Adek adek p.mat unnes tercinta yang nyari jawaban individual task 3.5 (divergent sequnce)

1. INDIVIDUAL TASK 3.5
AYU NITASARI (4101414150)
1. Give an example of a bounded sequence that is not a Cauchy sequence.
Solution:
Consider the sequence {(−1) 𝑛
}, 𝑛 = 1,2,3, . . .. It is bounded but not Cauchy
Notice that if 𝑛 is even then 𝑎 𝑛 = 1, and so 𝑎 𝑛+1 = −1. Choose𝜀0 = 2, and select any even
𝑛 such that 𝑛 ≥ 𝑁 ∈ 𝑁. Then choose 𝑚 = 𝑛 + 1. Therefore ∣ 𝑥𝑛 − 𝑥𝑚 ∣=∣ 1 − (−1) ∣=
2 ≥ 𝜀0 = 2. So ((−1) 𝑛
) is not Cauchy.
2. Show directly from the definition that the following are Cauchy sequences.
(a) (
𝑛+1
𝑛
)
(b) (1 +
1
2!
+ ⋯+
1
𝑛!
)
Solution:
a) Let 𝑋 = (
𝑛+1
𝑛
).
Will be proven 𝑋 = (
𝑛+1
𝑛
) is Cauchy sequence.
Proof:
Let 𝑥 𝑛 ≔
𝑛+1
𝑛
.
So we can chage 𝑥 𝑛 ≔
𝑛+1
𝑛
= 1 +
1
𝑛
.
If 𝑛 > 𝑚 then get | 𝑥 𝑛 − 𝑥 𝑚| = |
1
𝑛
−
1
𝑚
| ≤
1
𝑛
+
1
𝑚
.
If given 𝜀 > 0, tehn we choose 𝐻 = 𝐻( 𝜀) ∈ ℕ ∋ 𝐻 >
2
𝜀
.
If 𝑛, 𝑚 ≥ 𝐻 then
1
𝑛
,
1
𝑚
≤
1
𝐻
<
𝜀
2
, result | 𝑥 𝑛 − 𝑥 𝑚| <
𝜀
2
+
𝜀
2
= 𝜀.
So, we can conclude that X is Cauchy sequence.
b) Let 𝑋 = (1 +
1
2!
+ ⋯+
1
𝑛!
)
Will be proven 𝑋 = (1 +
1
2!
+ ⋯ +
1
𝑛!
) isCauchy sequence.
Proof:
From definotion Cauchy sequence: Sequence 𝑋 = (𝑥 𝑛)said Cauchy Sequence : if
∀ 𝜀 > 0 ∃ 𝐻( 𝜀) 𝜖 ℕ ∋ 𝑚, 𝑛 ≥ 𝐻( 𝜀), then 𝑥 𝑚 and 𝑥 𝑛 satisfies | 𝑥 𝑛 − 𝑥 𝑚| < 𝜀.
So if 𝜀 > 0 choose 𝐻 = 𝐻( 𝜀) ∈ ℕ ∋ 𝐻 >
2
𝜀

2. Then from definition 𝑚, 𝑛 ≥ 𝐻, then 𝑥 𝑚 and 𝑥 𝑛 satisfies | 𝑥 𝑛 − 𝑥 𝑚| = |
1
𝑛!
−
1
𝑚!
| ≤
1
𝑛!
+
1
𝑚!
<
𝜀
2
+
𝜀
2
= 𝜀
Because 𝜀 > 0, | 𝑥 𝑛 − 𝑥 𝑚| < 𝜀 then 𝑥 𝑛 = (
1
𝑛!
)is Cauchy sequence.
3. Show directly from the definition that the following are not Cauchy sequences.
(a) ((−1) 𝑛)
(b) (𝑛 +
(−1) 𝑛
𝑛
)
(c) (ln 𝑛)
Solution:
a) Let 𝑋 = ( 𝑥 𝑛) ≔ ((−1) 𝑛
)
Will be proven X is not Cauchy sequence.
Proof:
Obvious if n even number then 𝑥 𝑛 = 1 and 𝑥 𝑛+1 = −1.
If we choose 𝑛 > 𝐻 ∈ ℕ and let 𝑚 = 𝑛 + 1, then
| 𝑥 𝑛 − 𝑥 𝑚| = |1 − (−1)| = 1 + 1 = 2
So sequence (−1) 𝑛
) is not Cauchy sequence.
Another way:
Defined 𝑥 𝑛 ≔ (−1) 𝑛
.
Then if 𝑛 > 𝑚 then
| 𝑥 𝑛 − 𝑥 𝑚| = |(−1) 𝑛
− (−1) 𝑚| ≤ |(−1) 𝑛
+ (−1) 𝑚| = 2.
So, if we choose 𝜀0 = 2 then
∀ 𝐻( 𝜀0) ∈ ℕ∃ 𝑛 > 𝑚 > 𝐻( 𝜀0) ∋ | 𝑥 𝑛 − 𝑥 𝑚| ≤ 𝜀0 .
Because of that sequence (−1) 𝑛
) is not Cauchy sequence.
b) Let :𝑋 = (𝑛 +
(−1) 𝑛
𝑛
)
Will be proven X is not Cauchy sequence.
Proof :
Negation of Chaucy sequence is:
∃𝜀0 > 0 ∀𝐻 ∃ 𝑚𝑖𝑛. 1 𝑛 > 𝐻 𝑎𝑛𝑑 𝑚𝑖𝑛.1 𝑚 > 𝐻 ∋ | 𝑥 𝑛 − 𝑥 𝑚| ≥ 𝜀0 .
Let 𝑥 𝑛 = (𝑛 +
(−1) 𝑛
𝑛
)
If n even then 𝑥 𝑛 = (𝑛 +
(−1) 𝑛
𝑛
) = 𝑛 +
1
𝑛
=
𝑛2
+1
𝑛

3. And 𝑥 𝑛+1 = 1 +
(−1) 𝑛+1
𝑛
= 𝑛 +
(−1) 𝑛(−1)
𝑛
= 𝑛 −
1
𝑛
=
𝑛2
−1
𝑛
If we choose 𝜀0 =
2
𝑛
then for every 𝐻 we can choose 𝑛 > 𝐻 ∈ even number
And let
𝑚 ≔ 𝑛 + 1, m even and 𝑚 > 𝐻
| 𝑥 𝑛 − 𝑥 𝑚| = | 𝑥 𝑛 − 𝑥 𝑛+1| =
𝑛2
+ 1
𝑛
−
𝑛2
− 1
𝑛
=
𝑛2
+ 1 − 𝑛2
+ 1
𝑛
=
2
𝑛
= 𝜀0
So sequence (𝑋 𝑛) is not Cauchy sequence
c) (𝑙𝑛 𝑛) is not bounded so its certainly its not Cauchy sequence. This can be frooved using
the definition : pick 𝜀 = 1, we can find 𝐾 such that for any 𝑛, 𝑚 ≥ 𝐾 one has
|ln 𝑛 − ln 𝑚| = |ln
𝑛
𝑚
| < 1 ? No, take 𝑛 = 5𝑚 > 𝑚 ≥ 𝐾 the ln
6𝑚
𝑚
= ln 6 > 1
4. Show directly from the definition that if (𝑥 𝑛) and (𝑦 𝑛) are Cauchy sequences, then
( 𝑥 𝑛 + 𝑦 𝑛) and ( 𝑥 𝑛 𝑦 𝑛) are Cauchy sequences.
Solution:
Let ( 𝑥 𝑛) and (𝑦 𝑛) Cauchy sequence.
Will be proven ( 𝑥 𝑛 + 𝑦 𝑛) and (𝑥 𝑛 𝑦 𝑛) Cauchy sequence.
Proof:
Will be proven.( 𝑥 𝑛 + 𝑦 𝑛)Cauchy sequence.
𝜀 > 0 ⟹ 𝐻 = 𝐻( 𝜀0) ∈ ℕ ∋ 𝐻 >
2
𝜀
.
Becouse of ( 𝑥 𝑛) and ( 𝑦 𝑛) Cauchy sequence ∀𝑛, 𝑚 ≥ 𝐻 we get | 𝑥 𝑛 − 𝑥 𝑚| <
𝜀
2
and
| 𝑦 𝑛 − 𝑦 𝑚| <
𝜀
2
.
Let 𝑧 𝑛 ≔ 𝑥 𝑛 + 𝑦 𝑛 . Then ∀𝑛, 𝑚 ≥ 𝐻 we get
| 𝑧 𝑛 − 𝑧 𝑚| = | 𝑥 𝑛 + 𝑦 𝑛 − ( 𝑥 𝑚 + 𝑦 𝑚)|
= |( 𝑥 𝑛 − 𝑥 𝑚) + ( 𝑦 𝑛 − 𝑦 𝑚)| ≤ | 𝑥 𝑛 − 𝑥 𝑚| + | 𝑦 𝑛 − 𝑦 𝑚|
<
𝜀
2
+
𝜀
2
= 𝜀
So, sequence .( 𝑥 𝑛 + 𝑦 𝑛) 𝑖𝑠 Cauchy sequence.
Will be proven .( 𝑥 𝑛 𝑦 𝑛)Cauchy sequence.
𝜀 > 0 ⟹ 𝐻 = 𝐻( 𝜀0) ∈ ℕ ∋ 𝐻 >
2
𝜀
.
Becouse of ( 𝑥 𝑛) and ( 𝑦 𝑛) Cauchy sequence ∀𝑛, 𝑚 ≥ 𝐻 we get | 𝑥 𝑛 − 𝑥 𝑚| < √ 𝜀and
| 𝑦 𝑛 − 𝑦 𝑚| < √ 𝜀.

4. Let 𝑧 𝑛 ≔ 𝑥 𝑛 𝑦 𝑛 . Then ∀𝑛, 𝑚 ≥ 𝐻 we get
| 𝑧 𝑛 − 𝑧 𝑚| = |( 𝑥 𝑛 𝑦 𝑛) − ( 𝑥 𝑚 𝑦 𝑚)| ≤ | 𝑥 𝑛 𝑦 𝑛| + | 𝑥 𝑚 𝑦 𝑚|
≤ | 𝑥 𝑛 − 𝑥 𝑚|| 𝑥 𝑚 − 𝑦 𝑚| − | 𝑥 𝑚 𝑦 𝑛| − |𝑥 𝑛 𝑦 𝑚|
< √ 𝜀√ 𝜀 − | 𝑥 𝑚 𝑦 𝑛| − | 𝑥 𝑛 𝑦 𝑚|
< 𝜀 − | 𝑥 𝑚 𝑦 𝑛| − | 𝑥 𝑛 𝑦 𝑚|
So sequence.( 𝑥 𝑛 𝑦𝑛) 𝑖𝑠 Cauchy sequence.