Question

Suppose that the sequence {aₙ​} converges to A. Define the sequence {bₙ} by bₙ=(aₙ+aₙ+1)/2. Does the sequence {bₙ} converge? If so, specify the limit and prove your conclusion. Otherwise, give an example when this is not true.

Answer 1

The sequence {bₙ} defined as bₙ=(aₙ+aₙ+1)/2, where {aₙ} is a convergent sequence, also converges to the same limit as {aₙ}, which is A.

Step-by-step explanation:

Let's analyze the convergence of the sequence {bₙ} defined as bₙ=(aₙ+aₙ+1)/2, given that {aₙ} converges to A.

To prove the convergence of {bₙ}, we need to show that it is a Cauchy sequence. Let ε be a positive number.

1. Since {aₙ} converges to A, there exists an N such that for all n>N, |aₙ - A| < ε.
2. Now, for n>N, we have |bₙ - A| = |(aₙ+aₙ+1)/2 - A| = |(aₙ - A)/2 + (aₙ+1 - A)/2| ≤ (|aₙ - A| + |aₙ+1 - A|)/2 < ε/2 + ε/2 = ε.

This shows that for every ε>0, there exists an N such that for all n>N, |bₙ - A| < ε. Therefore, {bₙ} is a convergent sequence, and its limit is A.