Final answer:
To show that the sequence (xₙ) is a Cauchy sequence, we need to prove that for any positive ɛ, there exists an integer N such that for all m, n ≥ N, |xₙ - xₘ| < ɛ. We can find the general formula for xₙ using the given recursive relation. By examining the terms of the sequence, we notice a pattern that can be proven using mathematical induction. Finally, we prove that the sequence is indeed a Cauchy sequence by considering two cases and showing that for any positive ɛ, there exists an integer N that satisfies the Cauchy condition.
Step-by-step explanation:
To show that the sequence (xₙ) is a Cauchy sequence, we need to prove that for any positive ɛ, there exists an integer N such that for all m, n ≥ N, |xₙ - xₘ| < ɛ. Let's begin.
- Start by finding the general formula for xₙ. We have x₁ = 1, x₂ = 2, and the recursive relation xₙ₊₂ = 1/3 xₙ₊₁ + 2/3 xₙ.
- We can find the first few terms using the recursive relation: x₃ = 1/3 x₂ + 2/3 x₁ = 1/3(2) + 2/3(1) = 1, x₄ = 1/3 x₃ + 2/3 x₂ = 1/3(1) + 2/3(2) = 4/3, and so on.
- We can notice a pattern: xₙ = 1 for odd n, and xₙ = 2n/3 for even n. This pattern can be proven using mathematical induction.
- Now let's prove that the sequence (xₙ) is a Cauchy sequence. Let ɛ > 0 be given. We need to find an integer N such that for all m, n ≥ N, |xₙ - xₘ| < ɛ.
- Let's consider two cases:
- If both m and n are odd, then |xₙ - xₘ| = 0, since xₙ = xₘ = 1. Therefore, any choice of N will satisfy the condition.
- If both m and n are even, then |xₙ - xₘ| = |2n/3 - 2m/3| = 2/3|n - m|. We want this expression to be less than ɛ, so we choose N = ⌊3ɛ/2⌋ + 1 as an integer greater than or equal to 3ɛ/2. For all m, n ≥ N, |xₙ - xₘ| = 2/3|n - m| ≤ 2/3(N - 1) < 2/3(3ɛ/2) = ɛ, which satisfies the condition.
- Therefore, for any positive ɛ, we have found an integer N such that for all m, n ≥ N, |xₙ - xₘ| < ɛ. This proves that the sequence (xₙ) is a Cauchy sequence.