1. The document contains exercises on limit theorems from an introduction to real analysis course. It includes 4 problems asking the student to determine if sequences converge or diverge based on given formulas, provide examples of sequences whose sum and product converge but the individual sequences diverge, and prove statements about convergent sequences. 2. The solutions show work for determining convergence or divergence of sequences defined by formulas in problem 1. Examples are given in problem 2 where the sum and product of divergent sequences converge. Theorems are applied in problems 3 and 4 to prove relationships between convergent sequences.

1. INTRODUCTION TO REAL ANALYSIS 1
INDIVIDUAL TASK
EXERCISES 3.2
LIMIT THEOREMS
By:
Muhammad Nur Chalim
4101414101
MATHEMATICS DEPARTMENT
MATHEMATICS AND NATURAL SCIENCES FACULTY
SEMARANG STATE UNIVERSITY
2016

2. EXERCISES 3.2
Problem
1. For 𝑥 𝑛 given by the following formulas, establish either the convergence or the
divergence of the sequence 𝑋 = (𝑥 𝑛).
a. 𝑥 𝑛 ≔
𝑛
𝑛+1
b. 𝑥 𝑛 ≔
(−1) 𝑛
𝑛
𝑛+1
c. 𝑥 𝑛 ≔
𝑛2
𝑛+1
d. 𝑥 𝑛 ≔
2𝑛2
+3
𝑛2 +1
2. Give an example two divergence sequences 𝑋 and 𝑌 such that:
a. Their sum 𝑋 + 𝑌 converges.
b. Their product 𝑋𝑌 converges.
3. Show that if 𝑋 and 𝑌 are sequences such that 𝑋 and 𝑋 + 𝑌 are convergent, then 𝑌 is
convergent.
4. Show that if 𝑋 and 𝑌 are sequences such that 𝑋 converges to 𝑥 ≠ 0 and 𝑋𝑌 converges,
then 𝑌 converges.
Solution
1. Given the following formula of 𝑥 𝑛. establish either the convergence or the divergence of
the sequence 𝑋 = (𝑥 𝑛).
a. 𝑥 𝑛 ≔
𝑛
𝑛+1
We write
𝑥 𝑛 ≔
𝑛
𝑛 + 1
=
1
1 +
1
𝑛
Let 𝑌 = 1 dan 𝑍 = 1 +
1
𝑛
We use theorem 3.2.3(b). Such that lim 𝑌 = 1 and lim 𝑍 = 1 ≠ 0. We get
lim (
𝑛
𝑛+1
) =
1
1
= 1

3. since lim (
𝑛
𝑛+1
) =
1
1
= 1 (exist), then 𝑥 𝑛 =
𝑛
𝑛+1
convergence
b. 𝑥 𝑛 ≔
(−1) 𝑛
𝑛
𝑛+1
Let 𝑥 𝑛 convergence and lim( 𝑥 𝑛) = 𝑙 exist.
Take any 𝜀 =
1
2
. By the definition 3.1.3, ∃ 𝐾 ∈ ℕ such that
|
(−1) 𝑛
𝑛
𝑛 + 1
− 𝑙| <
1
2
; ∀𝑛 ≥ 𝐾
In specific
|
(−1)2𝑛(2𝑛)
2𝑛 + 1
− 𝑙| <
1
2
⇔ |
2𝑛
2𝑛 + 1
− 𝑙| <
1
2
; ∀𝑛 ∈ 𝐾
and
|
(−1)2𝑛+1(2𝑛 + 1)
(2𝑛 + 1) + 1
− 𝑙| <
1
2
⇔ |
−(2𝑛 + 1)
2𝑛 + 2
− 𝑙| <
1
2
⇔ |
(2𝑛 + 1)
2𝑛 + 2
+ 𝑙| <
1
2
; ∀𝑛 ∈ 𝐾
By Triangle Inequality, we get
1
2
+
1
2
= 1 > |
2𝑛
2𝑛 + 1
− 𝑙| + |
(2𝑛 + 1)
2𝑛 + 2
+ 𝑙| ≥ |
2𝑛
2𝑛 + 1
− 𝑙 +
(2𝑛 + 1)
2𝑛 + 2
+ 𝑙|
= |
2𝑛
2𝑛 + 1
+
(2𝑛 + 1)
2𝑛 + 2
| >
2𝑛
2𝑛 + 2
+
2
2𝑛 + 2
= 1 ; ∀𝑛 ∈ 𝐾
It means that 1 > 1, is a contradiction.
So, 𝑥 𝑛 is not convergence (divergence).
c. 𝑥 𝑛 =
𝑛2
𝑛+1
We write
𝑛2
𝑛 + 1
=
1
1
𝑛
+
1
𝑛2
Claimed that 𝑍 =
1
𝑛
+
1
𝑛2 and lim 𝑍 = 0.
Since lim Z = 0 then lim (
n2
n+1
) doesn’t exist

4. therefore 𝑥 𝑛 =
𝑛2
𝑛+1
divergence.
d. 𝑥 𝑛 ≔
2𝑛2
+3
𝑛2 +1
We write
2𝑛2
+3
𝑛2+1
=
2+
3
𝑛2
1+
1
𝑛2
We use theorem 3.2.3(b)
Claim that 𝑌 = 2 +
3
𝑛2 and 𝑍 = 1 +
1
𝑛2
So that lim 𝑌 = 2 and lim 𝑍 = 1
Thus lim
2𝑛2
+3
𝑛2 +1
=
2
1
= 2
Since lim
2𝑛2
+3
𝑛2 +1
exist, then 𝑥 𝑛 =
2𝑛2
+3
𝑛2 +1
convergence.
2. Give two divergence sequences 𝑋 and 𝑌
a. Their sum 𝑋 + 𝑌 converges,
Supposed 𝑋 ≔ ((−1) 𝑛)and 𝑌 ≔ ((−1) 𝑛+1); ∀𝑛 ∈ ℕ.
Note that
𝑋 ≔ ((−1) 𝑛);∀𝑛 ∈ ℕ ⇔ 𝑋 ≔ (−1,1,−1,1,… )
𝑌 ≔ ((−1) 𝑛+1);∀𝑛 ∈ ℕ ⇔ 𝑌 ≔ (1, −1,1, −1,… )
Obvious that 𝑋 and 𝑌 divergence. But
𝑋 + 𝑌 ≔ (0,0, 0, 0, … )
It means 𝑋 + 𝑌 convergence to 0.
b. Their product 𝑋𝑌 converges.
Supposed 𝑋 ≔ (0,1, 0, 1, … ) and 𝑌 ≔ (1,0, 1, 0, … )
Obvious that 𝑋 and 𝑌 divergence, but 𝑋𝑌 ≔ (0, 0, 0,0, … ). It means 𝑋𝑌 convergece to 0.
3. Show that if 𝑋 and 𝑌 are sequences such that 𝑋 and 𝑋 + 𝑌 are convergent, then 𝑌 is
convergent.
Proof:
Given 𝑋 and 𝑋 + 𝑌 convergence
Will be proved 𝑌 convergence

5. Since 𝑋 and 𝑋 + 𝑌 convergent and 𝑌 = ( 𝑋 + 𝑌) − 𝑌, then based on theorem 3.2.3(a), 𝑌
also convergence.
4. Show that if 𝑋 and 𝑌 are sequences such that 𝑋 converges to 𝑥 ≠ 0 and 𝑋𝑌 converges,
then 𝑌 converges.
Claim: ∃𝐾 ∋ ∀𝑛 ≥ 𝐾, 𝑥 𝑛 ≠ 0.
Proof:
 Case I: 𝑥 > 0
Take 𝜀 = 𝑥 > 0.
Because 𝑋 convergence to 𝑥, then lim( 𝑋) = 𝑥.
Based on definition ∃𝐾 ∈ ℕ
Such that | 𝑥 𝑛 − 𝑥| < 𝑥 ⇔ 𝑥 𝑛 > 𝑥 − 𝑥 = 0 ;∀𝑛 ∈ 𝐾.
 Case II: 𝑥 < 0
Take 𝜀 = −𝑥 > 0.
Because 𝑋 convergence to 𝑥, then lim( 𝑋) = 𝑥.
Based on definition ∃𝐾 ∈ ℕ
Such that | 𝑥 𝑛 − 𝑥| < −𝑥 ⇔ 𝑥 𝑛 < 𝑥 − 𝑥 = 0 ;∀𝑛 ∈ 𝐾.
Therefore ∃𝐾 ∈ ℕ, such that 𝑥 𝑛 ≠ 0, ∀𝑛 ∈ 𝐾.
Because 𝑋𝑌 convergence, 𝑋 ≔ ( 𝑥 𝑛) ≠ 0 convergence, and 𝑌 =
𝑋𝑌
𝑋
; 𝑋 ≠ 0
So, based on theorem 3.2.3 (a) means Y convergence.