Analisis Rill Tugas 3.5

Analisis Rill Tugas 3.5

• 1.
  INDIVIDUAL TASK 3.5 AYUNITASARI (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 definition𝑚, 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.