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.

• Are $a_n, b_n$ supposed to be positive? Commented Jun 7, 2022 at 16:59
• @GEdgar Yes they are Commented Jun 7, 2022 at 17:02
• $b_n = \ln\ln\ln\ln\ln(n)$ does not tend to zero. Commented Jun 7, 2022 at 17:07
• @TonyK Yes, I was talking about $1/lnlnln... :)$ Commented Jun 7, 2022 at 17:09

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 $$k, since $$\{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.
• Very nice. May I ask what was your intuition for constructing $a_n$? Commented Jun 7, 2022 at 17:11
• Really it just starts from the fact that we get to look at $b_n$ first before customizing $a_n$ to break the assertion. I also used the idea that we can use a "step function" for $\{a_n\}$ rather than needing some continuous underlying function, and that constant factors don't affect convergence so we only need to localize the $b_n$ values in dyadic intervals. That's already sufficient if $\{b_n\}$ is decreasing, and the rest is just accounting to deal with the fact that the $b_n$ could come in some different order. Commented Jun 7, 2022 at 17:30