The document provides an overview of sequences and the limit of a sequence. It defines an infinite sequence as a list of numbers in a definite order, with an denoting the nth term. The limit of a sequence is the number L such that the terms an can be made arbitrarily close to L by taking a sufficiently large n. Several properties of convergent sequences are described, including limit laws and the squeeze theorem. Special types of divergence like divergence to infinity are also covered. Examples are provided to illustrate key concepts.

1. Chapter 11: Sequences, Series, and Power Series
Section 11.1: Sequences
Alea Wittig
SUNY Albany

2. Outline
Sequences
The Limit of a Sequence
Properties of Convergent Sequences
Limit of an = f (n)
Limit Laws
Power Law
Squeeze Theorem
Absolute Values
Limit of f (an)
Limit of rn
Monotonic and Bounded Sequences

3. Sequences

4. ▶ An infinite sequence, or just a sequence, can be thought of
as a list of numbers in a definite order.
a1, a2, . . . , an, . . .
▶ a1 is the first term, a2 the second term, and in general an is
the nth term.
▶ {a1, a2, . . . , an, . . .} is also denoted by {an} or {an}∞
n=1.
▶ Unless otherwise stated we assume the sequence starts at
n = 1.
▶ A sequence can be viewed as a function whose domain is the
set of positive integers n = 1, 2, 3, . . . where f (n) = an
▶ Some sequences can be defined by giving a formula for the nth
term. Others don’t have a simple definining formula.

5. Example 1
(a) We can write out the first few terms of the sequence defined
by an = 1
2n as follows
a1 =
1
2
a2 =
1
22
=
1
4
a3 =
1
23
=
1
8
a4 =
1
24
=
1
16
This sequence can be denoted by any of the following notation:
n 1
2n
o
,
n 1
2n
o∞
n=1
, an =
1
2n
,
n1
2
,
1
4
,
1
8
,
1
16
,
1
32
, . . . ,
1
2n
, . . .
o

6. Example 1
(b) The notation
n n
n + 1
o∞
n=2
indicates that an =
n
n + 1
and we
start at n = 2
n n
n + 1
o∞
n=2
=
n2
3
,
3
4
,
4
5
, . . .
o
Notice that this same sequence could be written as
nn + 1
n + 2
o∞
n=1
=
n2
3
,
3
4
,
4
5
, . . .
o
ie, with an =
n + 1
n + 2
starting at n = 1.

7. Example 1
(c) The sequence {
√
3,
√
4,
√
5, . . .} can be denoted by
▶ {
√
n}∞
n=3
▶ an =
√
n, n ≥ 3
▶ {
√
n + 2}∞
n=1
▶ {
√
n + 2}
▶ an =
√
n + 2
There are often multiple ways to denote the same sequence of
numbers.

8. Example 1
(d) The definition
n
(−1)n n + 1
3n
o∞
n=0
generates the sequence
n
1, −
2
3
,
3
9
, −
4
27
,
5
81
, . . .
o
▶ You will often see factors of the form (−1)n and (−1)n+1,
(−1)n−1, etc.
▶ (−1)n = −1 for odd n and (−1)n = 1 for even n, so these
sequences will often alternate between negative and positive
values.

9. Example 2
Find a formula for the nth term of the sequence
n3
5
, −
4
25
,
5
125
, −
6
625
,
7
3125
, . . .
o
▶ Numerator starts at 3 and goes up by 1.
▶ We want the sequence to start at n = 1 and the first term has
numerator 3.
▶ In general, the nth
term has numerator n + 2.
▶ The denominators are powers of 5.
▶ The denominator of the first term is 51
, the second is 52
, . . ..
▶ In general, the nth
term has denominator 5n
.
▶ The signs are alternating so we multiply by a power of (−1)?.
▶ a1 is positive so we use (−1)n−1 or (−1)n+1.
an = (−1)n+1 n + 2
5n

10. Example 3
Some sequences don’t have simple defining formulas
(a) The sequence {pn} where pn is the population of the world
starting January 1 in the year n.
(b) Let an be the nth decimal place of the number e so
{7, 1, 8, 2, 8, 1, 8, 2, 8, 4, 5, . . .}
(c) The Fibonacci sequence is an example of a recursive
sequence. A recursive sequence is defined in terms of
preceding terms.
f1 = 1, f2 = 1, fn = fn−1 + fn−2 n ≥ 3
{1, 1, 2, 3, 5, 8, 13, 21, . . .}
This sequence is named after 13th century Italian
mathematician Fibonacci who was solving a problem
concerning the breeding of rabbits.

11. The Limit of a Sequence

12. The Limit of a Sequence
▶ We can view an = n
n+1 as a
function whose domain is
the set of positive integers.
▶ Its graph consists of
discrete points with
coordinates (1, a1), (2, a2),
(3, a3), etc.
▶ The terms of the sequence are strictly increasing,
ie, an < an+1 for all n
▶ And the terms of the sequence are all below 1,
ie, an < 1 for all n.
▶ What do you think the limit of the sequence is?

13. Limit of a Sequence
▶ Since the terms are strictly increasing and bounded above by 1
we might have an intuition that the limit is 1.
▶ The distance between an and 1 is 1
n+1 :
|1 − an| = 1 −
n
n + 1
=
1
n + 1
▶ We can make this distance as small as we like by taking
sufficiently large n.
▶ The Archimedean property of the real numbers means that for
any (arbitrarily small) positive real number ϵ > 0, there exists
a positive integer N large enough that 0 < 1
N+1 < ϵ.
lim
n→∞
n
n + 1
= lim
n→∞
1 +
1
n + 1
= 1 + lim
n→∞
1
n + 1
= 1

14. Intuitive Definition of Limit
A sequence {an} has the limit L and we write
lim
n→∞
an = L or an → L as n → ∞
if we can make the terms an as close to L as we like, by taking n
sufficiently large. If lim
n→∞
an exists then we say the sequence
converges (or is convergent). Otherwise we say the sequences
diverges (or is divergent).
Both of these sequences are convergent.

15. Precise Definition of Limit
A sequence {an} has the limit L and we write
lim
n→∞
an = L or an → L as n → ∞
if for every ϵ > 0 there is a corresponding integer N such that
if n > N then |an − L| < ϵ
A sequence diverges if its terms do not approach a single number.

16. Divergent Sequence
(a) Diverges because its terms oscillate between two different
numbers and do not approach a single number as n → ∞.
(b) Diverges because it increases without bound as n becomes
larger. In this case we say lim
n→∞
an = ∞ to indicate this
particular manner of divergence.

17. Precise Definition of Infinite Limit
▶ The notation
lim
n→∞
an = ∞
means that for every positive number M, there is an integer N
such that if n > N, then an > M.
eg, an = en
Let M be the largest positive number M you can think of.
en
> M ⇐⇒ n > ln M so take N = ln M
If n > N, then an > M
lim
n→∞
en
= ∞

18. Precise Definition of Infinite Limit
▶ Similarly we say
lim
n→∞
an = −∞
if that for every positive number M, there is an integer N such
that if n > N, then an < −M
eg, an = −n2
Let M be the largest positive number you can think of.
−n2
< −M ⇐⇒ n2
> M ⇐⇒ n >
√
M, so take N =
√
M
n > N =⇒ an < −M
lim
n→∞
−n2
= −∞

19. Properties of Convergent Sequences

20. Limit of an = f (n)

21. Limit of an = f (n)
▶ The only difference between the limit of a function
lim
x→∞
f (x) = L and lim
n→∞
an = L is that n is required to be an
integer.
▶ So all of the properties we have learned about limits of
functions at ∞ carry over to limits of infinite sequences.

22. Theorem 4
Theorem 4: If lim
x→∞
f (x) = L, and fn = an when n is an
integer, then lim
n→∞
an = L.
eg, an = 1
nr where r > 0
We know lim
x→∞
1
xr
= 0 where r > 0 from Calc I. So by
Theorem 4,
lim
n→∞
1
nr
= 0 where r > 0 (5)

23. Example 5
1/2
Calculate
lim
n→∞
ln n
n

24. Example 5
2/2
▶ Since lim
n→∞
ln n = ∞ and lim
n→∞
n = ∞ we have an
indeterminate form ∞
∞ .
▶ Apply L’Hospitals rule to the related function
f (x) =
ln x
x
lim
x→∞
ln x
x
= lim
x→∞
1
x
1
= lim
x→∞
1
x
= 0
▶ Thus by Theorem 4
lim
n→∞
ln n
n
= 0

25. Limit Laws

26. Limit Laws for Sequences
Suppose {an} and {bn} are convergent sequences, and let c be a
constant.
1. Sum Law :
lim
n→∞
an + bn = lim
n→∞
an + lim
n→∞
bn
2. Difference Law
lim
n→∞
an − bn = lim
n→∞
an − lim
n→∞
bn
3. Constant Multiple Law
lim
n→∞
can = c lim
n→∞
an
4. Product Law
lim
n→∞
anbn = lim
n→∞
an · lim
n→∞
bn
5. Quotient Law
lim
n→∞
an
bn
=
lim
n→∞
an
lim
n→∞
bn
h
if lim bn ̸= 0
i

27. Example 4
1/2
Find
lim
n→∞
n
n + 1
using the limit laws.
Hint: Divide top and bottom by highest power of n in denominator.
This method is found in section 2.6.

28. Example 4
2/2
▶ The highest power in the denominator is 1.
▶ So we divide top and bottom by n.
lim
n→∞
n
n + 1
= lim
n→∞
n
n
n
n + 1
n
= lim
n→∞
1
1 + 1
n
=
lim
n→∞
1
lim
n→∞
1 + lim
n→∞
1
n
=
1
1 + 0
= 1

29. Example 5
Is the sequence
an =
n
√
10 + n
convergent or divergent?
Try this on your own before going ahead for the solution.

30. Example 5
2/2
▶ The highest power in denominator is
√
n.
lim
n→∞
n
√
10 + n
= lim
n→∞
n
√
n
q
10
n + n
n
= lim
n→∞
√
n
q
10
n + 1
=
lim
n→∞
√
n
r
lim
n→∞
10
n
+ lim
n→∞
1
=
lim
n→∞
√
n
r
lim
n→∞
10
n
+ lim
n→∞
1
=
lim
n→∞
√
n
0 + 1
= ∞ so it is divergent.

31. Power Law

32. Power Law
Power Law:
lim
n→∞
ap
n =
h
lim
n→∞
an
ip
if p > 0 and an > 0
eg,
n n
n + 1
5o
lim
n→∞
ap
n = lim
n→∞
n
n + 1
5
=
lim
n→∞
n
n + 1
5
= 15
by example 5
= 1

33. Squeeze Theorem

34. Squeeze Theorem for Sequences
If an ≤ bn ≤ cn for all n ≥ n0 where n0 is some integer, and
lim
n→∞
an = lim
n→∞
cn = L
then lim
n→∞
bn = L.

35. Example 6
1/2
an =
n!
nn
▶ n! is called the factorial of n, we read as n factorial.
n! = 1 · 2 · 3 · . . . (n − 1) · n
eg, 5! = 1 · 2 · 3 · 4 · 5 = 120
3! = 1 · 2 · 3 = 6
▶ Now we can write the nth term of the sequence
an =
n!
nn
=
1 · 2 · 3 · . . . · n
n · n · n · . . . · n
=
1
n
2
n
|{z}
≤1
·
3
n
|{z}
≤1
. . .
n
n
|{z}
≤1

36. Example 6
2/2
▶ Since each factor in the parenthesis is ≤ 1, the whole product
in parenthesis is ≤ 1
an =
1
n
2 · 3 · . . .
n · n · . . .
!
| {z }
≤1
=⇒ an ≤
1
n
▶ So 0 an ≤ 1
n and we know that 1
n → 0 as n → ∞ so an → 0
by Squeeze theorem.

37. Absolute Values

38. Theorem 6
Theorem 6: If the limit of the absolute value of a series is 0
then the limit of the series is also 0.
lim
n→∞
|an| = 0 =⇒ lim
n→∞
an = 0 (6)
eg, an = (−1)n
n
lim
n→∞
|an| = lim
n→∞

41. (−1)n
n

44. = lim
n→∞
1
n
= 0 by example 4
=⇒ lim
n→∞
an = lim
n→∞
(−1)n
n
= 0

45. ▶ The graph of an = (−1)n
n is shown below.
▶ Notice how the terms jump above and below the x−axis
successively, but at each step the distance to the x−axis is
smaller.

46. Example 7
1/2
Is the sequence an = (−1)n convergent or divergent?

47. Example 7
2/2
{(−1)n
} = {−1, 1, −1, 1, −1, . . .}
▶ This sequence oscillates and does not get closer to any single
number therefore it is not convergent by definition of
convergence, so it is divergent.

48. Limit of f (an)

49. Theorem 7 - Limit of f (an)
If lim
n→∞
an = L and the function f is continuous at L, then
lim
n→∞
f (an) = f (L)
eg an = sin π
n
Since π
n → 0 as n → ∞, and sin x is continuous at x = 0:
lim
n→∞
sin
π
n
= sin
lim
n→∞
π
n
= sin 0 = 0

50. Limit of rn

51. Example 8
1/2
For what values of r is the sequence {rn} convergent?

52. Example 8
2/2
▶ From the graphs of exponential functions we know
▶ lim
x→∞
bx
= ∞ if b 1
▶ lim
x→∞
bx
= 0 if 0 b 1.
▶ Let b = r and using Theorem 4
lim
n→∞
rn
=
(
∞ if r 1
0 if 0 r 1
▶ Obviously lim
n→∞
1n
= 1 and lim
n→∞
0n
= 0. If −1 r 0 then
0 |r| 1 so
lim
n→∞
|rn
| = lim
n→∞
|r|n
= 0
▶ The other values of r : r ≤ −1 will oscillate between positive
and negative. And in each case it will be divergent.

53. Theorem 9 - Limit of rn
The sequence {rn} is convergent if −1 r ≤ 1 and divergent for
all other values of r
lim
n→∞
(
0 if −1 r 1
1 if r = 1

54. Monotonic and Bounded Sequences

55. Monotonic Sequence Definition
A sequence {an} is called increasing if an an+1 for all n ≥ 1, ie,
a1 a2 a3 . . .
A sequence {an} is called decreasing if an an+1 for all n ≥ 1, ie,
a1 a2 a3 . . .
A sequence is called monotonic if it is either increasing or
decreasing.

56. Bounded Sequence Definition
A sequence {an} is bounded above if there is a number M such
that
an ≤ M for all n ≥ 1
A sequence {an} is bounded below if there is a number m such
that
m ≤ an for all n ≥ 1
If a sequence is bounded above and below, then it is called a
bounded sequence (or just bounded.)

57. Example 9
(a) an = n is bounded below but not above.
It is monotonic (increasing) and is divergent lim
n→∞
n = ∞.
(b) an = (−1)n is bounded, but not monotonic.
It is not convergent.
(c) an = n
n+1 is bounded below by 0 and above by 1 so it is
bounded.
It is also monotonic (increasing) and converges to 1.

58. Example 10
1/3
Show that the following two sequences are decreasing
(a) { 3
n+5}
an an+1 ⇐⇒
3
n + 5
3
(n + 1) + 5
=
3
n + 6
⇐⇒
n + 6 n + 5✓ (cross multiply)
So an is decreasing.

59. Example 10
2/3
(b) an = n
n2+1
an an+1 ⇐⇒
n
n2 + 1
n + 1
(n + 1)2 + 1
⇐⇒
n((n + 1)2
+ 1) (n + 1)(n2
+ 1) ⇐⇒
n3
+ 2n2
+ 2n n3
+ n2
+ n + 1 ⇐⇒
n3
+ 2n2
+ 2n n3
+ n2
+ n + 1 ⇐⇒
n2
+ 2 1✓ (true for n ≥ 1)
So an is decreasing.

60. Example 10
3/3
(b) We can also see that an is decreasing by taking derivative of
associated function f (x) = x
x2+1
f ′
(x) =
(x2 + 1) · 1 − 2x · x
(x2 + 1)2
=
x2 − 2x2 + 1
(x2 + 1)2
=
−x2 + 1
(x2 + 1)2
0 ⇐⇒
1 − x2
0 ⇐⇒
1 x2
✓
So f (x) is decreasing on (1, ∞) and thus
f (n) = an f (n + 1) = an+1 for n ≥ 1.

61. Monotonic Sequence Theorem
▶ Every bounded, monotonic sequence is convergent.
▶ In particular,
▶ a sequence that is increasing and bounded above converges,
▶ a sequence that is decreasing and bounded below converges.
Proof : Completeness axiom says that if a set of real numbers has
an upper bound then it has a least upper bound. If it has a lower
bound then it has a greatest lower bound.

62. Example 11
1/4
Investigate the recurrence relation
a1 = 2 an+1 =
1
2
(an + 6)
a1 = 2 a2 = 4 a3 = 5
a4 = 11
2 a5 = 23
4 a6 = 47
8
▶ Notice that the values we have computed are increasing but
never exceed 6.
▶ We can prove our hunch that an is increasing using the
principle of mathematical induction.

63. Principle of Mathematical Induction
Principal of Mathematical Induction
Let Sn be a statement about the positive integer n. Suppose that
(1) S1 is true.
(2) Sk+1 is true whenever Sk is true.
Then Sn is true for all positive integers n.

64. Example 11
2/4
▶ Let Sn be the statement that an+1 an for integer n.
▶ The statement S1, that a2 a1, is clearly true.
▶ We assume Sk is true, ie, ak+1 ak and show that this
implies Sk+1 is true as well:
a1+k ak =⇒
a1+k + 6 ak + 6 =⇒
1
2
(a1+k + 6)
1
2
(ak + 6) =⇒
ak+2 ak+1
▶ Thus by induction we have that the sequence is increasing for
all n, ie, we proved Sn.

65. Example 11
3/4
▶ Now we can prove an it is bounded above by 6 by induction as
well.
▶ Let Sn be the statement an 6. S1, ie, a1 6 is clearly true.
▶ We assume Sk (ak 6) is true and show that this implies
Sk+1 (ak+1 6) is true as well.
ak 6 =⇒
ak + 6 12 =⇒
1
2
(ak + 6)
1
2
(12) = 6 =⇒
ak+1 6
▶ Therefore an is bounded above by 6.

66. Example 11
4/4
By Monotonic Sequence Theorem, an has a limit L.
lim
n→∞
an+1 = lim
n→∞
1
2
(an + 6) =
1
2
lim
n→∞
an + 6
= lim
n→∞
1
2
(L + 6)
since lim
n→∞
an+1 = lim
n→∞
an =⇒
L =
1
2
(L + 6) we have L = 6