Infinite Sequences
A sequence of real numbers is a function 𝑓(𝑛), whose domain is the set of positive integers. The values 𝑎𝑛 =
𝑓(𝑛) taken by the function are called the terms of the sequence.
The set of values 𝑎𝑛 = 𝑓(𝑛) is denoted by {𝑎𝑛 }.
• A sequence {𝑎𝑛 } has the limit 𝐿 if for every 𝜀 > 0 there exists an integer 𝑁 > 0 such that if 𝑛 ≥
𝑁, then |𝑎𝑛 − 𝐿| ≤ 𝜀. In this case we write:
lim 𝑎𝑛 = 𝐿.
𝑛→∞
• The sequence {𝑎𝑛 } has the limit ∞ if for every positive number 𝑀 there is an integer 𝑁 > 0 such that
if 𝑛 ≥ 𝑁 then 𝑎𝑛 > 𝑀. In this case we write
lim 𝑎𝑛 = ∞.
𝑛→∞
If the limit lim 𝑎𝑛 = 𝐿 exists and 𝐿 is finite, we say that the sequence converges. Otherwise the
𝑛→∞
sequence diverges.
Example : Write a formula for the nth term of an of the sequence and determine its limit (if it exists).
1 2 3 4 5
, , , , ,…
3 4 5 6 7
𝑛
Here 𝑎𝑛 =
. Then the limit is
𝑛+2
𝑛
𝑛+2−2
2
lim
= lim
= lim 1 −
= 1.
𝑛→∞ 𝑛 + 2
𝑛→∞ 𝑛 + 2
𝑛→∞
𝑛+2
Thus, the sequence converges to 1.
Example : Does the sequence
As L’Hospital’s rule yields
𝑛2
2𝑛
converge or diverge?
𝑛2
2𝑛
2
lim
= lim 𝑛
= lim 𝑛 2 = 0.
𝑛→∞ 2𝑛
𝑛→∞ 2 𝑙𝑛2
𝑛→∞ 2 ln 2
Since the limit is finite, the given sequence converges.
Squeezing Theorem.
Suppose that lim 𝑎𝑛 = lim 𝑏𝑛 = 𝐿 and {𝑐𝑛 } is a sequence such that 𝑎𝑛 ≤ 𝑐𝑛 ≤ 𝑏𝑛 for all 𝑛 > 𝑁, where 𝑁 is
𝑛→∞
𝑛→∞
a positive integer. Then
lim 𝑐𝑛 = 𝐿.
𝑛→∞
Example: Write a formula for the nth term of an of the sequence and determine its limit (if it exists).
2 3 4 5
1, − , , − , , …
2 4 8 16
We easily can see that 𝑛th term of the sequence is given by the formula 𝑎𝑛 =
−1 𝑛−1 𝑛 ≤ 𝑛, we can write:
𝑛
−1 𝑛−1 𝑛
𝑛
− 𝑛−1 ≤
≤ 𝑛−1 .
2
2𝑛−1
2ถ
𝑎𝑛
𝑛
Since lim 𝑎𝑛 = lim − 𝑛−1 = 0, lim 𝑏𝑛 = lim
𝑛→∞
𝑛→∞
2
𝑛→∞
𝑐𝑛
𝑛
𝑏𝑛
= 0,
−1 𝑛−1 𝑛
lim 𝑐𝑛 = lim
= 0.
𝑛→∞
𝑛→∞
2𝑛−1
𝑛→∞ 2𝑛−1
−1 𝑛−1 𝑛
. Since −𝑛 ≤
2𝑛−1
Example: Determine the limit (if it exists) of the sequence
𝑎𝑛 =
𝑛
2 + 2−𝑛 .
Since 2 < 2 + 2−𝑛 ≤ 3, we can write:
𝑛
ด2 ≤
𝑛
𝑐𝑛
𝑎𝑛
1
𝑛
𝑛
2 + 2−𝑛 ≤ ด3 .
𝑏𝑛
1
𝑛
lim 𝑎𝑛 = lim 2 = 1, lim 𝑏𝑛 = lim 3 = 1, then
𝑛→∞
𝑛→∞
𝑛→∞
𝑛→∞
lim 𝑐𝑛 = lim
𝑛→∞
𝑛→∞
𝑛
2 + 2−𝑛 = 1.
Example: Determine the limit (if it exists) of the sequence
1
1
1
𝑎𝑛 =
+
+⋯
.
2
2
2
1+𝑛
2+𝑛
𝑛+𝑛
Since
1
1
1
1
<
,⋯,
<
2
2
2
2+𝑛
1+𝑛
𝑛+𝑛
1 + 𝑛2
and
1
1
1
1
<
,⋯,
<
𝑛 + 𝑛2
1 + 𝑛2
𝑛 + 𝑛2
𝑛 − 1 + 𝑛2
we can write:
𝑛
𝑛 + 𝑛2
≤
1
1 + 𝑛2
+
1
𝑎𝑛
lim 𝑎𝑛 = lim
𝑛→∞
𝑛
𝑛→∞ 𝑛+𝑛
𝑛
𝑛→∞ 1+𝑛2
lim 𝑐𝑛 = lim
𝑛→∞
𝑛 + 𝑛2
≤
𝑛
1 + 𝑛2
𝑐𝑛
= 1, lim 𝑏𝑛 = lim
2
𝑛→∞
+⋯
2 + 𝑛2
1
𝑛→∞
𝑏𝑛
= 1, then
1
1 + 𝑛2
+
1
2 + 𝑛2
+⋯
1
𝑛 + 𝑛2
= 1.
.
Definition: The sequence {𝑎𝑛 } is bounded if there is a number 𝑀 > 0 such that |𝑎𝑛 | ≤ 𝑀 for every positive 𝑛.
Theorem: Every convergent sequence is bounded.
Remark: The reverse of this theorem is not always true.
Example: Let consider the sequence 𝑎𝑛 = { −1 𝑛 }. Easy to seen that
• If 𝑛 is even 𝑎𝑛 = {1} and
lim 𝑎𝑛 = 1.
𝑛→∞
• If 𝑛 is odd 𝑎𝑛 = {−1} and
lim 𝑎𝑛 = −1.
𝑛→∞
This is a contradiction. If the limit exists, it must be unique.
Definition: The sequence {𝑎𝑛 } is monotone increasing if 𝑎𝑛 ≤ 𝑎𝑛+1 for every 𝑛 ≥ 1. Similarly, the
sequence {𝑎𝑛 } is called monotone decreasing if 𝑎𝑛 ≥ 𝑎𝑛+1 for every 𝑛 ≥ 1. The sequence {𝑎𝑛 } is
called monotonic if it is either monotone increasing or monotone decreasing.
Example: Determine whether the sequence
5𝑛−7
3𝑛+4
is increasing, decreasing, or neither.
The (𝑛 + 1)𝑡ℎ term of the sequence is given by the formula
5 𝑛 + 1 − 7 5𝑛 − 2
𝑎𝑛+1 =
=
.
3 𝑛 + 1 + 4 3𝑛 + 7
Check the inequality 𝑎𝑛 ≤ 𝑎𝑛+1 :
5𝑛 − 7 5𝑛 − 2
5𝑛 − 7 5𝑛 − 2
41
≤
,⇒
−
≤ 0, ⇒ −
≤ 0.
3𝑛 + 4 3𝑛 + 7
3𝑛 + 4 3𝑛 + 7
3𝑛 + 4 3𝑛 + 7
The last inequality is obvious, since the numerator is negative and 3𝑛 + 4 ≥ 0 and 3𝑛 + 7 ≥ 0 for 𝑛 ≥
1. Therefore, this sequence is increasing.
Example: Determine whether the sequence
We have
Then the condition
𝑎𝑛
𝑎𝑛+1
2𝑛 +3
2𝑛 +1
2𝑛 + 3
𝑎𝑛 = 𝑛
,
2 +1
implies that
is increasing, decreasing, or not monotonic.
2𝑛+1 + 3
𝑎𝑛+1 = 𝑛+1
.
2
+1
2𝑛 + 3
𝑛 + 3 2𝑛+1 + 1
𝑛 2𝑛+1 + 3 ⋅ 2𝑛+1 + 2𝑛 + 3
𝑛
2
2
2 +1 =
= 𝑛 𝑛+1
≥ 1.
2𝑛+1 + 3
2𝑛 + 1 2𝑛+1 + 3
2 2
+ 3 ⋅ 2𝑛 + 2𝑛+1 + 3
2𝑛+1 + 1
Hence
𝑎𝑛
≥ 1, i.e.
𝑎𝑛+1
𝑎𝑛 ≥ 𝑎𝑛+1 .
We can conclude that the sequence is decreasing.
Theorem: If {𝑎𝑛 } is bounded and monotonic then {𝑎𝑛 } is convergent.
Example: Consider the sequence recursively defined by the conditions
𝑎𝑛−1 + 1
𝑎1 = 7 𝑎𝑛𝑑 𝑎𝑛 =
𝑓𝑜𝑟 𝑛 ≥ 2 .
4
This sequence is bounded below by zero, since all terms are positive.
Now let us chek the monotonicity.
Claim 𝑎𝑛+1 < 𝑎𝑛 for all 𝑛 ≥ 1. We will use induction.
• For 𝑛 = 1, we have 𝑎2 =
7+1
= 2 < 7 = 𝑎1 .
4
• Suppose that 𝑎𝑘+1 < 𝑎𝑘 for some 𝑘 ≥ 1. We want to show that 𝑎𝑘+2 < 𝑎𝑘+1 . This is true, because
𝑎𝑘+1 + 1
𝑎𝑘 + 1
𝑎𝑘+1 < 𝑎𝑘 ⇒ 𝑎𝑘+1 + 1 < 𝑎𝑘 + 1 ⇒ 𝑎𝑘+2 =
<
= 𝑎𝑘+1 .
4
4
Finally, since the sequences is bounded from below and decreasing, it is convergent
Example: Determine whether the sequence
2𝑛 ∞
∞
𝑎𝑛 𝑛=1 =
convergent or divergent.
𝑛! 𝑛=1
𝑎𝑛 > 0 for all 𝑛 ≥ 1, i.e. bounded below by zero.
Now let us chek the monotonicity.
2𝑛
𝑎𝑛
2𝑛 𝑛 + 1 ! 𝑛 + 1
𝑛!
=
=
=
≥ 1.
2𝑛+1
𝑎𝑛+1
2𝑛+1 𝑛!
2
𝑛+1 !
Hence
𝑎𝑛
≥ 1, 𝑖. 𝑒. 𝑎𝑛 ≥ 𝑎𝑛+1 .
𝑎𝑛+1
Since the sequences is bounded from below and decreasing, it is convergent
Example: Determine whether the sequence
𝑛
𝑎𝑛 ∞
𝑛=1 =
1 ∞
𝑛
σ𝑘=1
convergent or divergent.
𝑛+𝑘 𝑛=1
1
1
1
1
1
1
1
𝑛
෍
=
+
+ ⋯+
<
+
+ ⋯+
=
<1
𝑛+𝑘 𝑛+1 𝑛+2
2𝑛 𝑛 + 1 𝑛 + 1
𝑛+1 𝑛+1
𝑘=1
𝑎𝑛 < 1 for all 𝑛 ≥ 1, i.e. bounded above by one.
Now let us chek the monotonicity.
𝑛+1
𝑛
𝑘=1
𝑘=1
1
1
𝑎𝑛+1 − 𝑎𝑛 = ෍
−෍
𝑛+1+𝑘
𝑛+𝑘
1
1
1
1
1
1
1
1
=
+
+ ⋯+
+
+
−
+
+ ⋯+
𝑛+2 𝑛+3
2𝑛 2𝑛 + 1 2𝑛 + 2
𝑛+1 𝑛+2
2𝑛
=
1
1
1
1
1
+
−
=
−
>0
2𝑛 + 1 2𝑛 + 2 𝑛 + 1 2𝑛 + 1 2𝑛 + 2
Hence
𝑎𝑛+1 − 𝑎𝑛 > 0 𝑖. 𝑒. 𝑎𝑛+1 ≥ 𝑎𝑛 .
Since the sequences is bounded from above and increasing, it is convergent.