Given $\sum a_n$ converges and $a_n \ge 0$. Want to find $\{b_n\}$, such that $b_n\to \infty$ and $\sum a_nb_n$ converges

There are $N_1<N_2<N_3<\cdots$ such that

$$\sum_{n=N_k+1}^{N_{k+1}}a_n\le 2^{-N}.$$

Take $b_n=2^{N/2}$ for $N_k<n\le N_{k+1}$ (and anything you like for $n\le N_1$). Then

$$\sum_{n=N_k+1}^{N_{k+1}}a_n b_n \le 2^{-N/2}.$$

Therefore $\sum_{n=N_1+1}^\infty a_nb_n =\sum_{k=1}^\infty\sum_{n=N_k+1}^{N_{k+1}}a_n b_n$ converges.

We may assume $\sum_{n\in\mathbb{N}}a_n=1$ without loss of generality, by multiplying every element of $\{a_n\}_{n\in\mathbb{N}}$ by a suitable constant. Under such assumption, we may consider the sequence $\{b_n\}_{n\in\mathbb{N}}$ defined through: $$ b_0=c_0=1,\qquad b_n = \frac{1}{\sqrt{c_n}}, \qquad c_n = 1-\sum_{m=0}^{n-1} a_m. $$ Since $\sum_{n\in\mathbb{N}}a_n$ is convergent to $1$ and has non-negative terms, $\{c_n\}_{n\in\mathbb{N}}$ is a sequence with non-negative terms decreasing to $0$, hence $\{b_n\}_{n\in\mathbb{N}}$ is divergent to $+\infty$. We also have $a_n = c_n-c_{n+1}$, from which: $$ a_n b_n = \frac{c_n-c_{n+1}}{\sqrt{c_n}} = \frac{\left(\sqrt{c_n}-\sqrt{c_{n+1}}\right)\cdot\left(\sqrt{c_n}+\sqrt{c_{n+1}}\right)}{\sqrt{c_n}}\leq 2\left(\sqrt{c_n}-\sqrt{c_{n+1}}\right).$$ It follows that: $$ \sum_{n\leq N} a_n b_n \leq a_0 + 2\cdot\sum_{n\leq N}\left(\sqrt{c_n}-\sqrt{c_{n+1}}\right) = a_0 + 2\left(1-\sqrt{c_{N+1}}\right),$$ hence the series $\sum_{n\in\mathbb{N}}a_n b_n$ is convergent to some value $\leq\left(a_0+2\right)$.