The document defines sequences and provides examples of convergent and non-convergent sequences. It states that a sequence is an ordered list of infinitely many numbers that may or may not have a pattern. Examples are provided of sequences with patterns like odd numbers and square numbers, and sequences without obvious patterns. It further explains that a sequence converges if the limit of its terms is finite, and provides examples of sequences that converge, like terms approaching 0, -1, and 1. Criteria for determining if a sequence converges is given in terms of bounding all terms within any interval of the limit.

1. A sequence is an ordered list of infinitely many
numbers that may or may not have a pattern.
Sequences

2. A sequence is an ordered list of infinitely many
numbers that may or may not have a pattern.
Example A.
1, 3, 5, 7, 9,… is the sequence of odd numbers.
Sequences

3. A sequence is an ordered list of infinitely many
numbers that may or may not have a pattern.
Example A.
1, 3, 5, 7, 9,… is the sequence of odd numbers.
1, 4, 9, 16, 25, …is the sequence of square numbers.
Sequences

4. A sequence is an ordered list of infinitely many
numbers that may or may not have a pattern.
Example A.
1, 3, 5, 7, 9,… is the sequence of odd numbers.
1, 4, 9, 16, 25, …is the sequence of square numbers.
5, –2, , e2, –110, …is a sequence without an
obvious pattern.
Sequences

5. A sequence is an ordered list of infinitely many
numbers that may or may not have a pattern.
Example A.
1, 3, 5, 7, 9,… is the sequence of odd numbers.
1, 4, 9, 16, 25, …is the sequence of square numbers.
5, –2, , e2, –110, …is a sequence without an
obvious pattern.
Sequences
One way to describe a sequence is to give a formula
a(n) for its terms and list the sequence as {a(n)}n=1.
∞

6. A sequence is an ordered list of infinitely many
numbers that may or may not have a pattern.
Example A.
1, 3, 5, 7, 9,… is the sequence of odd numbers.
1, 4, 9, 16, 25, …is the sequence of square numbers.
5, –2, , e2, –110, …is a sequence without an
obvious pattern.
Sequences
One way to describe a sequence is to give a formula
a(n) for its terms and list the sequence as {a(n)}n=1.
We also write a(n) as an.
∞

7. A sequence is an ordered list of infinitely many
numbers that may or may not have a pattern.
Example A.
1, 3, 5, 7, 9,… is the sequence of odd numbers.
1, 4, 9, 16, 25, …is the sequence of square numbers.
5, –2, , e2, –110, …is a sequence without an
obvious pattern.
Sequences
One way to describe a sequence is to give a formula
a(n) for its terms and list the sequence as {a(n)}n=1.
We also write a(n) as an.
∞
Example B.
a. The sequence {an = 3n + 1}n=1 = {4, 7, 10, …}
with a1 = 4, a2 = 7, a3 = 10, …

8. A sequence is an ordered list of infinitely many
numbers that may or may not have a pattern.
Example A.
1, 3, 5, 7, 9,… is the sequence of odd numbers.
1, 4, 9, 16, 25, …is the sequence of square numbers.
5, –2, , e2, –110, …is a sequence without an
obvious pattern.
Sequences
One way to describe a sequence is to give a formula
a(n) for its terms and list the sequence as {a(n)}n=1.
We also write a(n) as an.
∞
Example B.
a. The sequence {an = 3n + 1}n=1 = {4, 7, 10, …}
with a1 = 4, a2 = 7, a3 = 10, …
From here on, it’s assumed
n → ∞ at the top

9. A sequence can have multiple representations.
b. The sequence 0, 0, 0, 0,.. may be listed as
{a(n) = 0}n=1 or as {sin(nπ)}n=1.
Example C. a. The sequence {(–1)n(2n – 1)}n=1
gives the alternating sequence –1, 3, –5, 7, –9, …
∞
b. Find a formula for, , , , ...with k = 0, 1, 2...
–4
9
2
3
6
27
–8
81
The formula 2(k + 1) gives the numerators,
3k+1 gives the denominators, and (–1)k switches
the ± signs so {(–1)k2(k+1)/ 3k+1}k=0 is one solution.∞
Sequences
The multiple factor (–1)n or (–1)n+1 alternate the signs
of the terms.
Sequences whose terms have alternating ± signs
are called alternating sequences.

10. Sequences
The sequence 1, –1/2, 1/3, –1/4.. → 0,
the sequence –2/1, –3/2, –4/3, –5/4,.. → –1,
and cos(1/2), cos(1/3), cos(1/4), .. → 1.
Convergent Sequences

11. Sequences
We say the sequence {an} converges (CG) or that
it’s a convergent (CG) sequence if lim an = L is finite.n∞
The sequence 1, –1/2, 1/3, –1/4.. → 0,
the sequence –2/1, –3/2, –4/3, –5/4,.. → –1,
and cos(1/2), cos(1/3), cos(1/4), .. → 1.
Convergent Sequences

12. Sequences
We say the sequence {an} converges (CG) or that
it’s a convergent (CG) sequence if lim an = L is finite.
All the above sequences are CG-sequences.
n∞
The sequence 1, –1/2, 1/3, –1/4.. → 0,
the sequence –2/1, –3/2, –4/3, –5/4,.. → –1,
and cos(1/2), cos(1/3), cos(1/4), .. → 1.
Convergent Sequences

13. Sequences
We say the sequence {an} converges (CG) or that
it’s a convergent (CG) sequence if lim an = L is finite.
All the above sequences are CG-sequences.
By lim an = L we mean that:n∞
n∞
The sequence 1, –1/2, 1/3, –1/4.. → 0,
the sequence –2/1, –3/2, –4/3, –5/4,.. → –1,
and cos(1/2), cos(1/3), cos(1/4), .. → 1.
Convergent Sequences

14. Sequences
We say the sequence {an} converges (CG) or that
it’s a convergent (CG) sequence if lim an = L is finite.
All the above sequences are CG-sequences.
L
No matter how small
n∞
By lim an = L we mean that: for every ϵ > 0,n∞
The sequence 1, –1/2, 1/3, –1/4.. → 0,
the sequence –2/1, –3/2, –4/3, –5/4,.. → –1,
and cos(1/2), cos(1/3), cos(1/4), .. → 1.
Convergent Sequences

15. Sequences
We say the sequence {an} converges (CG) or that
it’s a convergent (CG) sequence if lim an = L is finite.
All the above sequences are CG-sequences.
L
No matter how small
n∞
By lim an = L we mean that: for every ϵ > 0,
all except finitely many an’s, are inside the interval
(L – ϵ, L + ϵ),
n∞
The sequence 1, –1/2, 1/3, –1/4.. → 0,
the sequence –2/1, –3/2, –4/3, –5/4,.. → –1,
and cos(1/2), cos(1/3), cos(1/4), .. → 1.
Convergent Sequences

16. Sequences
We say the sequence {an} converges (CG) or that
it’s a convergent (CG) sequence if lim an = L is finite.
All the above sequences are CG-sequences.
LL–ϵ L+ϵ
No matter how small of an interval is
roped off around L,
n∞
By lim an = L we mean that: for every ϵ > 0,
all except finitely many an’s, are inside the interval
(L – ϵ, L + ϵ),
n∞
The sequence 1, –1/2, 1/3, –1/4.. → 0,
the sequence –2/1, –3/2, –4/3, –5/4,.. → –1,
and cos(1/2), cos(1/3), cos(1/4), .. → 1.
Convergent Sequences

17. Sequences
We say the sequence {an} converges (CG) or that
it’s a convergent (CG) sequence if lim an = L is finite.
All the above sequences are CG-sequences.
No matter how small of an interval is
roped off around L, all except finitely many
an’s, are inside the interval.
n∞
By lim an = L we mean that: for every ϵ > 0,
all except finitely many an’s, are inside the interval
(L – ϵ, L + ϵ),
n∞
The sequence 1, –1/2, 1/3, –1/4.. → 0,
the sequence –2/1, –3/2, –4/3, –5/4,.. → –1,
and cos(1/2), cos(1/3), cos(1/4), .. → 1.
Convergent Sequences
Lan’s L–ϵ L+ϵ an’s

18. Sequences
We say the sequence {an} converges (CG) or that
it’s a convergent (CG) sequence if lim an = L is finite.
All the above sequences are CG-sequences.
By lim an = L we mean that: for every ϵ > 0,
all except finitely many an’s, are inside the interval
(L – ϵ, L + ϵ),
Lan’s L–ϵ L+ϵ an’s
n∞
No matter how small of an interval is
roped off around L, all except finitely many
an’s, are inside the interval.
Finitely many an’s are outside
n∞
The sequence 1, –1/2, 1/3, –1/4.. → 0,
the sequence –2/1, –3/2, –4/3, –5/4,.. → –1,
and cos(1/2), cos(1/3), cos(1/4), .. → 1.
Convergent Sequences

19. Sequences
We say the sequence {an} converges (CG) or that
it’s a convergent (CG) sequence if lim an = L is finite.
All the above sequences are CG-sequences.
The sequence 1, –1/2, 1/3, –1/4.. → 0,
the sequence –2/1, –3/2, –4/3, –5/4,.. → –1,
and cos(1/2), cos(1/3), cos(1/4), .. → 1.
n∞
By lim an = L we mean that: for every ϵ > 0,
all except finitely many an’s, are inside the interval
(L – ϵ, L + ϵ),
which is equivalent to
saying that:
“for sufficiently large n,
we have l an – L l < ϵ”.
n∞
Convergent Sequences
Lan’s L–ϵ L+ϵ an’s
No matter how small of an interval is
roped off around L, all except finitely many
an’s, are inside the interval.
Finitely many an’s are outside

20. Sequences
Convergent Sequences

21. Sequences
The sequence of numbers 1/2, 2/3, 3/4, 4/5,..
are defined by {an= n/(n+1)}n=1 and they correspond
to the points on the graph of y = x/(x + 1) as shown.
∞
y = 1
y = x/(x + 1)(1,1/2)
(2,2/3)
(3,3/4) (4,4/5)
Convergent Sequences

22. Sequences
The sequence of numbers 1/2, 2/3, 3/4, 4/5,..
are defined by {an= n/(n+1)}n=1 and they correspond
to the points on the graph of y = x/(x + 1) as shown.
∞
y = 1
y = x/(x + 1)(1,1/2)
(2,2/3)
(3,3/4) (4,4/5)Since lim x/(x + 1) = 1 and
these points are on its graph,
we have lim an= 1 as n→∞.
x→∞
n∞
Convergent Sequences

23. Sequences
The sequence of numbers 1/2, 2/3, 3/4, 4/5,..
are defined by {an= n/(n+1)}n=1 and they correspond
to the points on the graph of y = x/(x + 1) as shown.
∞
y = 1
y = x/(x + 1)(1,1/2)
(2,2/3)
(3,3/4) (4,4/5)Since lim x/(x + 1) = 1 and
these points are on its graph,
we have lim an= 1 as n→∞.
x→∞
n∞
(Function-Sequence Convergent Theorem)
Given a sequence {fn} defined by y = f(x),
if lim f(x) = L is finite, then lim fn = L.x→∞
Convergent Sequences

24. Sequences
The sequence of numbers 1/2, 2/3, 3/4, 4/5,..
are defined by {an= n/(n+1)}n=1 and they correspond
to the points on the graph of y = x/(x + 1) as shown.
∞
y = 1
y = x/(x + 1)(1,1/2)
(2,2/3)
(3,3/4) (4,4/5)Since lim x/(x + 1) = 1 and
these points are on its graph,
we have lim an= 1 as n→∞.
x→∞
n∞
(Function-Sequence Convergent Theorem)
Given a sequence {fn} defined by y = f(x),
if lim f(x) = L is finite, then lim fn = L.
In the case that f(x) is differentiable, we may use the
L’Hopital Rule to see if f(x) converges; if the defining
function f(x) converges then the sequence converges.
x→∞
Convergent Sequences

25. Sequences
Example D. Is {an = n1/n = √n } a CG sequence?
Find lim an if it’s a CG sequence.
Lim x1/x = lim eLn(x)(1/x) = lim eLn(x)/x = e0 = 1 as x→∞ .
n∞
= 0 by L’Hopital Rule
Hence {an = n1/n } is a convergent sequence
and lim an = 1.
n
Let’s find the limit of the defining function x1/x.
Example E. a. Show that {an = Ln(n)/n1/2} →0.
Lim Ln(x)/x1/2 lim = lim 2x–1/2 = 0.x–1
½ x–1/2
Hence {an = Ln(n)/n1/2} is a CG sequence.
b. Show that {bn = n20/en} is a CG sequence.
L‘Hospital
Rule
Since [x20](21) = 0, so by applying the L‘Hospital Rule
21 times we’ve x20/ex → 0 so that {n20/en} converges.
→

26. Sequences
Example D. Is {an = n1/n = √n } a CG sequence?
Find lim an if it’s a CG sequence.
n

27. Sequences
Example D. Is {an = n1/n = √n } a CG sequence?
Find lim an if it’s a CG sequence.
n
Let’s find the limit of the defining function x1/x.

28. Sequences
Example D. Is {an = n1/n = √n } a CG sequence?
Find lim an if it’s a CG sequence.
Lim x1/x = lim eLn(x)(1/x)
n
Let’s find the limit of the defining function x1/x.

29. Sequences
Example D. Is {an = n1/n = √n } a CG sequence?
Find lim an if it’s a CG sequence.
Lim x1/x = lim eLn(x)(1/x) = lim eLn(x)/x
n
Let’s find the limit of the defining function x1/x.

30. Sequences
Example D. Is {an = n1/n = √n } a CG sequence?
Find lim an if it’s a CG sequence.
Lim x1/x = lim eLn(x)(1/x) = lim eLn(x)/x
= 0 by L’Hopital Rule
n
Let’s find the limit of the defining function x1/x.

31. Sequences
Example D. Is {an = n1/n = √n } a CG sequence?
Find lim an if it’s a CG sequence.
Lim x1/x = lim eLn(x)(1/x) = lim eLn(x)/x = e0 = 1 as x→∞ .
= 0 by L’Hopital Rule
n
Let’s find the limit of the defining function x1/x.

32. Sequences
Example D. Is {an = n1/n = √n } a CG sequence?
Find lim an if it’s a CG sequence.
Lim x1/x = lim eLn(x)(1/x) = lim eLn(x)/x = e0 = 1 as x→∞ .
n∞
= 0 by L’Hopital Rule
Hence {an = n1/n } is a convergent sequence
and lim an = 1.
n
Let’s find the limit of the defining function x1/x.

33. Sequences
Example D. Is {an = n1/n = √n } a CG sequence?
Find lim an if it’s a CG sequence.
Lim x1/x = lim eLn(x)(1/x) = lim eLn(x)/x = e0 = 1 as x→∞ .
n∞
= 0 by L’Hopital Rule
Hence {an = n1/n } is a convergent sequence
and lim an = 1.
n
Let’s find the limit of the defining function x1/x.
Example E. a. Show that {an = Ln(n)/n1/2} →0.

34. Sequences
Example D. Is {an = n1/n = √n } a CG sequence?
Find lim an if it’s a CG sequence.
Lim x1/x = lim eLn(x)(1/x) = lim eLn(x)/x = e0 = 1 as x→∞ .
n∞
= 0 by L’Hopital Rule
Hence {an = n1/n } is a convergent sequence
and lim an = 1.
n
Let’s find the limit of the defining function x1/x.
Example E. a. Show that {an = Ln(n)/n1/2} →0.
Lim Ln(x)/x1/2 lim x–1
½ x–1/2
L‘Hospital
Rule
→

35. Sequences
Example D. Is {an = n1/n = √n } a CG sequence?
Find lim an if it’s a CG sequence.
Lim x1/x = lim eLn(x)(1/x) = lim eLn(x)/x = e0 = 1 as x→∞ .
n∞
= 0 by L’Hopital Rule
Hence {an = n1/n } is a convergent sequence
and lim an = 1.
n
Let’s find the limit of the defining function x1/x.
Example E. a. Show that {an = Ln(n)/n1/2} →0.
Lim Ln(x)/x1/2 lim = lim 2x–1/2 = 0.x–1
½ x–1/2
L‘Hospital
Rule
→

36. Sequences
Example D. Is {an = n1/n = √n } a CG sequence?
Find lim an if it’s a CG sequence.
Lim x1/x = lim eLn(x)(1/x) = lim eLn(x)/x = e0 = 1 as x→∞ .
n∞
= 0 by L’Hopital Rule
Hence {an = n1/n } is a convergent sequence
and lim an = 1.
n
Let’s find the limit of the defining function x1/x.
Example E. a. Show that {an = Ln(n)/n1/2} →0.
Lim Ln(x)/x1/2 lim = lim 2x–1/2 = 0.x–1
½ x–1/2
Hence {an = Ln(n)/n1/2} is a CG sequence.
L‘Hospital
Rule
→

37. Sequences
Example D. Is {an = n1/n = √n } a CG sequence?
Find lim an if it’s a CG sequence.
Lim x1/x = lim eLn(x)(1/x) = lim eLn(x)/x = e0 = 1 as x→∞ .
n∞
= 0 by L’Hopital Rule
Hence {an = n1/n } is a convergent sequence
and lim an = 1.
n
Let’s find the limit of the defining function x1/x.
Example E. a. Show that {an = Ln(n)/n1/2} →0.
Lim Ln(x)/x1/2 lim = lim 2x–1/2 = 0.x–1
½ x–1/2
Hence {an = Ln(n)/n1/2} is a CG sequence.
b. Show that {bn = n20/en} is a CG sequence.
L‘Hospital
Rule
→

38. Sequences
Example D. Is {an = n1/n = √n } a CG sequence?
Find lim an if it’s a CG sequence.
Lim x1/x = lim eLn(x)(1/x) = lim eLn(x)/x = e0 = 1 as x→∞ .
n∞
= 0 by L’Hopital Rule
Hence {an = n1/n } is a convergent sequence
and lim an = 1.
n
Let’s find the limit of the defining function x1/x.
Example E. a. Show that {an = Ln(n)/n1/2} →0.
Lim Ln(x)/x1/2 lim = lim 2x–1/2 = 0.x–1
½ x–1/2
Hence {an = Ln(n)/n1/2} is a CG sequence.
b. Show that {bn = n20/en} is a CG sequence.
L‘Hospital
Rule
Applying the L‘Hospital Rule and differentiate 21 times
we have [x20](21) = 0, so that {n20/en} → 0 converges.
→

39. Sequences
As x→∞, the functions Ln(x), xp (p > 0), or ex go to ∞,
but at different paces as shown in example E.

40. Sequences
As x→∞, the functions Ln(x), xp (p > 0), or ex go to ∞,
but at different paces as shown in example E.
L’Hopital Rule gives us the following general results,
in short, as x→∞:

41. Sequences
As x→∞, the functions Ln(x), xp (p > 0), or ex go to ∞,
but at different paces as shown in example E.
L’Hopital Rule gives us the following general results,
in short, as x→∞:
* Ln(x) goes to ∞ “slower” than xp

42. Sequences
As x→∞, the functions Ln(x), xp (p > 0), or ex go to ∞,
but at different paces as shown in example E.
L’Hopital Rule gives us the following general results,
in short, as x→∞:
* Ln(x) goes to ∞ “slower” than xp and
* xp goes to ∞ “slower” than ex.

43. Sequences
As x→∞, the functions Ln(x), xp (p > 0), or ex go to ∞,
but at different paces as shown in example E.
L’Hopital Rule gives us the following general results,
in short, as x→∞:
* Ln(x) goes to ∞ “slower” than xp and
* xp goes to ∞ “slower” than ex.
(Sin(x) and cos(x) are bounded between –1 and 1.)

44. Sequences
As x→∞, the functions Ln(x), xp (p > 0), or ex go to ∞,
but at different paces as shown in example E.
L’Hopital Rule gives us the following general results,
in short, as x→∞:
* Ln(x) goes to ∞ “slower” than xp and
* xp goes to ∞ “slower” than ex.
(Sin(x) and cos(x) are bounded between –1 and 1.)
Here is a useful theorem for justifying convergence.

45. Sequences
As x→∞, the functions Ln(x), xp (p > 0), or ex go to ∞,
but at different paces as shown in example E.
L’Hopital Rule gives us the following general results,
in short, as x→∞:
* Ln(x) goes to ∞ “slower” than xp and
* xp goes to ∞ “slower” than ex.
(Sin(x) and cos(x) are bounded between –1 and 1.)
n∞ n∞
(The Sandwich Theorem)
Let 0 ≤ an ≤ bn for all but finitely many n’s
and that lim bn = 0, then lim an = 0.
Here is a useful theorem for justifying convergence.

46. Sequences
As x→∞, the functions Ln(x), xp (p > 0), or ex go to ∞,
but at different paces as shown in example E.
L’Hopital Rule gives us the following general results,
in short, as x→∞:
* Ln(x) goes to ∞ “slower” than xp and
* xp goes to ∞ “slower” than ex.
(Sin(x) and cos(x) are bounded between –1 and 1.)
n∞ n∞
(The Sandwich Theorem)
Let 0 ≤ an ≤ bn for all but finitely many n’s
and that lim bn = 0, then lim an = 0.
The Sandwich Theorem offers another method in
justifying convergence besides by the L’Hopital Rule.
Here is a useful theorem for justifying convergence.

47. Sequences
Recall that n-factorial n! = n(n – 1)(n – 1) ..(2)(1)

48. Sequences
n∞
Example F. Justify that lim an = 10n/n! = 0
Recall that n-factorial n! = n(n – 1)(n – 1) ..(2)(1)

49. Sequences
n∞
Example F. Justify that lim an = 10n/n! = 0
Compare bn = {1011/n} to an = {10n/n!}.
Recall that n-factorial n! = n(n – 1)(n – 1) ..(2)(1)

50. Sequences
n∞
Example F. Justify that lim an = 10n/n! = 0
We will show that bn > an (> 0) for sufficiently large n,
Compare bn = {1011/n} to an = {10n/n!}.
and since bn0 we must have an0.
Recall that n-factorial n! = n(n – 1)(n – 1) ..(2)(1)

51. Sequences
n∞
Example F. Justify that lim an = 10n/n! = 0
We will show that bn > an (> 0) for sufficiently large n,
Compare bn = {1011/n} to an = {10n/n!}.
and since bn0 we must have an0.
Specifically we claim that bn > an for 10 < n = 11,12,..
Recall that n-factorial n! = n(n – 1)(n – 1) ..(2)(1)

52. Sequences
n∞
Example F. Justify that lim an = 10n/n! = 0
We will show that bn > an (> 0) for sufficiently large n,
For n > 10, note the following blocks in their products
10n
n! = 10 * 10….. 10*10*10*10…10
n * (n–1)…11*10* 9 * 8…..1
Compare bn = {1011/n} to an = {10n/n!}.
and since bn0 we must have an0.
Specifically we claim that bn > an for 10 < n = 11,12,.. :
Recall that n-factorial n! = n(n – 1)(n – 1) ..(2)(1)
an =

53. Sequences
n∞
Example F. Justify that lim an = 10n/n! = 0
We will show that bn > an (> 0) for sufficiently large n,
For n > 10, note the following blocks in their products
10n
n! = 10 * 10….. 10*10*10*10…10
n * (n–1)…11*10* 9 * 8…..1
1
Compare bn = {1011/n} to an = {10n/n!}.
and since bn0 we must have an0.
Specifically we claim that bn > an for 10 < n = 11,12,.. :
Recall that n-factorial n! = n(n – 1)(n – 1) ..(2)(1)
an =

54. Sequences
n∞
Example F. Justify that lim an = 10n/n! = 0
We will show that bn > an (> 0) for sufficiently large n,
For n > 10, note the following blocks in their products
10n
n! = 10 * 10….. 10*10*10*10…10
n * (n–1)…11*10* 9 * 8…..1
10101
Compare bn = {1011/n} to an = {10n/n!}.
and since bn0 we must have an0.
Specifically we claim that bn > an for 10 < n = 11,12,.. :
Recall that n-factorial n! = n(n – 1)(n – 1) ..(2)(1)
an =

55. Sequences
n∞
Example F. Justify that lim an = 10n/n! = 0
We will show that bn > an (> 0) for sufficiently large n,
For n > 10, note the following blocks in their products
10n
n! = 10 * 10….. 10*10*10*10…10
n * (n–1)…11*10* 9 * 8…..1
10101
< 1011
n
Compare bn = {1011/n} to an = {10n/n!}.
and since bn0 we must have an0.
Specifically we claim that bn > an for 10 < n = 11,12,.. :
Recall that n-factorial n! = n(n – 1)(n – 1) ..(2)(1)
an = = bn

56. Sequences
n∞
Example F. Justify that lim an = 10n/n! = 0
We will show that bn > an (> 0) for sufficiently large n,
For n > 10, note the following blocks in their products
10n
n! = 10 * 10….. 10*10*10*10…10
n * (n–1)…11*10* 9 * 8…..1
10101
< 1011
n
n∞
Compare bn = {1011/n} to an = {10n/n!}.
and since bn0 we must have an0.
Specifically we claim that bn > an for 10 < n = 11,12,.. :
0
Recall that n-factorial n! = n(n – 1)(n – 1) ..(2)(1)
an = = bn

57. Sequences
n∞
Example F. Justify that lim an = 10n/n! = 0
We will show that bn > an (> 0) for sufficiently large n,
For n > 10, note the following blocks in their products
10n
n! = 10 * 10….. 10*10*10*10…10
n * (n–1)…11*10* 9 * 8…..1
10101
< 1011
n
So for large n, bn = 1011/n > an = 10n/n!.
n∞
Compare bn = {1011/n} to an = {10n/n!}.
and since bn0 we must have an0.
Specifically we claim that bn > an for 10 < n = 11,12,.. :
0
Recall that n-factorial n! = n(n – 1)(n – 1) ..(2)(1)
an = = bn

58. Sequences
n∞
Example F. Justify that lim an = 10n/n! = 0
We will show that bn > an (> 0) for sufficiently large n,
For n > 10, note the following blocks in their products
10n
n! = 10 * 10….. 10*10*10*10…10
n * (n–1)…11*10* 9 * 8…..1
10101
< 1011
n
So for large n, bn = 1011/n > an = 10n/n!. Since bn0,
therefore lim10n/n! = 0 by the sandwich theorem.
n∞
Compare bn = {1011/n} to an = {10n/n!}.
and since bn0 we must have an0.
Specifically we claim that bn > an for 10 < n = 11,12,.. :
0
n∞
Recall that n-factorial n! = n(n – 1)(n – 1) ..(2)(1)
an = = bn

59. Sequences
Properties of CG Sequences

60. Sequences
Properties of CG Sequences
Let lim an = L and lim bn = K be CG sequences,
1. If c is a constant, {can} converges and lim can = cL.

61. Sequences
Properties of CG Sequences
Let lim an = L and lim bn = K be CG sequences,
1. If c is a constant, {can} converges and lim can = cL.
2. {an± bn}, {an bn} and {an/bn} converge

62. Sequences
Properties of CG Sequences
Let lim an = L and lim bn = K be CG sequences,
1. If c is a constant, {can} converges and lim can = cL.
2. {an± bn}, {an bn} and {an/bn} converge with
lim (an ± bn) = L ± K, lim anbn = (lim an)(lim bn) = LK,
and lim (an / bn) = (lim an) / (lim bn) = L / K if K ≠ 0.

63. Sequences
Properties of CG Sequences
Let lim an = L and lim bn = K be CG sequences,
1. If c is a constant, {can} converges and lim can = cL.
2. {an± bn}, {an bn} and {an/bn} converge with
lim (an ± bn) = L ± K, lim anbn = (lim an)(lim bn) = LK,
and lim (an / bn) = (lim an) / (lim bn) = L / K if K ≠ 0.
Example G. Given that {an = 2 + 1/n} → 2,
{bn = 3cos(1/n)} → 3 as n →∞, then

64. Sequences
Properties of CG Sequences
Let lim an = L and lim bn = K be CG sequences,
1. If c is a constant, {can} converges and lim can = cL.
2. {an± bn}, {an bn} and {an/bn} converge with
lim (an ± bn) = L ± K, lim anbn = (lim an)(lim bn) = LK,
and lim (an / bn) = (lim an) / (lim bn) = L / K if K ≠ 0.
Example G. Given that {an = 2 + 1/n} → 2,
{bn = 3cos(1/n)} → 3 as n →∞, then
* {10an = 10(2 + 1/n)} → 10(2) = 20

65. Sequences
Properties of CG Sequences
Let lim an = L and lim bn = K be CG sequences,
1. If c is a constant, {can} converges and lim can = cL.
2. {an± bn}, {an bn} and {an/bn} converge with
lim (an ± bn) = L ± K, lim anbn = (lim an)(lim bn) = LK,
and lim (an / bn) = (lim an) / (lim bn) = L / K if K ≠ 0.
Example G. Given that {an = 2 + 1/n} → 2,
{bn = 3cos(1/n)} → 3 as n →∞, then
* {10an = 10(2 + 1/n)} → 10(2) = 20
* {an * bn = (2 + 1/n) * 3cos(1/n)}

66. Sequences
Properties of CG Sequences
Let lim an = L and lim bn = K be CG sequences,
1. If c is a constant, {can} converges and lim can = cL.
2. {an± bn}, {an bn} and {an/bn} converge with
lim (an ± bn) = L ± K, lim anbn = (lim an)(lim bn) = LK,
and lim (an / bn) = (lim an) / (lim bn) = L / K if K ≠ 0.
Example G. Given that {an = 2 + 1/n} → 2,
{bn = 3cos(1/n)} → 3 as n →∞, then
* {10an = 10(2 + 1/n)} → 10(2) = 20
* {an * bn = (2 + 1/n) * 3cos(1/n)} → (2)(3) = 6

67. Sequences
Properties of CG Sequences
Let lim an = L and lim bn = K be CG sequences,
1. If c is a constant, {can} converges and lim can = cL.
2. {an± bn}, {an bn} and {an/bn} converge with
lim (an ± bn) = L ± K, lim anbn = (lim an)(lim bn) = LK,
and lim (an / bn) = (lim an) / (lim bn) = L / K if K ≠ 0.
Example G. Given that {an = 2 + 1/n} → 2,
{bn = 3cos(1/n)} → 3 as n →∞, then
* {10an = 10(2 + 1/n)} → 10(2) = 20
* {an * bn = (2 + 1/n) * 3cos(1/n)} → (2)(3) = 6
* {an / bn = (2 + 1/n) / 3cos(1/n)} → 2/3.

68. Sequences
A non-convergent sequence is a called
a divergent (DG) sequence.
Divergent Sequences

69. Sequences
A non-convergent sequence is a called
a divergent (DG) sequence. In other words,
a sequence {an} diverges (DG) if lim an is UDF.n∞
Divergent Sequences

70. Sequences
A non-convergent sequence is a called
a divergent (DG) sequence. In other words,
a sequence {an} diverges (DG) if lim an is UDF.n∞
Divergent Sequences
A nice sequence {fn} might be defined by a badly
behaved function f(x).

71. Sequences
A non-convergent sequence is a called
a divergent (DG) sequence. In other words,
a sequence {an} diverges (DG) if lim an is UDF.n∞
Divergent Sequences
A nice sequence {fn} might be defined by a badly
behaved function f(x).
Example G.
The sequence 0, 0, 0,.. is defined by {fn= n*sin(n)}
which match the x-intercepts of
y = x*sin(x).

72. Sequences
A non-convergent sequence is a called
a divergent (DG) sequence. In other words,
a sequence {an} diverges (DG) if lim an is UDF.n∞
Divergent Sequences
A nice sequence {fn} might be defined by a badly
behaved function f(x).
Example G.
(1,0) (2,0) (3,0) (4,0)
y = x*sin(x)
The sequence 0, 0, 0,.. is defined by {fn= n*sin(n)}
which match the x-intercepts of
y = x*sin(x).

73. Sequences
A non-convergent sequence is a called
a divergent (DG) sequence. In other words,
a sequence {an} diverges (DG) if lim an is UDF.n∞
Divergent Sequences
A nice sequence {fn} might be defined by a badly
behaved function f(x).
Example G.
x→∞
(1,0) (2,0) (3,0) (4,0)
y = x*sin(x)
The sequence 0, 0, 0,.. is defined by {fn= n*sin(n)}
which match the x-intercepts of
y = x*sin(x).
but lim x*sin(x) is undefined.
Lim fn = 0 hence
{fn= n*sin(n)} converges,

74. Sequences
The convergence of the sequence does not imply the
convergence of the function because the sequence
corresponds to discrete points which contain a lot
less information than the graph itself.

75. Sequences
The convergence of the sequence does not imply the
convergence of the function because the sequence
corresponds to discrete points which contain a lot
less information than the graph itself.
Hence unlike the case for convergent sequences,
a divergent defining function f(x) is not enough to
justify that the sequence {fn} diverges.

76. Sequences
The convergence of the sequence does not imply the
convergence of the function because the sequence
corresponds to discrete points which contain a lot
less information than the graph itself.
Hence unlike the case for convergent sequences,
a divergent defining function f(x) is not enough to
justify that the sequence {fn} diverges.
We note the following two types of divergences,
the ones that go to ∞ and the ones that jump around

77. Sequences
The convergence of the sequence does not imply the
convergence of the function because the sequence
corresponds to discrete points which contain a lot
less information than the graph itself.
Hence unlike the case for convergent sequences,
a divergent defining function f(x) is not enough to
justify that the sequence {fn} diverges.
We note the following two types of divergences,
the ones that go to ∞ and the ones that jump around
* To-the-Infinity Divergence

78. Sequences
The convergence of the sequence does not imply the
convergence of the function because the sequence
corresponds to discrete points which contain a lot
less information than the graph itself.
Hence unlike the case for convergent sequences,
a divergent defining function f(x) is not enough to
justify that the sequence {fn} diverges.
We note the following two types of divergences,
the ones that go to ∞ and the ones that jump around
* To-the-Infinity Divergence
The function y = x → ∞ as x →∞,
so the sequence 1, 2, 3.. → +∞ as n →∞.

79. Sequences
The convergence of the sequence does not imply the
convergence of the function because the sequence
corresponds to discrete points which contain a lot
less information than the graph itself.
Hence unlike the case for convergent sequences,
a divergent defining function f(x) is not enough to
justify that the sequence {fn} diverges.
We note the following two types of divergences,
the ones that go to ∞ and the ones that jump around
* To-the-Infinity Divergence
The function y = x → ∞ as x →∞,
so the sequence 1, 2, 3.. → +∞ as n →∞.
The function y = –x2 → –∞ as x →∞,
so the sequence –1, –4, –9.. → –∞ as n →∞.

80. Sequences
(Divergence →±∞ )
If the defining function f(x) of fn goes to ±∞ as x →∞,
then lim fn → ±∞ as n →∞ and {fn} diverges.

81. Sequences
(Divergence →±∞ )
If the defining function f(x) of fn goes to ±∞ as x →∞,
then lim fn → ±∞ as n →∞ and {fn} diverges.
Specifically, by lim fn = ∞
we mean that for any
number C, we have that
fn > C for all except finitely
many fn’s as shown.

82. Sequences
(Divergence →±∞ )
If the defining function f(x) of fn goes to ±∞ as x →∞,
then lim fn → ±∞ as n →∞ and {fn} diverges.
Specifically, by lim fn = ∞
we mean that for any
number C, we have that
fn > C for all except finitely
many fn’s as shown.
C
(lim fn = ∞)
Given any number C,

83. Sequences
(Divergence →±∞ )
If the defining function f(x) of fn goes to ±∞ as x →∞,
then lim fn → ±∞ as n →∞ and {fn} diverges.
Specifically, by lim fn = ∞
we mean that for any
number C, we have that
fn > C for all except finitely
many fn’s as shown.
C
Given any number C,
all except finitely many fn’s, are > C.
(lim fn = ∞)

84. Sequences
(Divergence →±∞ )
If the defining function f(x) of fn goes to ±∞ as x →∞,
then lim fn → ±∞ as n →∞ and {fn} diverges.
Specifically, by lim fn = ∞
we mean that for any
number C, we have that
fn > C for all except finitely
many fn’s as shown.
C
Given any number C,
all except finitely many fn’s, are > C.
(lim fn = ∞)
a few fn’s < C

85. Sequences
(Divergence →±∞ )
If the defining function f(x) of fn goes to ±∞ as x →∞,
then lim fn → ±∞ as n →∞ and {fn} diverges.
We know the behavior of the terms of
a DG sequence that goes to →±∞. They blow up.
Specifically, by lim fn = ∞
we mean that for any
number C, we have that
fn > C for all except finitely
many fn’s as shown.
Ca few fn’s < C
Given any number C,
all except finitely many fn’s, are > C.
(lim fn = ∞)

86. Sequences
(Divergence →±∞ )
If the defining function f(x) of fn goes to ±∞ as x →∞,
then lim fn → ±∞ as n →∞ and {fn} diverges.
We know the behavior of the terms of
a DG sequence that goes to →±∞. They blow up.
Specifically, by lim fn = ∞
we mean that for any
number C, we have that
fn > C for all except finitely
many fn’s as shown.
Ca few fn’s < C
Given any number C,
all except finitely many fn’s, are > C.
(lim fn = ∞)
* Chaotic Divergence

87. Sequences
(Divergence →±∞ )
If the defining function f(x) of fn goes to ±∞ as x →∞,
then lim fn → ±∞ as n →∞ and {fn} diverges.
We know the behavior of the terms of
a DG sequence that goes to →±∞. They blow up.
Specifically, by lim fn = ∞
we mean that for any
number C, we have that
fn > C for all except finitely
many fn’s as shown.
Ca few fn’s < C
Given any number C,
all except finitely many fn’s, are > C.
(lim fn = ∞)
The sequence {sin(n)} = {sin(1)≈0.84, sin(2)≈0.90,
sin(3)≈0.14, ..} diverges because the terms jump
around between –1 and 1 without approaching a limit.
* Chaotic Divergence

88. Sequences
Note that for any number C between –1 and 1,
there are infinitely many sin(n)’s that are less than C
and there are also infinitely many sin(n)’s that are
more than C.

89. Sequences
Note that for any number C between –1 and 1,
there are infinitely many sin(n)’s that are less than C
and there are also infinitely many sin(n)’s that are
more than C. So unlike the case where lim fn = ∞,
we can’t predict the behavior of {sin(n)} as n →∞.

90. Sequences
Note that for any number C between –1 and 1,
there are infinitely many sin(n)’s that are less than C
and there are also infinitely many sin(n)’s that are
more than C. So unlike the case where lim fn = ∞,
we can’t predict the behavior of {sin(n)} as n →∞.
Hence {sin(n)} diverges chaotically.

91. Sequences
Note that for any number C between –1 and 1,
there are infinitely many sin(n)’s that are less than C
and there are also infinitely many sin(n)’s that are
more than C. So unlike the case where lim fn = ∞,
we can’t predict the behavior of {sin(n)} as n →∞.
Hence {sin(n)} diverges chaotically.
The oscillating behavior for chaotic divergence is
often caused by the periodic factors of sine or cosine
which equal to 0 periodically.

92. Sequences
Note that for any number C between –1 and 1,
there are infinitely many sin(n)’s that are less than C
and there are also infinitely many sin(n)’s that are
more than C. So unlike the case where lim fn = ∞,
we can’t predict the behavior of {sin(n)} as n →∞.
Hence {sin(n)} diverges chaotically.
The oscillating behavior for chaotic divergence is
often caused by the periodic factors of sine or cosine
which equal to 0 periodically.
Algebraic functions, Ln(x), ex, or xp (p > 0) are not
periodic so sequences defined by them converge,
or diverge to ±∞ (not chaotically).

93. For example,
{gn= n*sin(n/2)} = {1, 0, –3, 0, 5, 0, –7..}
does not converge so lim x*sin(x/2) must be UDF.
Sequences
Summary:
* a sequence {fn} inherits the “good” behaviors
such as lim f(x) = L or ±∞ of its defining function f(x).
* a function f(x) inherits the “bad” behavior of {fn}
so if lim f(n) doesn't exist then lim f(x) doesn't exist
For example, lim x/(x + 1) = 1 hence lim n/(n + 1) = 1
x→∞ n∞
* a CG sequence {fn} might be defined by
a non-convergent function f(x) as x →∞.
For example, the CG sequence 0, 0, 0.. is defined by
{fn= n*sin(n)} but y = x*sin(x) does not converge.