## Monday, March 5, 2007

### Precalculus 9.1 Sequences and Series

9.1 Sequences and Series

A sequence is a function whose domain is the set of positive integers.

Definition of Sequence:
An infinite sequence is a function whose domain is the set of positive integers. The function values
a1, a2, a3, a4, ... , an, ...
are the terms of the sequence. If the domain of the function consists of the first n positive integers only, the sequence is a finite sequence.

A. Finding the Terms of a Sequence:
Example 1: find the first five terms of an = 4n - 7
a1 = 4(1) - 7 = -3
a2 = 4(2) - 7 = 8 - 7 = 1
a3,= 4(3) - 7 = 12 - 7 = 5
a4 = 4(4) - 7 = 16 - 7 = 9
a5 = 4(5) - 7 = 20 - 7 = 13

Example 2: Find the 16th term of the sequence an = (-1)n-1(n(n-1))
a16 = (-1)16-1(16(16-1)) = (-1)15(16(15)) = (-1)(240) = -240

Example 3: Write the first five terms of the sequence defined recursively:
a1 = 15, ak+1 = ak+ 3
Let k = 1 so we have:
a1+1 = a2 = a1 + 3 = 15 + 3 = 18
Let k = 2 so we have:
a2+1 = a3 = a2 + 3 = 18 + 3 = 21
a3+1 = a4 = a3 + 3 = 21 + 3 = 24
a4+1 = a5 = a4 + 3 = 24 + 3 = 27

B. Finding the nth term of a Sequence
Example 4: Write an expression for the apparent nth term of the sequence
(assume n begins with 1): 3, 7, 11, 15, 19, ...

7 - 3 = 11 - 7 = 15 - 11 = 4

As you can see, the terms are going up by 4 and the first term is one less than 4 so
an = 4n - 1

Example 5: 1, 1/4, 1/9, 1/16, 1/25, ...
As you can see, the terms denominators are perfect squares so
an = 1/(n2) = n-2

C. The Fibonacci Sequence: A Recursive Sequence
The Fibonacci sequence is defined recursively as follows.
a0 = 1, a1 = 1, ak = ak-2 + ak-1 , where k is greater than or equal to 2
{1, 1, 2, 3, 5, 8, ...}

Example 6: Write the first five terms of the sequence defined recursively. Use this pattern to write the nth term of the sequence as a function of n.
a1 = 25, ak+1 = ak - 5
a2 = a1+1 = a1 -5 = 25 -5 = 20
a3 = a2+1 = a2 -5 = 20 - 5 = 15
a4 = a3+1 = a3 -5 = 15 - 5 = 10
a5 = a4+1 = a4 -5 = 10 - 5 = 5
Therefore an = 25 - 5n

C. Definition of Factorial:
In n is a positive integer, n factorial is defined by
n! = 1 x 2 x 3 x 4 x ... (n - 1) x n
As a special case, zero factorial is defined as 0! = 1

Example 7:
Simplify the ratio of factorials.
(4!)/(7!) = (4!) / (7 x 6 x 5 x 4!) = 1/ (7 x 6 x 5 ) = 1 / 210

Example 8: Simplify the ratio of factorials.
(n + 2)! / (n!) = (n + 2)(n + 1)(n!) / (n!) = (n + 2)(n + 1) = n2 + 3n + 2

D. Definition of Summation Notation:
The sum of the first n terms of a sequence is represented by

The summation of ai when i = 1 to i = n is equal to

a1 + a2 + a3 + a4 + ... + an

where i is called the index of summation, n is the upper limit of summation, and 1 is the lower limit of summation. (check out http://www.cs.fsu.edu/~cop4531/slideshow/chapter3/3-1.html)

Example 9: Find the sum of 3i - 1 when i = 1 to i = 6

3 - 1 + 6 - 1 + 9 - 1 + 12 - 1 + 15 - 1 + 18 - 1 = 57

To use your TI - 83, TI - 83 plus calculators:
Go to mode - change from function to sequential mode, then
2nd Stat arrow to math #5 Sum, 2nd Stat arrow to ops #5 seq then put in the sequence, then n to show the calculator the variable, then the lower bound number, then the upper bound number, then close the parenthesis twice, then press enter.
Your screen should look like this:
sum(seq(3n-1,n,1,6)) = 57

E. Properties of Sums - http://www.libraryofmath.com/summation-formulas.html

F. Series:

Definition of a Series:
Consider the infinite sequence a1, a2, a3, a4, ... , ai, ...
1. The sum of all terms of the infinite sequence is called an infinite series and is denoted by
a1 + a2 + a3 + a4 + ... + ai + ... = The summation of ai when i = 1 to i = infinity.

2. The sum of the first n terms of the sequence is called a finite series or the nth partial sum of the sequence and is denoted by
a1 + a2 + a3 + a4 + ... + an = The summation of ai when i = 1 to i = n.

Example 10: Find the sum of the partial sum of the series:

The summation of 8(-1/2)n when n = 1 to n = infinity, the 4th partial sum

= 8(-½ )1 + 8(-½ )2 + 8(-½ )3 + 8(-½ )4 = -4 + 2 + -1 + .5 = -5/2

(Calculator check: you can check your individual answers by putting you calculator in function mode, put the series into y1 = 8 (-.5)x and looking in your table to get your values.)

Example 11
: Find the sum of the infinite series:

The summation of 4(1/10)n when n = 1 to n = infinity

= 4 [ .1 + .01 + .001 + .0001 + ...] = 4 (.111111...) = 4 (1/9) = 4/9

Example 12: Find the sum of the infinite series:

The summation of 8(1/10)n when n = 1 to n = infinity is equal to

= 8 [ .1 + .01 + .001 + .0001 + ...] = 8 (.111111...) = 8 (1/9) = 8/9