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.

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

1 Answer 1


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.

  • $\begingroup$ Very nice. May I ask what was your intuition for constructing $a_n$? $\endgroup$
    – MStudent
    Commented Jun 7, 2022 at 17:11
  • 2
    $\begingroup$ 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. $\endgroup$ Commented Jun 7, 2022 at 17:30

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