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By "minimal divergent series", I mean a divergent series $a_n$ such that for any sequence $b_n$ with $\lim_{n \rightarrow \infty} b_n = 0$, $\sum^\infty_{n=0} a_nb_n$ converges; that is, any sequence which goes to $0$ faster than $a_n$ will converge. I suspect the answer is false, since $b_n$ can go to $0$ at an arbitrarily slow rate, but I'm not sure how to prove this.

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  • $\begingroup$ no there is no such thing as the minimial divergent series. You can always construct a series that is slower diverging. $\endgroup$ – MrYouMath Oct 27 '16 at 21:45
  • $\begingroup$ Somewhat related: math.stackexchange.com/questions/452053/… $\endgroup$ – Henricus V. Oct 27 '16 at 21:46
  • $\begingroup$ You just divide every term by a constant to get a sequence that is termwise less than $a_n$ but still diverges. $\endgroup$ – Nitin Oct 27 '16 at 22:10
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    $\begingroup$ By the Banach-Steinhaus theorem, if $(a_n)$ is a sequence such that $\sum a_n b_n$ is convergent for every sequence $(b_n)$ such that $b_n\to 0$, then $\sum \lvert a_n\rvert < +\infty$. $\endgroup$ – Daniel Fischer Oct 27 '16 at 22:49
  • $\begingroup$ And neither is there a maximal convergent series. $\endgroup$ – GEdgar Oct 27 '16 at 23:46
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If the $a_n$'s are all positive with $\sum a_n=\infty$, then there is an easy construction of $b_n>0$ such that $\sum a_nb_n=\infty$: Group the $a$'s into an infinite number of disjoint blocks, where each block has sum at least $1$. This is possible because the series $\sum a_n$ diverges. Now define $b_n$ to be $1$ for every $n$ in the first block of $a$'s, then $\frac12$ for every $n$ in the second block of $a$'s, $\ldots$, then $\frac 1k$ for every $n$ in the $k$th block of $a$'s. So $b_n\to0$, but the sum $\sum a_nb_n$ exceeds the harmonic series.

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$\newcommand{\eps}{\varepsilon}$As you suspect, no such $(a_{k})$ exists. For each $k$, put $$ a_{k}^{+} = \max(a_{k}, 0),\qquad a_{k}^{-} = \min(a_{k}, 0). $$ If $\sum_{k} a_{k}$ diverges, we may assume without loss of generality that $\sum_{k} a_{k}^{+}$ diverges. Construct $(b_{k})$ as follows:

Take $b_{k} = 0$ if $a_{k} < 0$.

Since the sequence $s_{n} = \sum_{0}^{n} a_{k}^{+}$ of partial sums is unbounded, there exists a sequence of indices $N_{1} < N_{2} < N_{3} < \cdots$ such that $$ 1 \leq \sum_{k=N_{j}}^{N_{j+1} - 1} a_{k}^{+}. $$ Now scale the $a_{k}^{+}$ at leisure, e.g., multiplying $a_{k}^{+}$ by $1$ for $1 \leq k < N_{1}$, multiplying by $\frac{1}{2}$ for $N_{1} \leq k < N_{3}$, multiplying by $\frac{1}{3}$ for $N_{3} \leq k < N_{6}$.

The partial sums of $\sum b_{k} a_{k}$ contain, by appropriate grouping, the partial sums of $$ 1 + \tfrac{1}{2} + \tfrac{1}{2} + \tfrac{1}{3} + \tfrac{1}{3} + \tfrac{1}{3} + \cdots. $$

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