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Mathematics · Ch 8 — Sequences and Series

Series

8.3

Series

Series: The Sum of a Sequence

When you have a sequence — an ordered list of numbers — the natural next step is to add its terms together. That sum, written as a1+a2+a3+⋯+an+…a_1 + a_2 + a_3 + \dots + a_n + \dots, is called a series.

The series inherits its nature from the sequence that generates it. If the sequence is finite, the series is finite; if the sequence goes on forever, the series is infinite. For instance, the finite sequence 1,3,5,71, 3, 5, 7 gives the finite series 1+3+5+71 + 3 + 5 + 7, while the infinite sequence 1,3,5,7,…1, 3, 5, 7, \dots gives the infinite series 1+3+5+7+…1 + 3 + 5 + 7 + \dots.

Watch out

A common confusion: the word "series" refers to the expression that indicates the sum — the string of terms with plus signs — not the numerical result of adding them. When we talk about the "sum of a series," we mean the number you get after actually performing the addition. For 1+3+5+71 + 3 + 5 + 7, the series is the expression itself, and its sum is 1616.

Sigma Notation: A Compact Way to Write Series

Writing out long sums term by term is tedious. Mathematicians use the Greek letter sigma (Σ\Sigma) as a shorthand for summation. The series a1+a2+a3+⋯+ana_1 + a_2 + a_3 + \dots + a_n is written compactly as:

∑k=1nak\sum_{k=1}^{n} a_k

The notation reads: "the sum of aka_k as kk runs from 11 to nn." The variable kk is called the index of summation; it starts at the lower limit (11) and increases by 11 each step until it reaches the upper limit (nn).

Working with Sequences and Series: Examples

The textbook demonstrates how to find terms of a sequence from a given rule, and then how to write the corresponding series. Let's walk through each example carefully.

Note

In all these examples, the key is to substitute the required value of nn into the formula for the nnth term, ana_n.

Example 1: Write the first three terms of the sequences defined by:

  1. an=2n+5a_n = 2n + 5
  2. an=n−34a_n = \frac{n-3}{4} Solution (i): For an=2n+5a_n = 2n + 5:
  • n=1n = 1: a1=2(1)+5=7a_1 = 2(1) + 5 = 7
  • n=2n = 2: a2=2(2)+5=9a_2 = 2(2) + 5 = 9
  • n=3n = 3: a3=2(3)+5=11a_3 = 2(3) + 5 = 11

The first three terms are 7,9,117, 9, 11.

Solution (ii): For an=n−34a_n = \frac{n-3}{4}:

  • n=1n = 1: a1=1−34=−24=−12a_1 = \frac{1-3}{4} = \frac{-2}{4} = -\frac{1}{2}
  • n=2n = 2: a2=2−34=−14=−14a_2 = \frac{2-3}{4} = \frac{-1}{4} = -\frac{1}{4}
  • n=3n = 3: a3=3−34=04=0a_3 = \frac{3-3}{4} = \frac{0}{4} = 0

The first three terms are −12,−14,0-\frac{1}{2}, -\frac{1}{4}, 0.

Example 2: Find the 20th term of the sequence defined by an=(n−1)(2−n)(3+n)a_n = (n-1)(2-n)(3+n).

Solution: Substitute n=20n = 20 directly into the formula:

a20=(20−1)(2−20)(3+20)=19×(−18)×23a_{20} = (20-1)(2-20)(3+20) = 19 \times (-18) \times 23

Now multiply step by step:

19×(−18)=−34219 \times (-18) = -342

Then (−342)×23=−7866(-342) \times 23 = -7866

So a20=−7866a_{20} = -7866.

Example 3: A sequence is defined recursively: a1=1a_1 = 1, and an=an−1+2a_n = a_{n-1} + 2 for n≥2n \ge 2. Find the first five terms and write the corresponding series.

Solution: A recursive definition gives each term based on the previous one. We start with a1=1a_1 = 1.

  • a2=a1+2=1+2=3a_2 = a_1 + 2 = 1 + 2 = 3
  • a3=a2+2=3+2=5a_3 = a_2 + 2 = 3 + 2 = 5
  • a4=a3+2=5+2=7a_4 = a_3 + 2 = 5 + 2 = 7
  • a5=a4+2=7+2=9a_5 = a_4 + 2 = 7 + 2 = 9

The first five terms are 1,3,5,7,91, 3, 5, 7, 9. The corresponding series is:

1+3+5+7+9+…1 + 3 + 5 + 7 + 9 + \dots

Tip

When a sequence is defined recursively, always start from the given initial term(s) and work your way forward one step at a time. Do not try to jump ahead — the pattern only emerges term by term.

The Fibonacci Sequence: A Special Recursive Definition

The textbook introduces the famous Fibonacci sequence in its exercise. It is defined by:

  • a1=1a_1 = 1, a2=1a_2 = 1 (two initial terms)
  • an=an−1+an−2a_n = a_{n-1} + a_{n-2} for n>2n > 2 …