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 conjectures about patterns governing the arrangement of terms in sequences is mathematical induction.

Example

• Ancestor counting with a sequence– two parents, four grandparents, eight great-

grandparents, 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

– the sequences a and b have the same terms and hence, are identical

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

Example

• Alternating Sequence– cj = (-1)j for all integers j≥0

– sequence has bound values for the term.– term {-1, 1}∈

c0 = (-1)0 = 1

c1 = (-1)1 = -1

c2 = (-1)2 = 1

c3 = (-1)3 = -1

c4 = (-1)4 = 1…

Example

• Find an explicit formula to fit given initial terms– sequence = 1, -1/4, 1/9, -1/16, 1/25, -1/36, …– What can we observe about this sequence?

• alternate in sign• numerator is always 1• denominator is a square

– 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

1/12 -1/22 1/32 -1/42 1/52 -1/62

a1 a2 a3 a4 a5 a6

Summation Notation

• Summation notation is used to create a compact form for summation sequences governed by a formula.

• 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.

€

ak∑k=1

n

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

Example

• Computing summations

– a1 = -2, a2 = -1, a3 = 0, a4 = 1, and a5 =2.

€

ak∑k=1

5

= a1 + a2 + a3 + a4 + a5

€

ak∑k=1

5

= −2 + −1+ 0 +1+ 2 = 0

Example

• Computing summation from sum form.

€

k 2

k=1

5

∑

€

k 2

k=1

5

∑ =12 + 22 + 32 + 42 + 52 = 55

Example

• Changing from Summation Notation to Expanded form

€

(−1)i

i +1i=0

n

∑

€

(−1)i

i +1i=0

n

∑ =(−1)0

0 +1+

(−1)1

1+1+

(−1)2

2 +1+ ...+

(−1)n

n +1

Example

• Changing from expanded to summation form.• Find a close form for the following:

€

1

n+

2

n +1+

3

n + 2+ ...+

n +1

2n

€

1

n+

2

n +1+

3

n + 2+ ...+

n +1

n + n=

k +1

n + kk=0

n

∑

Separating Off a Final Term

• A final term can be removed from the summation form as follows.

• Example of use: Rewrite the following separating the final term

€

akk=m

n

∑ = akk=m

n−1

∑ + an

€

1

k 2k=1

n

∑ =1

k 2k=1

n−1

∑ +1

n2

Example

• Combining final term

€

2k + 2n

k=0

n−1

∑

€

2k + 2n

k=0

n−1

∑ = 2k

k=0

n

∑

Telescoping Sum

• Telescoping sum can be evaluated to a closed form.

€

1

k−

1

k +1=

(k +1) − k

k(k +1)=

1

k(k +1)

€

1

k(k +1)k=1

n

∑ =1

k−

1

k +1

⎛

⎝ ⎜

⎞

⎠ ⎟

k=1

n

∑ =1

1−

1

2

⎛

⎝ ⎜

⎞

⎠ ⎟+

1

2−

1

3

⎛

⎝ ⎜

⎞

⎠ ⎟+

1

3−

1

4

⎛

⎝ ⎜

⎞

⎠ ⎟+ ...+

1

n −1−

1

n

⎛

⎝ ⎜

⎞

⎠ ⎟+

1

n−

1

n +1

⎛

⎝ ⎜

⎞

⎠ ⎟=1 −

1

n +1

Product Notation

€

akk=1

5

∏ = a1a2a3a4a5

€

akk=1

n

∏ = akk=1

n−1

∏ ⎛

⎝ ⎜

⎞

⎠ ⎟ an Recursive form

Example

• Compute the following products:

€

kk=1

5

∏ =1*2* 3* 4 *5 =120

k

k +1k=1

1

∏ =1

1+1=

1

2

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!

7!=

8* 7!

7!= 8

5!

2!*3!=

5* 4 * 3!

2!*3!=

5 * 4

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:

€

1. ak + bkk=m

n

∑k=m

n

∑ = ak +bk( )k=m

n

∑

2. c * akk=m

n

∑ = c * akk=m

n

∑

€

3. akk=m

n

∏ ⎛

⎝ ⎜

⎞

⎠ ⎟* bk

k=m

n

∏ ⎛

⎝ ⎜

⎞

⎠ ⎟= ak *bk( )

k=m

n

∏

Examples

• Let ak = k +1 and bk = k – 1 for all integers k

€

ak + 2⋅ bkk=m

n

∑k=m

n

∑

k +1+ 2⋅ k −1k=m

n

∑k=m

n

∑

k +1+ 2⋅ k −1( )k=m

n

∑

3k −1k=m

n

∑

Examples

• Let ak = k +1 and bk = k – 1 for all integers k

€

akk=m

n

∏ ⎛

⎝ ⎜

⎞

⎠ ⎟• bk

k=m

n

∏ ⎛

⎝ ⎜

⎞

⎠ ⎟= (k +1)

k=m

n

∏ ⎛

⎝ ⎜

⎞

⎠ ⎟• (k −1)

k=m

n

∏ ⎛

⎝ ⎜

⎞

⎠ ⎟

(k +1)(k −1) = k 2 −1k=m

n

∏∏

Transforming a Sum by Change of Variable

• Transform the following by changing the variable.– summation: – 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

€

1

k +1k=0

6

∑

€

1

( j −1) +1=

j=1

7

∑ 1

jj=1

7

∑