Martina run in the first 6 weeks of the program?
Step 1
Write an explicit rule for the sequence.
an=a1+d(n1)a_n=a_1+d(n-1)
The common difference is 2. The first term is 10.
an=10+2(n1)=8+2n\begin{aligned}a_{n}&=10+2(n-1)\\&=8+2n\end{aligned}
Step 2
The total miles she will run in the first 6 weeks is a sum of terms in a sequence. Write the series in summation notation.
n=16(8+2n)\sum_{n=1}^{6}(8+2n)
Step 3
Use the explicit rule to identify the sixth term.
an=8+2na6=8+2(6)a6=8+12a6=20\begin{aligned}a_{n}&=8+2n\\a_{6}&=8+2(6)\\a_{6}&=8+12\\a_{6}&=20\end{aligned}
Step 4

Substitute into the formula for a finite arithmetic series and simplify.

Substitute n=6n=6, ak=8+2na_k=8+2n, a1=10a_1=10, and an=20a_n=20.
k=1nak=n2(a1+an)k=16(8+2n)=62(10+20)k=16(8+2n)=62(30)k=16(8+2n)=1802k=16(8+2n)=90\begin{aligned}\sum_{k=1}^{n}a_k&= \frac{n}{2} \left(a_1+a_n \right)\\ \sum_{k=1}^{6}(8+2n)&=\frac{6}{2}(10+20)\\ \sum_{k=1}^{6}(8+2n)&=\frac{6}{2}(30)\\ \sum_{k=1}^{6}(8+2n)&=\frac{180}{2}\\ \sum_{k=1}^{6}(8+2n)&=90\end{aligned}
Solution
Martina will run a total of 90 miles in the first 6 weeks of the running program.

Finite Geometric Series

A formula can be used to find the sum of a finite geometric series.

A geometric series is the sum of terms in a geometric sequence. The sum can be found by adding the terms in the sequence, but this may be difficult for a large number of terms.

The formula for the sum of the first nn terms in a finite geometric series, a1+a2++ana_1+a_2+…+a_n, where a1a_1 is the first term and rr is the common ratio, is:
k=1nak=a1(1rn1r),r1\sum_{k=1}^{n}a_k= a_1\left(\frac{1-r^n}{1-r}\right)\!\!{,}\;\;r\neq1
Step-By-Step Example
Determining the Sum of a Finite Geometric Series
Suppose an employee has a starting salary of $40,000. The employee's salary increases by 3% each year for the first five years. What are the total earnings for the employee for the first five years?
Step 1
Write an explicit rule for the sequence.
an=a1rn1a_n=a_1 \cdot r^{n-1}
Each year, the salary is 100% of the previous year plus an increase of 3%. As a decimal, 103% is written as 1.03, so the common ratio is 1.03. The first term is 40,000.
an=40,000(1.03)n1a_{n}=40{,}000(1.03)^{n-1}
Step 2

Substitute into the formula for a finite geometric series and simplify. Round to the nearest hundredth.

Substitute n=5n=5, ak=40,000(1.03)n1a_k=40{,}000(1.03)^{n-1}, a1=40,000a_1=40{,}000, and r=1.03r=1.03.
k=1nak=a1(1rn1r)n=1540,000(1.03)n1=40,000(11.03511.03)n=1540,000(1.03)n140,000(0.1592740.03)n=1540,000(1.03)n1212,365\begin{aligned}\sum_{k=1}^{n}a_k&= a_1\left(\frac{1-r^n}{1-r}\right)\\ \sum_{n=1}^{5}40{,}000(1.03)^{n-1}&=40{,}000\left(\frac{1-1.03^5}{1-1.03}\right)\\ \sum_{n=1}^{5}40{,}000(1.03)^{n-1}&\approx40{,}000\left(\frac{-0.159274}{-0.03}\right)\\ \sum_{n=1}^{5}40{,}000(1.03)^{n-1}&\approx 212{,}365\end{aligned}
Solution
The employee earned approximately $212,365 in the first 5 years.

Infinite Series

The sum of an infinite series may be found by using partial sums.
An infinite series is the sum of the terms in an infinite sequence. Since an infinite sequence has infinitely many terms, the sum can be found by using partial sums. The sum of a specified number of terms in a sequence is called a partial sum, written as SnS_n.
S1=a1S2=a1+a2S3=a1+a2+a3S4=a1+a2+a3+a4S5=a1+a2+a3+a4+a5S6=a1+a2+a3+a4+a5+a6S7=a1+a2+a3+a4+a5+a6+a7S8=a1+a2+a3+a4+a5+a6+a7+a8S9=a1+a2+a3+a4+a5+a6+a7+a8+a9\begin{aligned}S_1&=a_1\\S_2&=a_1+ a_2\\S_3&=a_1+ a_2+ a_3\\S_4&=a_1+ a_2+ a_3+ a_4\\S_5&=a_1+ a_2+ a_3+ a_4+ a_5\\S_6&=a_1+ a_2+ a_3+ a_4+ a_5+ a_6\\S_7&=a_1+ a_2+ a_3+ a_4+ a_5+ a_6+ a_7\\S_8&=a_1+ a_2+ a_3+ a_4+ a_5+ a_6+ a_7+ a_8\\S_9&=a_1+ a_2+ a_3+ a_4+ a_5+ a_6+ a_7+ a_8+ a_9\end{aligned}
As nn increases, the sum SnS_n may get closer to a fixed number without reaching that number. When SnS_n approaches a fixed number, that number is the sum of the infinite series. If SnS_n does not approach a fixed number, then the sum is undefined.

In an infinite arithmetic series, if the common difference is positive, each term is greater than the previous term, so the sum becomes greater and greater and does not approach a fixed number. It is said to approach \infty. If the common difference is negative, each term is less than the previous term, so the sum approaches -\infty. When the sequence of partial sums does not approach a fixed number, the series is called a divergent series. Since the sum of an infinite arithmetic series does not approach a fixed number, it is undefined.

In an infinite geometric series with r<1\left | r \right | <1, the partial sums get very close to a fixed number. The series is called a convergent series, which means that the sequence of partial sums approaches that number. The formula for the sum of an infinite geometric series, where a1a_1 is the first term and rr is the common ratio, that converges is:
k=1ak=a11r\sum_{k=1}^{\infty}a_k= \frac{a_1}{1-r}
In an infinite geometric series with r1\left|r\right|\geq 1, the sequence of partial sums does not approach a fixed number, so the series is said to diverge. Therefore, the sum of an infinite geometric series with r1\left|r\right|\geq 1 is undefined.
Step-By-Step Example
Identifying the Sum of an Infinite Geometric Series
Identify the sum of the series:
40+30+22.5+16.875+40+30+22.5+16.875+…
Step 1
Write an explicit rule for the sequence.
an=a1rn1a_n=a_1 \cdot r^{n-1}
The first term is 40. The common ratio is 0.75 because:
30÷40=0.7522.5÷30=0.7516.875÷22.5=0.75\begin{gathered}30\div40=0.75\\22.5\div30=0.75\\16.875\div22.5=0.75\end{gathered}
Since 0.75<1\left | 0.75 \right |<1, the infinite series converges and has a sum.
an=40(0.75)n1a_{n}=40(0.75)^{n-1}
Step 2

Substitute into the formula for an infinite geometric series with r<1\left | r \right | <1 and simplify.

Substitute ak=40(0.75)n1a_k=40(0.75)^{n-1}, a1=40a_1=40, and r=0.75r=0.75.
k=1ak=a11rn=140(0.75)n1=4010.75n=140(0.75)n1=160\begin{aligned}\sum_{k=1}^{\infty}a_k&= \frac{a_1}{1-r}\\ \sum_{n=1}^{\infty}40(0.75)^{n-1}&=\frac{40}{1-0.75}\\ \sum_{n=1}^{\infty}40(0.75)^{n-1}&=160\end{aligned}
Solution
The sum of the infinite geometric series is 160.