Question

Let (a_n) and (b_n) be sequences of real numbers such that
\(|a_n-a_{n+1}|=\frac{1}{2^n}\) and \(|b_n-b_{n+1}|=\frac{1}{\sqrt{n}}\) for n ∈ \(\N\).
Then

• A. both (a_n) and (b_n) are Cauchy sequences
• B. (a_n) is a Cauchy sequence but (b_n) need NOT be a Cauchy sequence
• C. (a_n) need NOT be a Cauchy sequence but (b_n) is a Cauchy sequence
• D. both (a_n) and (b_n) need NOT be Cauchy sequences

Answer 1

The Correct Option is B

To determine whether the sequences \((a_n)\) and \((b_n)\) are Cauchy sequences, we need to understand the definition of a Cauchy sequence:

• A sequence \((x_n)\) is called a Cauchy sequence if for every \(\epsilon > 0\), there exists an integer \(N\) such that for all \(m, n \geq N\), \(|x_m - x_n| < \epsilon\).

Now, let's consider each sequence separately: 

1. Sequence \((a_n)\):
   • We are given that \(|a_n - a_{n+1}| = \frac{1}{2^n}\).
   • For a Cauchy sequence, we need \(|a_m - a_n| < \epsilon\) for sufficiently large \(n\) and \(m\).
   • The term \(|a_n - a_{n+1}| = \frac{1}{2^n}\) implies that the differences between consecutive terms decrease exponentially.
   • Since \(\sum \frac{1}{2^n}\) is a convergent series, this implies that for any \(\epsilon > 0\), we can find an \(N\) so that for all \(m, n \geq N\), \(|a_m - a_n| < \epsilon\).
2. Sequence \((b_n)\):
   • We are given that \(|b_n - b_{n+1}| = \frac{1}{\sqrt{n}}\).
   • Observe that the terms \(\frac{1}{\sqrt{n}}\) decrease slowly compared to \(\frac{1}{2^n}\).
   • The series \(\sum \frac{1}{\sqrt{n}}\) is a well-known divergent series.
   • Due to this divergence, for some \(\epsilon > 0\), no finite \(N\) can be found such that for all \(m, n \geq N\), \(|b_m - b_n| < \epsilon\). Consequently, the sequence does not satisfy the Cauchy condition.

Conclusively, the correct choice is that \((a_n)\) is a Cauchy sequence but \((b_n)\) need NOT be a Cauchy sequence.