This document contains exercises and solutions related to monotone sequences, convergent subsequences, and the Bolzano-Weierstrass theorem from an introduction to real analysis course. It includes 4 problems examining properties of specific sequences, showing whether they are bounded/monotone and finding their limits, as well as examples of an unbounded sequence with a convergent subsequence and sequences that diverge. The solutions provide detailed proofs of the properties of each sequence using induction and algebraic manipulations.

1. INTRODUCTION TO REAL ANALYSIS 1
INDIVIDUAL TASK
EXERCISES 3.3 MONOTONE SEQUENCES
EXERCISES 3.4 SUBSEQUENCES AND THE BOLZANO-WEIERSTRASS
THEOREM
By:
Muhammad Nur Chalim
4101414101
MATHEMATICS DEPARTMENT
MATHEMATICS AND NATURAL SCIENCES FACULTY
SEMARANG STATE UNIVERSITY
2016

2. EXERCISES 3.3
Problem
1. Let 𝑥1 ≔ 8 and 𝑥 𝑛+1 ≔
1
2
𝑥 𝑛 + 2 for 𝑛 ∈ 𝑁. Show that (𝑥 𝑛) is bounded and monotone.
Find the limit.
2. Let 𝑥1 > 1 and 𝑥 𝑛+1 ≔ 2 −
1
𝑥 𝑛
for 𝑛 ∈ 𝑁. Show that (𝑥 𝑛) is bounded and monotone.
Find the limit.
3. Let 𝑥1 ≥ 2 and 𝑥 𝑛+1 ≔ 1 + √ 𝑥 𝑛 − 1 for 𝑛 ∈ 𝑁. show that (𝑥 𝑛) is decreasing and
bounded below by 2. Find the limit
4. Let 𝑥1 ≔ 1 and 𝑥 𝑛+1 ≔ √2 + 𝑥 𝑛 for 𝑛 ∈ 𝑁. Show that (𝑥 𝑛) converges and find the
limit.
Solution 3.3
1. Let 𝑥1 ≔ 8 and 𝑥 𝑛+1 ≔
1
2
𝑥 𝑛 + 2 for 𝑛 ∈ ℕ. Show that ( 𝑥 𝑛) is bounded and monotone.
Find the limit.
Proof:
First, let’s show that it is monotone decreasing.
Note that 𝑥1 = 8 > 𝑥2 = 6. For induction, assume that 𝑥 𝑘 > 𝑥 𝑘+1.
Next notice
𝑥 𝑘 > 𝑥 𝑘+1
⇔
1
2
𝑥 𝑘 >
1
2
𝑥 𝑘+1
⇔
1
2
𝑥 𝑘 + 2 >
1
2
𝑥 𝑘+1 + 2
⇔ 𝑥 𝑘+1 > 𝑥 𝑘+2
Since the sequence is decreasing and 𝑥 𝑛 > 0 for all 𝑛, we can be assured that the upper
bound is 8 and we have a lower bound of 0.
To find the limit, we use the same technique as the examples.
1
2
𝑥 𝑘 + 2 =
1
2
𝑥 𝑘+1 + 2

3. ⇔ 𝑥 =
1
2
𝑥 + 2
⇔ 𝑥 −
1
2
𝑥 = 2
⇔
1
2
𝑥 = 2
⇔ 𝑥 =
2
1
2
= 4 both limits must be 𝑥.
2. Let 𝑥1 > 1 and 𝑥 𝑛+1 ≔ 2 −
1
𝑥 𝑛
for 𝑛 ∈ ℕ. Show that ( 𝑥 𝑛) is bounded and monotone.
Find the limit.
Proof:
Since 𝑥1 > 1, we have that
1
𝑥1
< 1. It means that 𝑥2 = 2 −
1
𝑥1
> 1.
We would like to show that 𝑥 𝑛 > 1 for all 𝑛. By induction, we see that if 𝑥 𝑘 > 1, then
𝑥 𝑘+1 = 2 −
1
𝑥1
> 1.
From this, we obtain that at this point we are bounded above by 2 and below by 1.
Now, let’s show that it is monotone (decreasing). To do this, we want to start by looking
at
0 < (𝑥1 − 1)2
⇔ 0 < 𝑥1
2
− 2𝑥1 + 1
⇔ 2𝑥1 < 𝑥1
2
+ 1 (multiplied by 1/𝑥1)
⇔ 2 < 𝑥1 +
1
𝑥1
⇔ 2 −
1
𝑥1
< 𝑥1
⇔ 𝑥2 < 𝑥1
That only takes care of the first case. Now we use induction to finish the monotone.
𝑥 𝑘+1 < 𝑥 𝑘
⇔
1
𝑥 𝑘+1
>
1
𝑥 𝑘
⇔ −
1
𝑥 𝑘+1
< −
1
𝑥 𝑘
⇔ 2 −
1
𝑥 𝑘+1
< 2 −
1
𝑥 𝑘

4. ⇔ 𝑥 𝑘+2 < 𝑥 𝑘+1
To find the limit, we use the same technique as the examples in the book
𝑥 𝑛+1 = 2 −
1
𝑥 𝑛
⇔ 𝑥 = 2 −
1
𝑥
⇔ 𝑥 +
1
𝑥
= 2 (multiplied by 𝑥)
⇔ 𝑥2
+ 1 = 2
⇔ 𝑥2
+ 1 − 2 = 0
⇔ (𝑥 + 2)(𝑥 − 1) = 0
𝑥 = 1 or 𝑥 = −2
3. Let 𝑥1 ≥ 2 and 𝑥 𝑛+1 ≔ 1 + √ 𝑥 𝑛 − 1 for 𝑛 ∈ ℕ. Show that ( 𝑥 𝑛) is decreasing and
bounded below by 2. Find the limit.
Proof:
Since 𝑥1 ≥ 2, we have that 𝑥1 − 1 ≥ 1
𝑥1 − 1 ≥ 1
⇔ √𝑥1 − 1 ≥ √1
⇔ √𝑥1 − 1 ≥ 1
⇔ 1 + √𝑥1 − 1 ≥ 1 + 1
⇔ 1 + √ 𝑥1 − 1 ≥ 2 (since given that 𝑥 𝑛+1 ≔ 1 + √ 𝑥 𝑛 − 1)
⇔ 𝑥2 ≥ 2
Using induction and a similar argument, we find 𝑥 𝑛+1 ≥ 2. That takes care of bounded.
We now need to work on monotone. To do this we start by noticing that if 𝑥1 − 1 > 1,
then 𝑥1 − 1 > √ 𝑥1 − 1 = 𝑥1 > √ 𝑥1 − 1 + 1 = 𝑥2.
That is our base case, we then assume that 𝑥 𝑘 > 𝑥 𝑘+1
𝑥 𝑘+1 < 𝑥 𝑘
⇔ 𝑥 𝑘+1 − 1 < 𝑥 𝑘 − 1
⇔ √𝑥 𝑘+1 − 1 < √𝑥 𝑘 − 1
⇔ 1 + √𝑥 𝑘+1 − 1 < 1 + √𝑥 𝑘 − 1

5. ⇔ 𝑥 𝑘+2 < 𝑥 𝑘+1
Again, we use the same technique as the examples in the book
𝑥 𝑛+1 = 2 −
1
𝑥 𝑛
⇔ 𝑥 = 2 −
1
𝑥
⇔ 𝑥 = 1
4. Let 𝑥1 ≔ 1 and 𝑥 𝑛+1 ≔ √2 + 𝑥 𝑛 for 𝑛 ∈ ℕ. Show that ( 𝑥 𝑛) converges and find the
limit.
Proof:
Like the first problem, let’s show taht it is monotone increasing first. Note that 𝑥1 = 1 <
𝑥2 = √3. For induction, assume that 𝑥 𝑘 < 𝑥 𝑘+1.
𝑥 𝑘 < 𝑥 𝑘+1
⇔ 2 + 𝑥 𝑘 < 2 + 𝑥 𝑘+1
⇔ √2 + 𝑥 𝑘 < √2 + 𝑥 𝑘+1
⇔ 𝑥 𝑘+1 < 𝑥 𝑘+2
Since the sequence is increasing and 𝑥 𝑛 > 0 for all 𝑛, we now want to obtain an upper
bound on the sequence. To do this, note that 𝑥1 = 1 < 𝑥2 = √3 < 4. Using induction
that 𝑥 𝑘 < 4, we get
𝑥 𝑘+1 = √2 + 𝑥 𝑘
⇔ 𝑥 𝑘+1 < √2 + 4
⇔ 𝑥 𝑘+1 = √6
⇔ 𝑥 𝑘+1 < 2
This again, allows us to use MCT to say that the swquence converges. Lasty, to find the
limit, we do what we have done for the first three problems.
𝑥 𝑛+1 = √2 + 𝑥 𝑛
⇔ 𝑥 = √2 + 𝑥
⇔ 𝑥 = 2

6. EXERCISES 3.4
Problem
1. Give an example of an unbounded sequence that has a convergent subsequence.
2. Use the method of Example 3.4.3(b) to show that if 0 < 𝑐 < 1, then lim (𝑐
1
𝑛 ) = 1
3. Let (𝑓𝑛) be the Fibonacci sequence of Example 3.1.2(d), and let 𝑥 𝑛 ≔
𝑓𝑛+1
𝑓𝑛
. Given that
lim( 𝑥 𝑛) = 𝐿 exists, determine the value of 𝐿.
4. Show that the following sequences are divergent.
(a) (1 − (−1) 𝑛
+
1
𝑛
),
(b) (sin
𝑛𝜋
4
)
Solution 3.4
1. Let 𝑥2𝑛−1 ∶= 2𝑛 − 1,
So we will obtain 𝑥2𝑛 ∶=
1
2𝑛
;
Its mean that (𝑥 𝑛) = (1 ,
1
2
,3,
1
4
,5,
1
6
, . . . ).
We can find an example of an unbounded sequence that has a convergent subsequence is
(𝑥 𝑛) = (1 ,
1
2
, 3,
1
4
,5,
1
6
, . . . ).
2. We assume that 𝑥 𝑛 ∶= 𝑐
1
𝑛 ,
where 0 < 𝑐 < 1,
then (𝑥 𝑛)is increasing and bounded,
so it has a limit x.
Since 𝑥2𝑛 = √ 𝑥 𝑛,
the limit satisfies 𝑥 = √ 𝑥, 𝑠𝑜 𝑥 = 0 𝑜𝑟 𝑥 = 1.

7. Since 𝑥 = 0 is impossible ,
we have 𝑥 = 1.
2. Since 𝑥 𝑛 ≥ 1 for all 𝑛 ∈ 𝑁, 𝐿 > 0.
Further, we have 𝑥 𝑛 =
1
𝑥 𝑛−1
+ 1 → 𝐿 =
1
𝐿
+ 1 → 𝐿2
− 𝐿 − 1 = 0 → 𝐿 =
1
2
(1 + √5)
4. (a) 𝑥2𝑛 → 0 𝑎𝑛𝑑 𝑥2𝑛+1 → 2.
(b) 𝑥8𝑛 = 0 and 𝑥8𝑛+1 = 𝑠𝑖𝑛(
𝜋
4
) =
1
√2
for all 𝑛 ∈ 𝑁