Question

Which of the following statements is/are TRUE ?

• A. \(\sum\limits_{n=1}^{\infin}n\log(1+\frac{1}{n^3})\) is convergent
• B. \(\sum\limits_{n=1}^{\infin}(1-\cos(\frac{1}{n}))\log n\) is convergent
• C. \(\sum\limits_{n=1}^{\infin}n^2 \log(1+\frac{1}{n^3})\) is convergent
• D. \(\sum\limits_{n=1}^{\infin}(1-\cos(\frac{1}{\sqrt n}))\log n\) is convergent

Answer 1

The Correct Option is A, B

- (A) The term \( \log \left( 1 + \frac{1}{n^3} \right) \) behaves like \( \frac{1}{n^3} \) for large \( n \), so we can approximate: \[ n \log \left( 1 + \frac{1}{n^3} \right) \sim n \cdot \frac{1}{n^3} = \frac{1}{n^2}. \] The series \( \sum_{n=1}^{\infty} \frac{1}{n^2} \) converges, and hence, the series in (A) is convergent. - (B) The term \( 1 - \cos \left( \frac{1}{n} \right) \) behaves like \( \frac{1}{2n^2} \) for large \( n \) because: \[ 1 - \cos x \approx \frac{x^2}{2} \text{ for small } x. \] Thus, for large \( n \), \[ \left( 1 - \cos \left( \frac{1}{n} \right) \right) \log n \sim \frac{1}{2n^2} \log n. \] Since \( \sum_{n=1}^{\infty} \frac{\log n}{n^2} \) converges, the series in (B) is convergent. - (C) The term \( n^2 \log \left( 1 + \frac{1}{n^3} \right) \sim n^2 \cdot \frac{1}{n^3} = \frac{1}{n} \), and \( \sum_{n=1}^{\infty} \frac{1}{n} \) diverges. Therefore, the series in (C) is divergent. - (D) The term \( 1 - \cos \left( \frac{1}{\sqrt{n}} \right) \) behaves like \( \frac{1}{2n} \) for large \( n \), so the series behaves like: \[ \left( 1 - \cos \left( \frac{1}{\sqrt{n}} \right) \right) \log n \sim \frac{1}{2n} \log n. \] Since \( \sum_{n=1}^{\infty} \frac{\log n}{n} \) diverges, the series in (D) is divergent. Therefore, the correct answers are (A) and (B).