Does there exists a sequence $b_n$, s.t. $\lim b_n = 0$ and for every divergent series $\sum a_n$, The series $\sum a_n b_n$ also diverges? I am trying to detemine whether the following statement is true or false:

Does there exists a positive sequence $b_n$, s.t. $\lim b_n = 0$ and for any positive
divergent series $\sum a_n$, The series $\sum a_n b_n$ also diverges?

At first I believed the statement is false, Because it does make sense to me that we can always find a divergent series $\sum a_n$ such that $a_n$ will need a "little enough push" so that the series will do converge.
when looking at some very slow growing sequences such as $b_n =1/ \ln\ln\ln\ln\ln(n)$, I was not able to find such $a_n$.
So the question is, Is it possible to find such $b_n$ which converge to $0$ slow enough?
Any hints will be appericiated.
 A: The answer is no. Let $\{b_n\}$ be a sequence with $\lim_{n\to\infty} b_n=0$; we need to construct a sequence $\{a_n\}$ such that $\sum_{n=1}^\infty a_n$ diverges yet $\sum_{n=1}^\infty a_nb_n$ converges. We may assume that infinitely many of the $b_n$ are nonzero (or else any divergent series $\sum_{n=1}^\infty a_n$ will suffice).
We will use the fact that for any positive number $x$ there is a unique integer $k$ such that $2^{-k} \le x < 2^{1-k}$. Let $N_k$ be the set of positive integers $n$ such that $2^{-k} \le |b_n| < 2^{1-k}$; this set is finite since $\lim_{n\to\infty} b_n=0$. Moreover, there exists some integer $K$ such that $N_k=\emptyset$ for all $k<K$, since $\{b_n\}$ is bounded.
For every $n$, define $a_n=0$ if $b_n=0$ and otherwise
$$
a_n = \frac1{\#N_k} \text{ if } 2^{-k} \le |b_n| < 2^{1-k}.
$$
In particular, $\{a_n\}$ is nonnegative and therefore we may evaluate the series $\sum_{n=1}^\infty a_n$ in any order. Since
$$
\sum_{n\in N_k} a_n = \sum_{n\in N_k} \frac1{\#N_k} = 1,
$$
and there must be infinitely many $k$ such that $N_k\ne\emptyset$ (here we use the assumption that infinitely many of the $b_n$ are nonzero), the series $\sum_{n=1}^\infty a_n$ is larger than the sum of infinitely many $1$s and thus diverges.
On the other hand, we can show that $\sum_{n=1}^\infty a_nb_n$ converges absolutely. Again we can evaluate $\sum_{n=1}^\infty |a_nb_n|$ in any order, and we may ignore any terms for which $b_n=0$. The resulting series can be written as
$$
\sum_{k\ge K} \sum_{n\in N_k} |a_nb_n| = \sum_{k\ge K} \sum_{n\in N_k} \frac1{\#N_k}|b_n| < \sum_{k\ge K} \sum_{n\in N_k} \frac1{\#N_k} 2^{1-k} = \sum_{k\ge K} 2^{1-k} = 2^{2-K}
$$
by the geometric series formula, and thus $\sum_{n=1}^\infty |a_nb_n|$ converges by the comparison test.
