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

Answer: (A)

$\sum\limits^{\infin}_{n=3}\frac{1}{a_n}$ is convergent but $\sum\limits_{n=3}^{\infin}\frac{1}{b_n}$ is divergent

Answer 2

The correct option is (A) : $\sum\limits^{\infin}_{n=3}\frac{1}{a_n}$ is convergent but $\sum\limits_{n=3}^{\infin}\frac{1}{b_n}$ is divergent.