Answer
$a_n=\Sigma_{n=1}^{\infty} \dfrac{1}{n^2}$ and $b_n=\Sigma_{n=1}^{\infty} \dfrac{1}{n^3}$ and the series $\Sigma a_nb_n$ converges by the p-series test.
Work Step by Step
Proof: Let us suppose that $a_n=\Sigma_{n=1}^{\infty} \dfrac{1}{n^2}$ and $b_n=\Sigma_{n=1}^{\infty} \dfrac{1}{n^3}$
Then $\Sigma_{n=1}^{\infty} a_n b_n=\Sigma_{n=1}^{\infty} \dfrac{1}{n^5}$
We can notice that $\dfrac{1}{n}\leq \dfrac{1}{n^2}$ because $\dfrac{1}{n^2} \gt 0$
Also, by the p-series test ($\Sigma_{n=1}^{\infty} \dfrac{1}{n^p}\implies p \gt 1;$ series converges), so the series $\Sigma a_nb_n$ converges because $p=5 \gt 1$.
