Tenth Grade Mathematics: Introduction to Sequences and Series

Patterns and orderly repetition are omnipresent in nature, from the arrangement of seeds in a sunflower to the growth of financial investments over time. In mathematics, we formalize the concept of an ordered list of numbers governed by a definite rule as a sequence. When the terms of a sequence are cumulatively added together, they form a series. Developing an understanding of how sequences are generated, how their general terms are formulated, and how series behave forms the essential bridge between basic arithmetic and the continuous mathematics of calculus.

Definition of a Sequence

A sequence is an ordered list of numbers arranged according to a specific, well-defined mathematical rule:

Formal Definition:
A sequence is a function whose domain is the set of natural numbers N = {1, 2, 3, ...}
or a subset of non-negative integers:
                     f: N ---> R
The values f(1), f(2), f(3), ..., f(n) are called the terms of the sequence,
customarily denoted by:
                     a1, a2, a3, ..., an

The subscript n indicates the position (or index) of the term. The term an is called the n-th term, or the general term of the sequence.

Contrasting Sets and Sequences:
In a set, order does not matter and elements are not repeated: {1, 2, 3} = {3, 1, 2}.
In a sequence, order is paramount: (1, 2, 3) != (3, 2, 1).
Repeated numbers represent distinct indexed positions: (1, 1, 1, 1, ...).

Finite vs. Infinite Sequences

A sequence is called a finite sequence if it contains a bounded, countable number of terms ending at some terminal term ak: Examples: (2, 4, 6, 8, 10) consists of exactly 5 terms.

A sequence is called an infinite sequence if its terms continue endlessly without termination: Examples: (1, 1/2, 1/4, 1/8, ...) continues indefinitely.

Generating Sequences from the General Term

When a formula for the n-th term an is provided, any individual term can be evaluated directly by substituting the positive integer index n:

Examples of Explicit Term Generation:
1. an = 2*n + 5:
   a1 = 2*(1) + 5 = 7
   a2 = 2*(2) + 5 = 9
   a3 = 2*(3) + 5 = 11
   Sequence: (7, 9, 11, 13, ...)

2. an = (-1)^(n - 1) * 5^n:
   a1 = (-1)^0 * 5^1 = 5
   a2 = (-1)^1 * 5^2 = -25
   a3 = (-1)^2 * 5^3 = 125
   Sequence: (5, -25, 125, -625, ...) (Alternating Sequence)

3. an = n / (n + 1):
   a1 = 1/2, a2 = 2/3, a3 = 3/4, a4 = 4/5
   Sequence: (1/2, 2/3, 3/4, 4/5, ...)

Recursive Definitions and the Fibonacci Sequence

Rather than defining an explicitly in terms of n alone, many sequences are defined recursively. In a recursive sequence, the initial terms are specified, and each subsequent term is defined as an algebraic function of its preceding terms.

The Celebrated Fibonacci Sequence:
Defined by the recurrence relation:
  a1 = 1
  a2 = 1
  an = a_(n-1) + a_(n-2),   for all n >= 3

Computing Successive Terms:
  a3 = a2 + a1 = 1 + 1 = 2
  a4 = a3 + a2 = 2 + 1 = 3
  a5 = a4 + a3 = 3 + 2 = 5
  a6 = a5 + a4 = 5 + 3 = 8
  a7 = a6 + a5 = 8 + 5 = 13
Fibonacci Sequence: (1, 1, 2, 3, 5, 8, 13, 21, 34, ...)

Definition of a Series and Sigma Notation

If a1, a2, a3, ..., an is a sequence of numbers, then the indicated sum of its terms:

S = a1 + a2 + a3 + ... + an

is called the series associated with the sequence. Series are compactly represented using the Greek capital letter Sigma (summation symbol):

                     n
     Sn = Sum(ak) = Sum a_k = a1 + a2 + a3 + ... + an
                    k=1

In this notation, k is the index of summation, 1 is the lower limit of summation, and n is the upper limit of summation.

Chapter Exercises

Exercise 1: Write the first five terms of the sequences whose n-th terms are given by: (a) an = n*(n + 2), (b) an = n / (n^2 + 1), and (c) an = (-1)^(n-1) * 2^n.

Exercise 2: A sequence is defined recursively by a1 = 3, and an = 3*a_(n-1) + 2 for all n > 1. Find the first four terms of this sequence and write down the corresponding series.

Exercise 3: Find the 15th and 20th terms of the sequence whose n-th term is defined piecewise as: an = n^2 if n is odd, and an = (n - 1) / (n + 1) if n is even.

Solutions and Explanations

Solution 1: For (a) an = n(n + 2): a1 = 1(3) = 3; a2 = 2(4) = 8; a3 = 3(5) = 15; a4 = 4(6) = 24; a5 = 5(7) = 35. The terms are 3, 8, 15, 24, 35. For (b) an = n / (n^2 + 1): a1 = 1/(1 + 1) = 1/2; a2 = 2/(4 + 1) = 2/5; a3 = 3/(9 + 1) = 3/10; a4 = 4/(16 + 1) = 4/17; a5 = 5/(25 + 1) = 5/26. The terms are 1/2, 2/5, 3/10, 4/17, 5/26. For (c) an = (-1)^(n-1) * 2^n: a1 = (-1)^0 * 2^1 = 2; a2 = (-1)^1 * 2^2 = -4; a3 = (-1)^2 * 2^3 = 8; a4 = (-1)^3 * 2^4 = -16; a5 = (-1)^4 * 2^5 = 32. The terms are 2, -4, 8, -16, 32.

Solution 2: The sequence starts with a1 = 3. Using the recurrence relation an = 3a_(n-1) + 2: For n = 2: a2 = 3a1 + 2 = 3(3) + 2 = 9 + 2 = 11. For n = 3: a3 = 3a2 + 2 = 3(11) + 2 = 33 + 2 = 35. For n = 4: a4 = 3a3 + 2 = 3*(35) + 2 = 105 + 2 = 107. The first four terms of the sequence are 3, 11, 35, 107. The corresponding series is the sum of these terms: 3 + 11 + 35 + 107.

Solution 3: We evaluate an according to the parity of n. To find the 15th term, observe that n = 15 is an odd integer. Applying the odd rule: a15 = 15^2 = 225. To find the 20th term, observe that n = 20 is an even integer. Applying the even rule: a20 = (20 - 1) / (20 + 1) = 19 / 21. Thus, the 15th term is 225 and the 20th term is 19/21.