Sequences and Series: Definitions and Terms

Fundamentals of Sequences

• Sequence Definition: A sequence is a mathematical function whose domain is the collection of all integers greater than or equal to a given integer $m$ (usually $0$ or $1$).
• Notation: A sequence is usually denoted by $a_n$.
• Terms of a Sequence: The functional values $a_1, a_2, a_3, \dots, a_n, \dots$ are called the terms of a sequence.
• General Term / $n^{\text{th}}$ Term: The functional value $a_n$ is called the general term, or the $n^{\text{th}}$ term of the sequence.

Classification of Sequences

Sequences are classified into two types depending on the existence of a last term:

• Finite Sequence:

  • Definition: A sequence that has a last term.
  • Domain: The domain of a finite sequence is the finite set of integers $\{1, 2, 3, \dots, n\}$.

• Infinite Sequence:

  • Definition: A sequence that does not have a last term.
  • Domain: The domain of an infinite sequence is the set of natural numbers, denoted by $\mathbb{N}$ or $\{1, 2, 3, \dots\}$.

Listing Terms of a Sequence

To find the terms of a sequence given its general term $a_n$, substitute successive positive integer values ($n = 1, 2, 3, 4, 5, \dots$) into the expression for the general term.

Example 1(a): General Term $a_n = 2n - 1$

• Task: List the first five terms of the sequence where $n$ is a positive integer.
• Calculation of Terms:
  • First term ($n = 1$): $a_1 = 2(1) - 1 = 1$
  • Second term ($n = 2$): $a_2 = 2(2) - 1 = 3$
  • Third term ($n = 3$): $a_3 = 2(3) - 1 = 5$
  • Fourth term ($n = 4$): $a_4 = 2(4) - 1 = 7$
  • Fifth term ($n = 5$): $a_5 = 2(5) - 1 = 9$
• Result: The first five terms are $1, 3, 5, 7,$ and $9$.

Example 1(b): General Term $a_n = \left(-\frac{1}{3}\right)^{n-1}$

• Task: List the first five terms of the sequence where $n$ is a positive integer.
• Calculation of Terms:
  • First term ($n = 1$): $a_1 = \left(-\frac{1}{3}\right)^{1-1} = \left(-\frac{1}{3}\right)^0 = 1$
  • Second term ($n = 2$): $a_2 = \left(-\frac{1}{3}\right)^{2-1} = \left(-\frac{1}{3}\right)^1 = -\frac{1}{3}$
  • Third term ($n = 3$): $a_3 = \left(-\frac{1}{3}\right)^{3-1} = \left(-\frac{1}{3}\right)^2 = \frac{1}{9}$
  • Fourth term ($n = 4$): $a_4 = \left(-\frac{1}{3}\right)^{4-1} = \left(-\frac{1}{3}\right)^3 = -\frac{1}{27}$
  • Fifth term ($n = 5$): $a_5 = \left(-\frac{1}{3}\right)^{5-1} = \left(-\frac{1}{3}\right)^4 = \frac{1}{81}$
• Result: The first five terms are $1, -\frac{1}{3}, \frac{1}{9}, -\frac{1}{27},$ and $\frac{1}{81}$.

Example 1(c): General Term $a_n = \frac{1}{n}$

• Task: List the first five terms of the sequence where $n$ is a positive integer.
• Calculation of Terms:
  • First term ($n = 1$): $a_1 = \frac{1}{1} = 1$
  • Second term ($n = 2$): $a_2 = \frac{1}{2}$
  • Third term ($n = 3$): $a_3 = \frac{1}{3}$
  • Fourth term ($n = 4$): $a_4 = \frac{1}{4}$
  • Fifth term ($n = 5$): $a_5 = \frac{1}{5}$
• Result: The first five terms are $1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4},$ and $\frac{1}{5}$.