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> 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 $2$).
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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](https://math.stackexchange.com/questions/3110574/a-problem-with-sequences-with-composition-of-logs).



A problem with sequences with composition of $\log$s
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. 


