Question

Let \( a_1 = 1 \), \( a_{n+1} = a_n \left( \frac{\sqrt{n} + \sin n}{n} \right) \), and \( b_n = a_n^2 \) for all \( n \in \mathbb{N} \). Then which of the following statements is/are correct?

• A. the series \( \sum_{n=1}^{\infty} a_n \) converges
• B. the series \( \sum_{n=1}^{\infty} b_n \) converges
• C. the series \( \sum_{n=1}^{\infty} a_n \) converges but the series \( \sum_{n=1}^{\infty} b_n \) does NOT converge
• D. neither the series \( \sum_{n=1}^{\infty} a_n \) nor the series \( \sum_{n=1}^{\infty} b_n \) converges

Answer 1

The Correct Option is A, B

Step 1: Expressing \( a_n \). We are given the recurrence relation: \[ a_{n+1} = a_n \left(\sqrt{n} + \frac{\sin n}{n}\right), \quad a_1 = 1. \] Expanding iteratively, we can express \( a_n \) as: \[ a_n = \prod_{k=1}^{n-1} \left(\sqrt{k} + \frac{\sin k}{k}\right). \] For large \( k \), \[ \sqrt{k} + \frac{\sin k}{k} \sim \sqrt{k}, \] so the dominant growth of \( a_n \) is approximately: \[ a_n \sim \prod_{k=1}^{n-1} \sqrt{k} = \sqrt{\prod_{k=1}^{n-1} k} = \sqrt{(n-1)!}. \] Thus, \( a_n \) grows very quickly. Step 2: Convergence of \( \sum_{n=1}^\infty a_n \). From the above expression, \( a_n \) grows much faster than \( n^{-p} \) for any \( p > 0 \). Hence, \( a_n \to 0 \) as \( n \to \infty \) at a sufficiently fast rate, ensuring that the series \( \sum_{n=1}^\infty a_n \) converges. Step 3: Convergence of \( \sum_{n=1}^\infty b_n \). Since \( b_n = a_n^2 \), its growth rate is approximately: \[ b_n \sim \left(\sqrt{(n-1)!}\right)^2 = (n-1)!. \] The factorial growth of \( b_n \) is too rapid for \( \sum_{n=1}^\infty b_n \) to converge. Thus, \( \sum_{n=1}^\infty b_n \) diverges