“Analytical Principles of Engineering”

WS 2020/21

3.2.4 Convergence Theorems:

Remark: The term “for almost all….” Will be used very often during this
section. It means that the relation not nesessrly have to be valid for all values
of 𝑛 as shown in the next example.

Example:

Consider the sequences 𝑎𝑛 = 𝑛3 and 𝑏𝑛 = 𝑛!. We will assume that

𝑎𝑛 ≤ 𝑏𝑛 is valid even though this is not the case for 𝑛 = 0 to 5 as shown in the
table:

n 𝒏𝟑 n!

0 0 1
1 1 1
2 8 2
3 27 6
4 64 24
5 125 120
6 216 720
7 343 5040
8 512 40320
9 729 362880
10 1000 3628800
11 1331 39916800
12 1728 479001600
13 2197 6227020800
14 2744 8,717829120E10
15 3375 1,307674368E12

Instructor: Ibrahim Bader 1

Theorem:

Let (𝑎𝑛 ) and (𝑏𝑛 ) be convergent sequences with: Lim (𝑎𝑛 ) = 𝑎 and
𝑛→∞

Lim (𝑏𝑛 ) = 𝑏, if (for almost all) 𝑛: 𝑎𝑛 ≤ 𝑏𝑛 , then 𝑎 ≤ 𝑏.

𝑛→∞

Theorem: (Squeeze od Sandwich Therorem)

Let (𝑎𝑛 ) and (𝑏𝑛 ) be convergent sequences with: Lim (𝑎𝑛 ) = 𝑎 and
𝑛→∞

Lim (𝑏𝑛 ) = 𝑎, if (for almost all) 𝑎𝑛 ≤ 𝑐𝑛 ≤ 𝑏𝑛 , then Lim (𝑐𝑛 ) = 𝑎.

𝑛→∞ 𝑛→∞

Note:

𝑎𝑛 ≤ 𝑐𝑛 ≤ 𝑏𝑛

↓ ↓ ↓

𝑎 𝒂 𝑎

Example:
(cos 𝑛)2
Compute Lim , where 𝑛 ≥ 1
𝑛→∞ 3𝑛

Since -1≤ cos 𝑛 ≤ 1 and (cos 𝑛)2 always ≥ 0, it follows 0 ≤ (cos 𝑛)2 ≤ 1.

Since 𝑛 ≥ 1, we could divide the inequality by 3𝑛 and to get

(cos 𝑛)2 1
0≤ ≤
3𝑛 3𝑛
1
Since Lim 0 = Lim = 0, applying the squeeze theorem will result in
𝑛→∞ 𝑛→∞ 3𝑛

(cos 𝑛)2
Lim = 0.
𝑛→∞ 3𝑛

Monotonic Sequences:

Definition: (Monotonic Sequences)

A sequence (𝑎𝑛 ) of real numbers is called:

Instructor: Ibrahim Bader 2

 1. Monotonic increasing ⟺ 𝑎𝑛+1 ≥ 𝑎𝑛 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑛.

Strict monotonic increasing ⟺ 𝑎𝑛+1 > 𝑎𝑛 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑛.

2. Monotonic decreasing ⟺ 𝑎𝑛+1 ≤ 𝑎𝑛 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑛.

Strict monotonic decreasing ⟺ 𝑎𝑛+1 < 𝑎𝑛 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑛.

3. Monotonic if it is either monotonic increasing or decreasing.

Definition: (Bounded Sequences)

A sequence (𝑎𝑛 ) of real numbers is called:

1. Bounded below if there exists an 𝑚 𝜖 𝑹 with 𝑚 ≤ 𝑎𝑛 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑛.

2. Bounded above if there exists an 𝑀 𝜖 𝑹 with 𝑎𝑛 ≤ 𝑀 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑛.

3. Bounded if bounded above and bounded below.

Example:
1
Let 𝑎𝑛 = , 𝑛≥1
𝑛

1 1 1 1 1 1
1, , , , , , ….
2 3 4 5 6 7
0 ≤ 𝑎𝑛 ≤ 1 bounded below by 0 and above by 1.

Definition: (Subsequence)

A subsequence is an infinite ordered subset of a sequence.

Theorem:

Any subsequence of a convergent sequence is convergent to the same limit.

Example:
1 1
Let 𝑎𝑛 = , the sequence 𝑏𝑛 = is a subsequence of 𝑎𝑛
𝑛 𝑛2

Instructor: Ibrahim Bader 3

 1 1
Lim ( ) = 0 ⟹ lim = 0.
𝑛→∞ 𝑛 𝑛→∞ 𝑛2

Theorem:

Any bounded sequence has a convergent subsequence.

Example:

The sequence 𝑎𝑛 = (−1)𝑛 is bounded: −1 ≤ 𝑎𝑛 ≤ 1, the subsequences (𝑏𝑛 )

and (𝑐𝑛 ) given by:

Let 𝑏𝑛 = (−1)2𝑛

(𝑏𝑛 ) = 1,1,1, ….

Let 𝑐𝑛 = (−1)2𝑛+1

(𝑐𝑛 ) = −1, −1, −1, ….

Both constant subsequences are convergent with:

Lim (𝑏𝑛 ) = 1 and Lim (𝑐𝑛 ) = −1

𝑛→∞ 𝑛→∞

Corollary:

If for a sequence (𝑎𝑛 ), there exists seubsequences (𝑏𝑛 ) and (𝑐𝑛 ) with

Lim (𝑏𝑛 ) = 𝑏 and Lim (𝑐𝑛 ) = 𝑐, with 𝑏 ≠ 𝑐, or if either (𝑏𝑛 ) or (𝑐𝑛 ) is not
𝑛→∞ 𝑛→∞
convergent, then the sequence (𝑎𝑛 ) is not convergent (diveregent).

Example:

The sequence 𝑎𝑛 = (−1)𝑛 is divergent, since the subsequences (𝑏𝑛 ) and (𝑐𝑛 )
from the previous example given by:

𝑏𝑛 = (−1)2𝑛

𝑐𝑛 = (−1)2𝑛+1

converege to +1 and −1 respectivelly. Since +1 ≠ −1, by the previous corollary

⟹ the sequence must be diverging.

Instructor: Ibrahim Bader 4

Theorem:

Any bounded and almost always monotonic sequence is convergent.

Example:
1
𝑎𝑛 = 1 +
𝑛
(𝑎𝑛 ) is bounded (1 ≤ 𝑎𝑛 ≤ 2) and monotonic decreasing ⟹ (𝑎𝑛 ) is convergent.

Theorem:

The seuence 𝑎𝑛 = 𝑟 𝑞 converges for −1 < 𝑟 ≤ 1 and diverges for all other values
of 𝑟. Further:
0 𝑓𝑜𝑟 − 1 < 𝑟 < 1
Lim (𝑟 𝑞 ) = {
𝑛→∞ 1 𝑓𝑜𝑟 𝑟 = 1

Theorem:

Every convergent sequence is bounded.

Note: Not all bounded sequences are convergent. For example 𝑎𝑛 = (−1)𝑛 is
bounded by -1 1nd 1, but it is not convergent.

Which of these sequences is bounded? If the sequence is bounded determine

the lower and upper bounds.

Answer 1. Bounded?
1 1
i. 𝑎𝑛 = Yes bounded by 0 ≤ ≤ 1 ∶= 𝑎1
𝑛 𝑛
2𝑛+1 1 2𝑛+1
ii. 𝑎𝑛 = ⟹ 𝑎𝑛 = 2 + Yes bounded 2 ≤ ≤ 3 ∶= 𝑎1
𝑛 𝑛 𝑛
𝑛−1 𝑛+1−2 2
iii. 𝑎𝑛 = ⟹ 𝑎𝑛 = =1− . This means it is never ≥
𝑛+1 𝑛+1 𝑛+1
1since 𝑛 ≥ 1. Further 𝑎𝑛 is always > 0⟹ bounded 𝑎1 ∶= 0 ≤ 𝑎𝑛 ≤ 1
Instructor: Ibrahim Bader 5
 𝑛2 +6𝑛+8 (𝑛+2)(𝑛+4) (𝑛+3)+1 1
iv. Since 𝑎𝑛 ≥ 0 and 𝑎𝑛 = = (𝑛+2)(𝑛+3) = = 1 + (𝑛+3) it
𝑛2 +5𝑛+6 (𝑛+3)
5 5
follows that 𝑎𝑛 ≥ 1 and at 𝑎1 ∶= , so 1 ≤ 𝑎𝑛 ≤ .
4 4

𝑛!
v. 𝑎𝑛 = . Bounded 0 ≤ 𝑎𝑛 ≤ 1 ∶= 𝑎1
𝑛𝑛
(−1)𝑛
vi. 𝑎𝑛 = 1 + . For all 𝑛 > 1:
𝑛2

1 5 5
𝑖𝑓 𝑛 = 2𝑘 (𝑒𝑣𝑒𝑛) ⟹ 𝑎2 = 1 + = ⟹ 0 ≤ 𝑎2𝑘 ≤
4 4 4
−1
𝑖𝑓 𝑛 = 2𝑘 + 1 (𝑜𝑑𝑑) ⟹ 𝑎1 = 1 + = 0 ⟹ 0 ≤ 𝑎2𝑘+1 ≤ 1
1
5
From both inequalities, it follows that (𝑎𝑛 ) is bounded with: 0 ≤ 𝑎𝑛 ≤ .
4

1. Which of these sequences is convegent? Which is divergent? In case of

convergenvce evaluate the limit.

Answer 2. Converging?

We use the monotone convergence theorem, the sequence is convergence

when it is both bounded (from above and below) and the sequence is
monotonic. According to what we found in the previous question and question
4 in Worksheet #9, we decide as follows:

i. converging to 0
ii. converging to 𝑛→∞ 1
𝑛→∞ 𝑛→∞
1
lim (2 + 𝑛) = lim 2 + lim (𝑛) = 2 + 0 = 0.

iii. Converges to 1 since 𝑛→∞ 2

𝑛→∞ 𝑛→∞
2
lim (1 − 𝑛+1) = lim 1 − lim (𝑛+1) = 1 + 0 = 1.
1
iv. Multiplying and dividing by ⟹ to 𝑛→∞
lim (𝑎)𝑛 = 1.
𝑛2
v. Converges, the limit is 0. It could be proved using the squeeze theorem
(compare examples in class!).

Instructor: Ibrahim Bader 6

 vi. Converges to 1.
𝑛
(−1)
Since 𝑛→∞
lim
𝑛2
= 0 (proof using the absolute value theorem:

(−1)𝑛 1
| |=
𝑛2 𝑛2𝑛

1
lim =0
𝑛→∞ 𝑛2𝑛

Instructor: Ibrahim Bader 7