### Chapter 4 - People Server at UNCW

```Chapter 4
Sequences and Mathematical
Induction
4.1
Sequences
Sequences
• The main mathematical structure used to
study repeated processes is the sequence.
• The main mathematical tool used to verify
arrangement of terms in sequences is
mathematical induction.
Example
• Ancestor counting with a sequence
– two parents, four grandparents, eight greatgrandparents, etc.
– Number of ancestors can be represented as 2position
– Example: 23 = 8 (great grandparents), therefore
parents removed three generations are great
grandparents for which you have a total of 8.
Sequences
• Sequence is a set of elements written in a row
as illustrated on prior slide. (NOTE: a sequence
can be written differently)
• Each element of the sequence is a term.
• Example
– am, am+1, am+2, am+3, …, an
– terms a sub m, a sub m+1, a sub m+2, etc.
– m is subscript of initial term
– n is subscript of final term
Example
• Finding terms of a sequence given explicit
formulas
– ak = k/(k+1) for all integers k ≥ 1
– bi = (i-1)/i for all integers i ≥ 2a
a1 = 1/(1+1) = ½
b2 = (2-1)/2 = ½
a2 = 2/(2+1) = 2/3
b3 = (3-1)/3 = 2/3
a3 = 3/(3+1) = 3/4
b4 = (4-1)/4 = 3/4
a4 = 4/5
b5 = 4/5
– the sequences a and b have the same terms and
hence, are identical
Example
• Alternating Sequence
– cj = (-1)j for all integers j≥0
c0 = (-1)0 = 1
c1 = (-1)1 = -1
c2 = (-1)2 = 1
c3 = (-1)3 = -1
c4 = (-1)4 = 1
…
– sequence has bound values for the term.
– term ∈ {-1, 1}
Example
• Find an explicit formula to fit given initial terms
– sequence = 1, -1/4, 1/9, -1/16, 1/25, -1/36, …
• alternate in sign
• numerator is always 1
• denominator is a square
1/12
-1/22
1/32
-1/42
1/52
-1/62
a1
a2
a3
a4
a5
a6
– ak = ±1 / k2 (from the previous example we know how to
create oscillating sign sequence, odd negative and even
positive.
– ak = (-1)k+1 / k2
Summation Notation
• Summation notation is used to create a compact
form for summation sequences governed by a
formula.
n
a
k 1
k
 a1  a2  a3  a4  ...  an
• the sequence is governed by k which has lower
limit (1) and a upper limit of n.
 • This sequence is finite because it is bounded on
the lower and upper limits.
Example
• Computing summations
5
a
k
 a1  a2  a3  a4  a5
k 1
– a1 = -2, a2 = -1, a3 = 0, a4 = 1, and a5 =2.

5
a
k
k 1
 2  1 0 1 2  0
Example
• Computing summation from sum form.
5
k
2
k 1
5


k
k 1
2
12  22  32  4 2  52  55
Example
• Changing from Summation Notation to
Expanded form
n
(1) i
 i 1
i0
n


(1)i (1) 0 (1)1 (1)2
(1)n
 i 1  0 1  11  2 1  ... n 1
i0
Example
• Changing from expanded to summation form.
• Find a close form for the following:
1
2
3
n 1


 ...
n n 1 n  2
2n
1
2
3
n 1
k 1


 ...

n n 1 n  2
n  n k 0 n  k
n


Separating Off a Final Term
• A final term can be removed from the
summation form as follows.
n 1
n
a  a
k
k m
k
 an
k m
• Example
of use: Rewrite the following

separating the final term
1 n 1 1
1


 k2  k2 n2
k 1
k 1
n
Example
• Combining final term
n 1
2
k
 2n
k 0
n 1


2
k 0
n
k
 2n  2k
k 0
Telescoping Sum
• Telescoping sum can be evaluated to a closed
form.
1
1
(k 1)  k
1



k k 1
k(k 1)
k(k 1)
n
1
 1
1
1  1 1  1 1  1 1 
1  1
1 
1
   
      ... 
1 
 k(k 1)  k  k 1 1  2 
 2 3 3 4 
n 1 n  n n 1
n 1
k 1
k 1
n


Product Notation
5
a
k
 a1a2 a3 a4 a5
k 1
n 1 
 ak   ak  an
k 1 
k 1
n

Recursive form
Example
• Compute the following products:
5
 k  1* 2 * 3* 4 * 5  120
k 1
1
k
1
1
 k 1  11  2
k 1

Factorial
• Factorial is for each positive integer n, the
quantity n factorial denoted n! is defined to
be the product of all the integers from 1 to n:
– n! = n * (n-1) *…*3*2*1
• Zero factorial denoted 0! is equal to 1.
Example
• Computing Factorials
8! 8 * 7!

8
7!
7!
5!
5 * 4 * 3! 5 * 4


2!*3!
2!*3!
2!

Properties of Summations and
Products
• Theorem 4.1.1
– If am, am+1, am+2, … and bm, bm+1, bm+2, … are
sequences of real numbers and c is any real
number, then the following equations hold for any
integer n≥m:
n
1.
n
n
 a  b  a
k
k m
k
k m
k
 bk 
k m
n
n
k m
k m
n
2. c *  ak  c * ak
 n  
 n
3.  ak * bk  ak * bk 
k m  k m  k m

Examples
• Let ak = k +1 and bk = k – 1 for all integers k
n
a
n
k
 2
k m
b
k
k m
n
n
k m
k m
 k 1 2  k 1
n
 k 1 2 k 1
k m
n
 3k 1
k m

Examples
• Let ak = k +1 and bk = k – 1 for all integers k
 n   n   n
  n

 ak   bk   (k 1)  (k 1)
k m  k m  k m
 k m

n
(k 1)(k 1)   k
k m

2
1
Transforming a Sum by Change of
Variable
• Transform the following by changing the
variable.
6
– summation:  k 1
1
k 0
– change of variable: j = k+1
– Solution:

• compute the new limits:
– lower: j=k+1, j=0+1=1
– upper: j=k+1, j=6+1=7
7
7
1
1

 ( j 1) 1  j
j 1
j 1
```