Question

Define the sequences \(\left\{a_n\right\}^{\infin}_{n=3}\) and \(\left\{b_n\right\}^{\infin}_{n=3}\) as
\(a_n=(\log n+\log\ \log n)^{\log n}\) and \(b_n=n^{(1+\frac{1}{\log n})}\).
Which one of the following is TRUE ?

• A. \(\sum\limits^{\infin}_{n=3}\frac{1}{a_n}\) is convergent but \(\sum\limits_{n=3}^{\infin}\frac{1}{b_n}\) is divergent
• B. \(\sum\limits^{\infin}_{n=3}\frac{1}{a_n}\) is divergent but \(\sum\limits_{n=3}^{\infin}\frac{1}{b_n}\) is convergent
• C. Both \(\sum\limits_{n=3}^{\infin}\frac{1}{a_n}\) and \(\sum\limits_{n=3}^{\infin}\frac{1}{b_n}\) are divergent
• D. Both \(\sum\limits_{n=3}^{\infin}\frac{1}{a_n}\) and \(\sum\limits_{n=3}^{\infin}\frac{1}{b_n}\) are convergent

Answer 1

The Correct Option is A

- For \( a_n = (\log n + \log \log n)\log n \): We simplify \( a_n \) as: \[ a_n \approx (\log n) \log n = (\log n)^2 \quad \text{for large } n. \] Thus, the series \( \sum_{n=3}^{\infty} \frac{1}{a_n} \) behaves like the series \( \sum_{n=3}^{\infty} \frac{1}{(\log n)^2} \). Since \( \frac{1}{(\log n)^2} \) decays faster than \( \frac{1}{n} \), the series \( \sum_{n=3}^{\infty} \frac{1}{a_n} \) converges. - For \( b_n = n^{(1 + \frac{1}{\log n})} \): We analyze the behavior of \( b_n \): \[ b_n = n^{1 + \frac{1}{\log n}} = n \cdot n^{\frac{1}{\log n}}. \] As \( n \) grows large, \( n^{\frac{1}{\log n}} \) approaches 1, so \( b_n \approx n \). Therefore, the series \( \sum_{n=3}^{\infty} \frac{1}{b_n} \) behaves like the harmonic series \( \sum_{n=3}^{\infty} \frac{1}{n} \), which is divergent. Thus, the correct answer is (A): \( \sum_{n=3}^{\infty} \frac{1}{a_n} \) is convergent but \( \sum_{n=3}^{\infty} \frac{1}{b_n} \) is divergent.