Cauchy Sequences

Definition of a Cauchy Sequence: A sequence {an } is called a Cauchy sequence if, for every ε > 0, there exists an N ∈ N such that for all n, m ≥ N , we have: |an − am | < ε. This definition means that as the sequence progresses (i.e., as n, m → ∞), the terms of the sequence become arbitrarily close to each other. — Definition of a Subsequence: A subsequence {akℓ } of a sequence {an } is a sequence formed by selecting specific terms of {an } while preserving their order. Formally, a subsequence is defined by a strictly increasing sequence of indices {kℓ } such that: k1 < k2 < k3 < . . . Examples: 1. If {an } = 1 , a possible subsequence is:   1 1 1 , , ,... . {akℓ } = {a2 , a4 , a6 , . . . } = 2 4 6 n 2. Another subsequence of the same sequence is:  {akℓ } = {a1 , a3 , a5 , . . . } = 1 1 1, , , . . . 3 5  . — Key Idea of the Proof: If {an } is a Cauchy sequence, any subsequence {akℓ } must also satisfy the Cauchy property. This holds because: 1. A subsequence retains the order of the original sequence, so the indices kℓ are part of the original sequence {an }. 2. The Cauchy condition, which ensures the closeness of terms for sufficiently large indices, applies universally to all terms of the sequence, including those in the subsequence. — Intuition Behind Cauchy Subsequences: Imagine a Cauchy sequence {an } as a ”path” where the steps between consecutive terms get shorter and shorter. A subsequence {akℓ } selects specific steps along this path but does not change the overall behavior. Since the steps in the original sequence are already close together, the selected steps in the subsequence will also be close. — Examples to Illustrate the Concept 1. **Example 1: A Simple Cauchy Sequence** Consider the sequence {an } = n1 . For any ε > 0, choose N > 1ε . Then for all n, m ≥ N , we have: |an − am | = 1 1 1 − ≤ < ε. n m N 1 Thus, {an } is a Cauchy sequence. Now consider the subsequence {akℓ } = {a2 , a4 , a6 , . . . }. For this subsequence, the same argument applies because {akℓ } is a part of {an }. 2. **Example 2: A Constant Sequence** The sequence {an } = c, where c is a constant, is trivially a Cauchy sequence because: |an − am | = |c − c| = 0 for all n, m. Any subsequence of {an } is also constant and therefore Cauchy. — Proof: Every Subsequence of a Cauchy Sequence is Cauchy Let {an } be a Cauchy sequence. By definition, for every ε > 0, there exists an N ∈ N such that for all n, m ≥ N , we have: |an − am | < ε. Let {akℓ } be a subsequence of {an }, where kℓ < kℓ+1 for all ℓ ∈ N. This implies that for all ℓ ∈ N, kℓ ≥ ℓ (since kℓ is strictly increasing). For every ℓ, p ∈ N with ℓ, p ≥ N , we have kℓ , kp ≥ N because kℓ ≥ ℓ ≥ N and kp ≥ p ≥ N . Therefore: |akℓ − akp | < ε. Since ε > 0 was arbitrary, this proves that {akℓ } is a Cauchy sequence. — Practice Problems and Solutions Problem 1: Show that the sequence {an } = n1 is Cauchy. Then find a subsequence {akℓ } and prove that it is also Cauchy. Solution: 1. For ε > 0, choose N > 1ε . For all n, m ≥ N , we have: 1 1 1 ≤ |an − am | = − < ε. n m N Hence, {an } is a Cauchy sequence. 2. Let {akℓ } = {a2 , a4 , a6 , . . . }. For ε > 0, choose N > 1ε . For all ℓ, p ≥ N , we have kℓ , kp ≥ 2N . Therefore: 1 |akℓ − akp | ≤ < ε. 2N Thus, {akℓ } is also Cauchy. — Problem 2: Is the sequence {an } = (−1)n Cauchy? Why or why not? Solution: The sequence {an } = (−1)n alternates between 1 and −1. For n, m of opposite parity (e.g., n even and m odd), we have: |an − am | = |1 − (−1)| = 2. 2 Thus, the terms do not get arbitrarily close as n, m → ∞, so {an } is not Cauchy. — Problem 3: Consider the sequence {an } = 1 + n1 . Prove that it is Cauchy and find a subsequence {akℓ } that converges. Solution: 1. For ε > 0, choose N > 1ε . For all n, m ≥ N , we have:     1 1 1 1 |an − am | = 1 + − 1+ − = < ε. n m n m Thus, {an } is Cauchy. 2. Let {akℓ } = {a2 , a3 , a4 , . . . }. As kℓ → ∞, akℓ → 1. Therefore, the subsequence converges to 1. — 3