If $(a_n)_{n \ge 1}$ is a non-negative sequence s.t., $$\sum\limits_{n = c_k}^\infty \frac{a_n}{\log^{(k)} n} < \infty, \, \forall k \ge 1 \overset{?}{\implies} \sum\limits_{n \ge 1} a_n < \infty$$ where, $\log^{(k)} = \underbrace{\log \circ \log \circ \cdots \circ \log}_{k \text{ times }}$ and $c_k$ is large say $2^{2^{ \cdots ^{2}}}$ ($k$ power tower of $2$).

Perhaps it is more pertinent to ask some sort of characterization of a subset $S$ of sequences in $c_0$ , s.t., if $\displaystyle \sum\limits_{n \ge 1} a_nu_n < \infty$ for all $(u_n) \in S$ implies $(a_n) \in \ell^1$.

A softer related question is as follows, if $(a_n)_{n \ge 1}$ is a non-negative sequence s.t., $$\sum\limits_{n=1}^\infty \frac{a_n}{n^s} < \infty, \, \forall s > 0 \overset{?}{\implies} \sum\limits_{n \ge 1} a_n < \infty$$

i.e, if the abscissa of convergence of the Dirichlet series $\displaystyle F(s) = \sum\limits_{n=1}^\infty \frac{a_n}{n^s}$ is $\sigma_c = 0$, then does it imply $\displaystyle \sum\limits_{n \ge 1} a_n < \infty$?

Apparently Landau's theorem guarantees that if $\sigma_c \ge 0$ then $\displaystyle \limsup\limits_{x \to \infty} \frac{\log |A(x)|}{\log x} = \sigma_c$ where, $\displaystyle A(x) = \sum\limits_{n < x} a_n$ is the summatory function.

Also asked on m.se.

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    $\begingroup$ I believe it is not difficult to find counterexamples to your questions. $\endgroup$ Feb 13, 2019 at 17:17

1 Answer 1


Let $l_k:=\ln^{(k)}$. Let $(n_k)$ be any strictly increasing sequence of natural numbers such that $l_k(n_k)>k^2$. Let $a_n:=1$ if $n=n_k$ for some $k$ and let $a_n:=0$ otherwise. Then $\sum_n a_n=\infty$, but $\sum_n a_n/l_j(n)=\sum_k 1/l_j(n_k)<\infty$ for all $j$, since $l_j(n_k)\ge l_k(n_k)>k^2$ for $k\ge j$.

So, the implication in your main question is false in general. As for the "softer related question", just let $a_n=1/n$ to disprove that implication as well.


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