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