I was working with a problem from functional analysis. I reduced the problem to the following problem:

> 
> Let $b\_k>0$ be a decreasing sequence converging to $0$. Does there exist a non-negative sequence $a\_k$ such that
>  $$\sum\_{k\geq 1} b\_ka\_k<\infty$$ 
>  while at the same time
>  $$\sum\_{k\geq 1} a\_k = \infty$$
> 
> 
> 

If I know what $b\_k$ looks like, then I guess it is not hard to find out how to choose $a\_k$. However we don't know what $b\_k$ looks like and neither in which rate it decays. I really don't know what to do except that I have tried something like choosing for $n>m$
\begin{align}
\sum\_{k=m}^n a\_k = \frac 1 {\sqrt {b\_mb\_n^\varepsilon}}
\end{align}
for some $\varepsilon>0$, so that the sequence is not Cauchy, that leads to the divergence of $\sum\_k a\_k$ which is nice. At the same time I get
\begin{align}
\sum\_{k=m}^n b\_ka\_k \leq b\_m \sum\_{k=m}^na\_k =\sqrt{\frac{b\_m}{b\_n^\varepsilon}}
\end{align}
This tells us that, if there is $\varepsilon$ such that 
$$\sqrt{\frac{b\_m}{b\_n^\varepsilon}}\to 0 \ \ \ \text{ as } n,m\to\infty$$
we are done. **However**, I don't think that will work, since $\varepsilon$ is fixed...

> 
> **Question.** How can the problem be solved?
> 
> 
>