Question

Let \(\{a_n\}_{n \ge 1}\) be a sequence of real numbers such that \(a_n \ge 1\), for all \(n \ge 1\). Then, which of the following conditions imply the divergence of \(\{a_n\}_{n \ge 1}\)?

• A. \(\{a_n\}_{n \ge 1}\) is non-increasing
• B. \(\sum_{n=1}^{\infty} b_n\) converges, where \(b_1 = a_1\) and \(b_n = a_{n+1} - a_n\) for all \(n>1\)
• C. \(\lim_{n \to \infty} \frac{a_{2n+1}}{a_{2n}} = \frac{1}{2}\)
• D. \(\{\sqrt{a_n}\}_{n \ge 1}\) converges

Hint:

For any convergent sequence \(\{a_n\}\), the ratio of consecutive terms must approach 1. If it approaches any other constant, the sequence diverges.

Answer 1

The Correct Option is C

Step 1: Analyze the condition in (C).
Given that \[ \lim_{n \to \infty} \frac{a_{2n+1}}{a_{2n}} = \frac{1}{2}, \] it implies that for large \(n\), the odd-indexed terms are roughly half of the even-indexed terms. This means the sequence keeps halving every two steps, indicating it cannot settle to a finite nonzero limit.
Step 2: Examine convergence behavior.
If \(\{a_n\}\) were convergent to \(L\), then the ratio \[ \lim_{n \to \infty} \frac{a_{2n+1}}{a_{2n}} = \frac{L}{L} = 1. \] However, since the limit is \(1/2 \ne 1\), this contradicts convergence. Thus, \(\{a_n\}\) diverges.
Step 3: Check other options.
(A) Non-increasing and bounded below (\(a_n \ge 1\)) implies convergence. (B) Convergent series of differences implies \(\{a_n\}\) converges. (D) Convergence of \(\{\sqrt{a_n}\}\) implies convergence of \(\{a_n\}\). Hence, only (C) indicates divergence. Final Answer: \[ \boxed{\lim_{n \to \infty} \frac{a_{2n+1}}{a_{2n}} = \frac{1}{2} \text{ implies divergence.}} \]