Existence of $a_k$ such that $\sum_k a_kb_k<\infty$ and $\sum_k a_k=\infty$ given $b_k\to 0$

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?

Since $$b_k\to 0$$, there is a strictly increasing subsequence $$(k_n)_n$$ such that $$|b_{k_n}|<\frac{1}{2^{n}}.$$ Now, for any $$k\in \mathbb{N}$$, let $$a_k=\begin{cases}1 &\text{if k=k_n}\\ 0 &\text{otherwise}\end{cases}.$$
• You didn't even use the fact that $b_n$ is decreasing. Jan 10 '19 at 20:39
Define $$a_k=\begin{cases}1&,\quad b_k<{1\over 2^k},\not\exists ufor example if $$b_n={1\over n}$$ then we have $$a_n=\begin{cases}1&,\quad n=1,2,4,8,16,\cdots \\0&,\quad \text{elsewhere}\end{cases}$$