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

Question

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?

Answer 1

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 $$a_k=\begin{cases}1 &\text{if $k=k_n$}\\ 0 &\text{otherwise}\end{cases}.$$

Answer 2

Define $$a_k=\begin{cases}1&,\quad b_k<{1\over 2^k},\not\exists u<k,b_{u}<{1\over 2^{k}}\\0&,\quad \text{elsewhere}\end{cases}$$for example if $b_n={1\over n}$ then we have $$a_n=\begin{cases}1&,\quad n=1,2,4,8,16,\cdots \\0&,\quad \text{elsewhere}\end{cases}$$