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# 7.2: Summation Notation

Difficulty Level: At Grade Created by: CK-12

## Learning objectives

• Write a finite sum using sigma notation.
• Read sigma notation and evaluate a sum.
• Write an infinite sum using sigma notation.
• Find terms of a sequence and find sums using a graphing calculator.

## Introduction

Consider the following situation:

You are participating in a 100-mile walk for a charity. In order to complete the entire walk, you first have to go halfway, or 50 miles. After you complete the first half of the walk, you again have to go halfway from where you are, or 25 more miles. Continuing with this pattern, you will always be going halfway. So will you ever finish the 100-mile walk?

This situation may seem silly, but it is actually a version of a problem that was first proposed thousands of years ago, by Zeno of Elea, a Greek philosopher who wanted to challenge accepted ideas about space and time. In today’s world, most adults have an intuitive understanding of distance, rate and time, that allows us to answer the above question: yes, you will finish the walk, though it depends on how fast you walk.

We can also answer this question if we add up the sequence of distances you would walk each time you went half of the previous distance: 50+25+12.5+... even though this is the sum of infinitely many numbers, we can actually find the value of this expression. By the end of the chapter, you will be able to do this type of calculation.

An infinite sum is known as an infinite series. In this lesson you will learn about series, both finite and infinite. You will learn to use a particular kind of notation that allows you to write a sum succinctly.

## Summation Notation

Consider for example a sequence defined by an=3n . If we write out the sum of the first 4 terms, we have 3+6+9+12=30 . But what if we want to write out terms for a larger sum?

Summation notation is a method of writing sums in a succinct form. To write the sum 3+6+9+12=30 , we use the Greek letter Sigma, as follows:

n=143n\begin{align*}\sum_{n=1}^4 3n\end{align*}

The expression 3n is called the summand, the 1 and the 4 are referred to as the limits of the summation, and the n is called the index of the sum. Here we have used a “sigma” to write a sum. We can also read a sigma, and determine the sum. For example, we can read the above sigma notation as “find the sum of the first four terms of the series, where the nth term is 3n.” We always read the limits from the bottom to the top. The bottom number tells you which term to start with, and the top limit tells you which term is the final term to add. We could then write out the sigma above as:

n=143n\begin{align*}\sum_{n=1}^4 3n\end{align*} =3(1)+3(2)+3(3)+3(4)
=3+6+9+12=30

In general, we can either rewrite a given series in sigma notation, or we can read sigma notation in order to find the value of the sum.

Example 1: Write the sum using sigma notation:

2+4+6+8+10+12+14+16+19+20

Solution: n=1102n\begin{align*}\sum_{n=1}^{10} 2n\end{align*}

Every term is a multiple of 2. The first term is 2 × 1, the second term is 2 × 2 , and so on. So the summand of the sigma is 2n. There are 10 terms in the sum. Therefore the limits of the sum are 1 and 10.

## Properties of Sigma

Notice that we can write the sum 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 19 + 20 as 2(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) Therefore n=1102n=2n=110n\begin{align*}\sum_{n=1}^{10} 2n = 2 \sum_{n=1}^{10} n\end{align*}. In general, we can factor a coefficient out of a sum:

n=1kcan\begin{align*}\sum_{n=1}^{k}ca_n \end{align*} =n=1kan\begin{align*} = \sum_{n=1}^{k} a_n \end{align*}

We can also break a sum into two or more sums.

Example 2: Write out the terms of n=25(n+7)\begin{align*}\sum_{n=2}^{5}(n+7) \end{align*} and evaluate the sum.

Solution: The sum is 37.

n=25(n+7)\begin{align*}\sum_{n=2}^{5}(n+7) \end{align*} =(2 + 7) + (3 + 7) + (4 + 7) + (5 + 7)
=2 + 3 + 4 + 5 + 7 + 7 + 7 + 7
=2 + 3 + 4 + 5 + 7 × 4 = 14 + 28 = 42

Notice we could have written n=25(n+7)\begin{align*}\sum_{n=2}^{5}(n+7) \end{align*} as n=25n+n=257\begin{align*}\sum_{n=2}^{5}n+ \sum_{n=2}^{5}7\end{align*} . Also, the second sum does not depend on the index of the sum (i.e. it stays at 7 regardless of the index), only that there are 4 terms to add together. Seeing this can make a sum easier to evaluate.

In general, we can write a sum as a sum of sums: i1n(ai+bi)=i=1n(ai)+i=1n(bi)\begin{align*}\sum_{i-1}^{n}(a_i+b_i) =\sum_{i=1}^{n}(a_i)+\sum_{i=1}^{n}(b_i)\end{align*}.

Example 3: Write out the terms of i=0432(14)n\begin{align*}\sum_{i=0}^{4}32\left ( \frac{1}{4} \right )^n\end{align*} and evaluate the sum.

Solution: The sum is 4258\begin{align*}42\frac{5}{8}\end{align*}

i=0432(14)n\begin{align*}\sum_{i=0}^{4}32 \left ( \frac{1}{4} \right)^n\end{align*} =32(14)0+32(14)1+32(14)2+32(14)3+32(14)4\begin{align*}= 32 \left ( \frac{1}{4} \right )^0+32 \left ( \frac{1}{4} \right )^1+32 \left ( \frac{1}{4} \right )^2+32 \left ( \frac{1}{4} \right )^3+32 \left ( \frac{1}{4} \right )^4\end{align*}
=321+3214+32116+32164+321256\begin{align*}= 32 \cdot 1+32\cdot \frac{1}{4}+32 \cdot \frac{1}{16}+ 32\cdot \frac{1}{64}+32 \cdot \frac{1}{256}\end{align*}
=32+8+2+12+18=4258\begin{align*}= 32+8+2+\frac{1}{2}+\frac{1}{8}=42 \frac{5}{8}\end{align*}

The properties of sigma we have used in the past 3 examples will be useful as we continue to work with sigma in this chapter, as well as in your future study of calculus. The same is true of the next topic, the infinite sum.

## Infinite sums

Now let’s return to the sum at the beginning of the lesson: 50 + 25 + 12.5 +...

We can write this sum using sigma notation if we identify the form of the nth term. Notice that each term is half of the previous term. We can use this pattern to find the nth term:

a1=50\begin{align*}a_1 = 50\end{align*}
a2=a112\begin{align*}a_2 = a_1 \cdot \frac{1} {2}\end{align*}
a3=a212=a11212=a1(12)2\begin{align*}a_3 = a_2 \cdot \frac{1} {2} = a_1 \cdot \frac{1} {2} \cdot \frac{1} {2} = a_1 \left(\frac{1} {2} \right)^2\end{align*}
a4=a1(12)3\begin{align*}a_4 = a_1 \left(\frac{1} {2} \right)^3\end{align*}
an=a1(12)n1\begin{align*}a_n = a_1 \left( \frac{1} {2} \right)^{n - 1}\end{align*}

The terms form a geometric sequence, so we refer to the sum as an infinite geometric series. We can write this series as: n=150(12)n1\begin{align*} \sum_{n = 1}^\infty 50 \left(\frac{1} {2} \right)^{n - 1}\end{align*} . And, while this sum is the sum of infinitely many numbers, you can actually find the value of this sum. Remember: it is possible to complete the 100-mile charity walk. Therefore the sum is 100.

When an infinite sum has a finite value, we say the sum converges. Otherwise, the sum diverges. A sum converges only when the terms get closer to 0 after each step, but this is not a sufficient criterion for convergence. For example, the sum \begin{align*}\sum_{n - 1}^\infty \frac{1} {n} = 1 + \frac{1} {2} + \frac{1} {3} + \frac{1} {4} + ....\end{align*} does not converge. Later in this chapter, you will learn how to find the value certain of infinite sums that converge.

### Using a graphing calculator

To generate either a sequence or a series, you can use a graphing calculator. The TI-83/84 series gives you several options. You can, for example, work in sequence mode, which allows you to define a sequence and find terms. If you have an explicit formula for a sequence, you can keep your calculator in function mode. For example, consider the sum \begin{align*}\sum_{n = 1}^9 n^2\end{align*} . It would be time consuming to write out the first 9 squares. The calculator is faster. To generate the 9 terms, press <TI font_2nd> [LIST], then select OPS, then option 5, seq(. This takes you back to the main screen. You should see seq(. After this, enter x^2, x, 1, 9, 1). (The x tells the calculator that x is the input. The 1 and the 9 tell it the limits of the sum. The second 1 tells the calculator to go up in increments of 1.)

Press <TI font_ENTER>, and you should see the list of squares. Scroll to the right to see all of them. The scrolling will end when you reach 81.

If you want to find the sum of the terms, first store the sequence in a list (see screen below), then <TI font_2nd> [LIST], then select MATH, then option 5, sum(.Then enter the name of the list and press <TI font_)><TI font_ENTER>. You should get 285.

## Summary

In this lesson we have introduced Sigma notation, which allows us to write a sum without having to write out terms. We can use sigma notation to write a sum, and we can find the value of a sum, given its sigma notation. We can also represent an infinite sum using this notation. In some cases, an infinite sum has a finite value. Understanding the concepts in this lesson will help you to evaluate such sums later in the chapter.

## Points to Consider

1. Consider the 100-mile walk example: how can infinitely many distances add up to 100?

2. What kind of infinite sum will converge? What will the terms look like?

## Review Questions

For questions 1 – 3, express the sum using sigma notation

1. 1 + 3 + 5 + 7 + 9
2. 2 + 6 + 18 + 54
3. \begin{align*}1 + \frac{1} {2} + \frac{1} {3} + \frac{1} {4} + ... + \frac{1} {10}\end{align*}
4. For questions 4 – 6, expand the sigma, and find the sum.
5. \begin{align*}\sum_{n = 1}^7 (2n - 3) \end{align*}
6. \begin{align*}\sum_{n = 3}^6 (n^2 - 5) \end{align*}
7. \begin{align*}\sum_{n = 1}^5 \left (\frac{1} {3n} \right) \end{align*}
8. Express the sum using sigma notation: 1 + 3 + 9 + 27 + ...
9. Write the sum in question 4 as 2 sums.
10. Consider the sums \begin{align*}\sum_{n = 1}^5 (n + 1)\end{align*} and \begin{align*}\sum_{n = 1}^5 (n - 4)\end{align*}. Is the product \begin{align*}\left (\sum_{n = 1}^5 (n + 1) \right) \left (\sum_{n = 1}^5 (n - 4) \right)\end{align*} equal to \begin{align*}\sum_{n = 1}^5 (n + 1) (n - 4)\end{align*} ?
11. Consider the sum \begin{align*}\sum_{n = 1}^\infty \left (\frac{1} {5} \right)^n\end{align*}. a. Find the sum of the first 10 terms using a graphing calculator. b. Find the sum of the first 20 terms using a graphing calculator c. Do you think the series converges? Explain.

1. \begin{align*}\sum_{n = 1}^5 2n - 1\end{align*}
2. \begin{align*}\sum_{n = 1}^4 2 \left(3^{n - 1} \right)\end{align*} or \begin{align*}\sum_{n = 0}^3 2 \left(3^n \right)\end{align*}
3. \begin{align*}\sum_{n = 1}^{10} \left(\frac{1} {n} \right)\end{align*}
4. \begin{align*}\sum_{n = 1}^7 (2n - 3)\end{align*} \begin{align*}= (2 \times 1 - 3) + (2 \times 2 - 3) + (2 \times 3 - 3) + (2 \times 4 - 3) + (2 \times 5 - 3) + (2 \times 6 - 3) + (2 \times 7 - 3)\end{align*}
\begin{align*}= (-1) + (1) + (3) + (5) + (7) + (9) + (11)\end{align*}
\begin{align*}= 35\end{align*}
5. \begin{align*}\sum_{n = 3}^6 (n^2 - 5)\end{align*} \begin{align*}= (3^2 - 5) + (4^2 - 5) + (5^2 - 5) + (6^2 - 5)\end{align*}
\begin{align*}= (4) + (11) + (20) + (31)\end{align*}
\begin{align*}= 66\end{align*}
6. \begin{align*}\sum_{n = 1}^5 \left (\frac{1} {3n} \right)\end{align*} \begin{align*}= \left (\frac{1} {3} \right) + \left (\frac{1} {6} \right) + \left (\frac{1} {9} \right) + \left (\frac{1} {12} \right) + \left (\frac{1} {15} \right)\end{align*}
\begin{align*}= \frac{60} {180} + \frac{30} {180} + \frac{20} {180} + \frac{15} {180} + \frac{12} {180}\end{align*}
\begin{align*}= \frac{137} {180}\end{align*}
7. \begin{align*}\sum_{n = 1}^\infty 3^{n - 1}\end{align*} or \begin{align*}\sum_{n = 0}^\infty 3^n\end{align*}
8. \begin{align*}\sum_{n = 1}^7 (2n - 3) = \sum_{n = 1}^7 \left(2n \right) + \sum_{n = 1}^7 \left(-3 \right)\end{align*} or \begin{align*}2 \sum_{n = 1}^7 \left(n \right) - \sum_{n = 1}^7 \left (3 \right)\end{align*}
9. The product of the sums is not equal to the sum of the product. \begin{align*}\sum_{n = 1}^5 (n + 1) = 20\end{align*} and \begin{align*}\sum_{n = 1}^5 (n - 4) = -5\end{align*}, so the product is \begin{align*}-100\end{align*} \begin{align*}\sum_{n = 1}^5 (n + 1) (n - 4) = \sum_{n = 1}^5 \left (n^2 - 3n - 4 \right) = -10\end{align*}
10. a. 0.2499999744 b. 0.25 c. It likely converges to 0.25. If you find sums beyond the 20th sum, the sum is still .25.

## Vocabulary

Converge
If the limit of the partial sums of series exists and is finite, the series converges.
Diverge
If the limit of the partial sums of series does not exist or is infinite, then the series diverges.
Index
The index of the sum is the variable in the sum.
Limits of the summation
The limits of a summation are the starting and ending points of the sum.
Sigma
Sigma (\begin{align*}\Sigma \end{align*}) is the Greek letter used to represent a sum.
Summand
The summand of a sigma is the expression being summed.

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