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.
Thanks in advance! 
 A: (Contraposition.) Suppose that $a_n\geq 0$ for each $n$ and $\sum\limits_{n=1}^\infty a_n=+\infty$.  Define $n_1<n_2<n_3<\cdots$ such that $\sum\limits_{n=n_{k}}^{n_{k+1}-1}a_n>1$ for each $k$ (with $n_0=1$).  Define $b_n=\frac{1}{k}$ for $n_{k-1}\leq n<n_k$.  Then $\sum\limits_{n=1}^\infty a_nb_n$ diverges.
A: $\sum a_nb_n$ converges for every $b_n$ which goes monotonically to 0.  Let $b_n$ be such a sequence such that $\sum b_n$ diverges. If $ \sum a_n b_n$ is to converge, then "on average" the {$a_n$}  < the {$b_n$}, even though for any particular n it could be that $a_n > b_n$.  One way to write this is
$ \sum_{1}^n a_n$ < $\sum_{1}^n b_n$ for "most" n.  More precisely as $n \rightarrow \infty$ we have $\sum a_n$ < $ \sum b_n$.  That is,  $\sum a_n$ is strictly less than $\sum b_n$ whenever b is divergent. 
We can define a series {$b_n$} as being divergent in this way:  that for any number R > 0 there exists a number $N_R$ such that $ \sum_{n =1}^{N_R} b_n> R$.
Let {$b_n$} be a divergent series and pick R.  The $N_R$ that works for {$b_n$} will not work for {$a_n$}. The reason is that for every positive integer k, $\sum_{n=1}^\infty a_n$ < $ \frac{1}{k} \sum _{n=1}^\infty b_n$ ,(which is divergent) and there is always a k such that the chosen $N_R$ is too small, unless $\sum a_n$ converges to a number > R. 
Since $ \frac{1}{k} \sum _{n+1}^\infty b_n$  does diverge, there is a new $N_{R,k}$ which works for this series, and if we take k large enough  we have  $N_{R,k}$ > $N_R$ where the inequality is strict.
However, no matter what k we pick $N_{R,k}$ if $\sum a_n$ does not converge to a number > R it is not large enough that  $\sum_{n = 1}^{N_{R,k}} a_n> R$, because there is always a larger k, thus a smaller divergent series, and thus a bigger $N_{R,k}$ for that series.
If for any R there can be no  ${N_R}$ for $\sum a_n$ then that sum cannot be divergent. If it is not divergent, it must converge.
