Answer
\[\sum_{n=1}^{\infty}a_n=\sum_{n=1}^{\infty}\frac{1}{n^2}\]
and \[\sum_{n=1}^{\infty}b_n=\sum_{n=1}^{\infty}\frac{1}{n}\]
Work Step by Step
Let \[\sum_{n=1}^{\infty}a_n=\sum_{n=1}^{\infty}\frac{1}{n^2}\]
and \[\sum_{n=1}^{\infty}b_n=\sum_{n=1}^{\infty}\frac{1}{n}\]
\[\Rightarrow a_n=\frac{1}{n^2},\;\;\;b_n=\frac{1}{n}\]
\[\Rightarrow \frac{a_n}{b_n}=\frac{\frac{1}{n^2}}{\frac{1}{n}}=\frac{1}{n}\]
\[\Rightarrow \lim_{n\rightarrow \infty}\frac{a_n}{b_n}=\lim_{n\rightarrow\infty}\frac{1}{n}=0\]
[P-Series Test: $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^P}$ is convergent if and only if P>1]
$\displaystyle\Rightarrow \sum_{n=1}^{\infty}b_n=\sum_{n=1}^{\infty}\frac{1}{n}$ is divergent by P-Series Test and $\displaystyle\sum_{n=1}^{\infty}a_n=\sum_{n=1}^{\infty}\frac{1}{n^2}$ is convergent by P-Series Test.
