Question

For \( n \in \mathbb{N} \), let \[a_n = \sqrt{n} \sin^2\left(\frac{1}{n}\right) \cos n,\] and \[b_n = \sqrt{n} \sin\left(\frac{1}{n^2}\right) \cos n.\] Then

• A. The series \( \sum_{n=1}^{\infty} a_n \) converges, but the series \( \sum_{n=1}^{\infty} b_n \) does not converge.
• B. The series \( \sum_{n=1}^{\infty} a_n \) does not converge, but the series \( \sum_{n=1}^{\infty} b_n \) converges.
• C. Both the series \( \sum_{n=1}^{\infty} a_n \) and \( \sum_{n=1}^{\infty} b_n \) 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 C

1. Analyze \( a_n \): - Expand \( \sin^2\left(\frac{1}{n}\right) \) using the small-angle approximation \( \sin(x) \approx x \) for \( x \to 0 \): \[ \sin^2\left(\frac{1}{n}\right) \approx \left(\frac{1}{n}\right)^2. \] - Substituting into \( a_n \): \[ a_n \approx \sqrt{n} \cdot \frac{1}{n^2} \cdot \cos n = \frac{\cos n}{n^{3/2}}. \] - The term \( \frac{\cos n}{n^{3/2}} \) decays rapidly enough for the series \( \sum_{n=1}^\infty a_n \) to converge by comparison with a \( p \)-series where \( p = \frac{3}{2} > 1 \). 2. Analyze \( b_n \): - Expand \( \sin\left(\frac{1}{n^2}\right) \) using \( \sin(x) \approx x \) for \( x \to 0 \): \[ \sin\left(\frac{1}{n^2}\right) \approx \frac{1}{n^2}. \] - Substituting into \( b_n \): \[ b_n \approx \sqrt{n} \cdot \frac{1}{n^2} \cdot \cos n = \frac{\cos n}{n^{3/2}}. \] - As in the case of \( a_n \), the term \( \frac{\cos n}{n^{3/2}} \) decays rapidly enough for the series \( \sum_{n=1}^\infty b_n \) to converge. 3. Conclusion: - Both series \( \sum_{n=1}^\infty a_n \) and \( \sum_{n=1}^\infty b_n \) converge due to the rapid decay of their terms