Answer
a) See proof
b) i) See proof
ii) See proof
Work Step by Step
We are given the series $\sum a_n$ and $\sum b_n$:
1) $a_n\geq 0,b_n\geq 0$
2) $\sum b_n$ divergent
3) $\displaystyle{\lim_{n \to \infty}}\dfrac{a_n}{b_n}=\infty$
Because $\displaystyle{\lim_{n \to \infty}}\dfrac{a_n}{b_n}=\infty$, we have $a_n\geq b_n$ for $n$ sufficiently large.
Using the Comparison Test, because $\sum b_n$ divergent, we get that $\sum a_n$ is divergent.
b) i) We are given the series:
$\sum_{n=2}^{\infty} \dfrac{1}{\ln n}$
Consider $a_n=\dfrac{1}{\ln n}$, $b_n=\dfrac{1}{n}$.
The series $\sum b_n$ is divergent.
Determine $\displaystyle{\lim_{n \to \infty}}\dfrac{a_n}{b_n}=\displaystyle{\lim_{n \to \infty}}\dfrac{\dfrac{1}{\ln n}}{\dfrac{1}{n}}=\displaystyle{\lim_{n \to \infty}}\dfrac{n}{\ln n}$
$=\displaystyle{\lim_{n \to \infty}}\dfrac{1}{\dfrac{1}{n}}=\displaystyle{\lim_{n \to \infty}} n=\infty$
Using part a), we conclude that the series $\sum a_n$ is divergent.
ii) We are given the series:
$\sum_{n=1}^{\infty} \dfrac{\ln n}{n}$
Consider $a_n=\dfrac{\ln n}{n}$, $b_n=\dfrac{1}{n}$.
The series $\sum b_n$ is divergent.
Determine $\displaystyle{\lim_{n \to \infty}}\dfrac{a_n}{b_n}=\displaystyle{\lim_{n \to \infty}}\dfrac{\dfrac{\ln n}{n}}{\dfrac{1}{n}}=\displaystyle{\lim_{n \to \infty}}\ln n=\infty$
Using part a), we conclude that the series $\sum a_n$ is divergent.
