Q:

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

The question is as in the title.  
I think $\{b_n\}$ should be some function of $a_n$. I was thinking about 
$$b_n = \frac{1}{n\sqrt{a_n}},$$ 
but I don't know if $b_n$ approaches infinity.  Can someone give some hint about what $b_n$ should be? 

A:

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.