Question

Let \( x_1 = 1 \). For \( n \in \mathbb{N} \), define \[ x_{n+1} = \left( \frac{1}{2} + \frac{\sin^2 n}{n} \right) x_n. \] Then, which one of the following is TRUE?

• A. \( \sum_{n=1}^{\infty} x_n \) converges
• B. \( \sum_{n=1}^{\infty} x_n \) does NOT converge
• C. \( \sum_{n=1}^{\infty} x_n^2 \) does NOT converge
• D. \( \sum_{n=1}^{\infty} x_n x_{n+1} \) does NOT converge

Hint:

For recurrences involving exponential decay, check the decay rate of \( x_n \) to determine the convergence of the series. If the decay is fast enough, the series will converge.

Answer 1

The Correct Option is A

Step 1: Analyze the recurrence.
The recurrence relation is: \[ x_{n+1} = \left( \frac{1}{2} + \frac{\sin^2 n}{n} \right) x_n. \] As \( n \to \infty \), the term \( \frac{\sin^2 n}{n} \to 0 \), so the recurrence behaves asymptotically like: \[ x_{n+1} \approx \frac{1}{2} x_n. \] This suggests that \( x_n \) decays exponentially. Step 2: Check the convergence of the series.
Since \( x_n \) decays exponentially, the series \( \sum_{n=1}^{\infty} x_n \) converges, as it rapidly approaches zero. Final Answer: \[ \boxed{\sum_{n=1}^{\infty} x_n \text{ converges.}} \]