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

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.

• I believe it is not difficult to find counterexamples to your questions. Feb 13, 2019 at 17:17

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.