My question is

Let $\sum a_n $ be a series such that $\sum a_nb_n$ converges absolutely for every sequence $\{b_n\}$ that is bounded. Prove that $\sum a_n$ is absolutely convergent.

My thoughts were as follow.

We know that $\sum a_nb_n$ is absolutely convergent,

$$\sum_{i=1}^n |a_nb_n|$$ is bounded for some $M$.

Since the $b_n$'s are also bounded, then there exists some $K$ such that $K\ge b_n$ for any $n$.

Thus we can write $a_nb_n$ being bounded by $a_n K$, and because of the comparison theorem, since $a_nb_n$ converges absolutely and $a_nK$ converges absolutely, then its must be true that $\sum a_n$ is absolutely convergent.

Is this correct?

  • 2
    $\begingroup$ $b_n=1 \forall n$ $\endgroup$
    – Aloizio Macedo
    Sep 25 '15 at 3:09
  • $\begingroup$ Wouldn't this be the same thing I had said? That because of the comparison test. since anbn converges abs, then an must also converge abs. $\endgroup$
    – mjo
    Sep 25 '15 at 3:12
  • 2
    $\begingroup$ No. What I said provides an immediate answer, perhaps showing that maybe something is wrong with your hypotheses. $\endgroup$
    – Aloizio Macedo
    Sep 25 '15 at 3:14
  • 1
    $\begingroup$ Note that there is a (minor) gap: You know $a_n b_n \le a_n K$ and $\sum a_n b_n$ converges absolutely, that does not force $a_n K$ converges absolutely (You do not have the correct inequality to apply comparison test). To show that probably you need to take a constant sequence $b_n$, as suggested by Aloizio. $\endgroup$
    – user99914
    Sep 25 '15 at 3:17
  • $\begingroup$ What if we take bn to be the max {b1,b1,b3...,bn}, this would not only show that anmax {b1,b1,b3...,bn} converges abs, but since its definately greater than an, would show that an converges absolutely? $\endgroup$
    – mjo
    Sep 25 '15 at 3:24

Even if you just ask $∑a_nb_n$ to converge (not necessarily uniformly) for every bounded $b_n$, you can take $$b_n = \begin{cases}+1 & a_n \geq 0 \\ -1 & a_n < 0\end{cases} \implies a_nb_n = |a_n|$$

this then says that $\sum |a_n| $ converges, QED.


Responding to question: Taking $b_n=1 ~ \forall n$ makes the result immediate.

Responding to the question of whether your thought is correct: As pointed out by John Ma, you have the "wrong-sided" inequality. For instance, if you took an arbitrary bounded sequence in order to prove the statement, it is bound to fail: the sequence $b_n \equiv 0$ is bounded, makes $\sum a_n b_n$ converge absolutely, but provides no information on absolute convergence of $\sum a_n$.

  • $\begingroup$ Taking $b_n=1$ for all $n$ yields that $\sum a_n$ is convergent, but not that it is absolutely convergent. $\endgroup$
    – bangs
    Sep 14 '18 at 18:44

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