Question

Let (sn) be a sequence such that |sₙ₊₁ - sₙ| < 2⁻ⁿ for all n ∊ N. Prove (sn) is a Cauchy sequence and hence a convergent, sequence.

Answer 1

To prove that the sequence (sn) is a Cauchy sequence, we need to show that for any positive epsilon, there exists a positive integer N such that for all n, m ≥ N, |sn - sm| < epsilon.

Step-by-step explanation:

To prove that the sequence (sn) is a Cauchy sequence, we need to show that for any positive epsilon, there exists a positive integer N such that for all n, m ≥ N, |sn - sm| < epsilon. In this case, we are given that |sₙ₊₁ - sₙ| < 2⁻ⁿ for all n.

Let's choose an epsilon > 0. We can rewrite the given inequality as: |sₙ₊₁ - sₙ| < epsilon, since 2⁻ⁿ < epsilon for all n.

Now, we can consider the difference |sₙ₊₁ - sₙ| as a geometric series. Since the common ratio is less than 1 (2⁻ⁿ), the series converges, and the sequence (sn) is a Cauchy sequence.

Therefore, (sn) is a convergent sequence.