Convergence of $\sum_n a_nb_n$ for all $b_n\searrow 0$ implies convergence of $\sum_n a_n$

I need a hint for a practice problem:

Let $a_n \geq 0$. Show that if $\displaystyle\sum_{n=1}^\infty a_nb_n$ converges for every monotonically decreasing sequence $b_n \to 0$, then $\displaystyle\sum_{n=1}^\infty a_n$ converges.

I've been trying to use the fact that $\sum\limits_{n=1}^\infty a_n r^n \leq M < \infty$ for all $r \in [0,1)$ iff $\displaystyle\sum_{n=1}^\infty a_n$ converges, but I can't seem to get it, so I'm not sure that's the right way to go about it.

• Sorry, fixed it I think. $b_n \geq 0$ and decreasing monotonically to $0$. – user93370 Sep 6 '13 at 2:47
• The first thing to notice is that your condition does imply that a$_n$ goes to zero. To see this consider that $\sum 1/n$ – Betty Mock Sep 6 '13 at 3:37
• continued $\sum 1/n$ does not converge, so if $\sum a_n b_n$ converges you must have $a_n < 1/n$. That shows $a_n$ goes to zero, which is a necessary condition for convergence. An no matter fast $\sum b_n$ diverges, $a_n$ is small enough to force convergence. That is not a proof yet, but a pointer in a good direction. – Betty Mock Sep 6 '13 at 3:48