Answer
$a_{n}=\frac{1}{n^{2}}$, $b_{n}=\frac{1}{n}$
Work Step by Step
For a positive convergent series, we can use $a_{n}=\frac{1}{n^{2}}$ (p-series with $p\gt 1$)
For a positive divergent series, we can use $b_{n}=\frac{1}{n}$ (p-series with $p= 1)$
Then we have
$\lim\limits_{n \to \infty}\frac{a_{n}}{b_{n}}=\lim\limits_{n \to \infty}\frac{\frac{1}{n^{2}}}{\frac{1}{n}}$
$=\lim\limits_{n \to \infty}\frac{1}{n}$
$=0$
Hence, $a_{n}=\frac{1}{n^{2}}$, $b_{n}=\frac{1}{n}$
